Dropzone always return 'empty($_FILES)' as true - javascript

I need to add drag and drop to my web page. I'm trying to get dropped image and upload them to my database.
HTML
<form action="parser.php" id="file-up" class="dropzone">
<input type="submit" value="Upload" id="file-up_btn"/>
</form>
JS
Dropzone.autoDiscover = false;
$('#file-up_btn').click(function(e){
var myDropzone = new Dropzone("#myId", {
autoQueue: false,
url: "/parser.php",
});
console.log("Uploading");
});
Dropzone shows that the uploading finished successfully. But it does not show any files in the php file.
<?php
if(!empty($_FILES)){
echo 'Inside';
$temp = $_FILES['file']['tmp_name'];
$dir_seperator = DIRECTORY_SEPARATOR;
$folder = "uploads";
//$destination_path = dirname(__FILE_).$dir_seperator.$folder.$dir_seperator;
$destination_path = tempnam(sys_get_temp_dir(), 'Tux');
echo '<h1>Uploading section $destination_path</h1>';
$target_path = $destination_path.$_FILES['file']['name'];
move_uploaded_file($temp,$target_path);
echo 'Updated';
}else{
echo '<h1>No files</h1>';
}
?>
it always returns No files. I am new to DropZone. Please help me on this. Thanks in advance. :)

Two thinks to check:
id isn't the same.
in js: "#myId"
in html: #file-up or #file-up_btn
There no myid in html to refer so img isn't added to form.
from logical behaviour:
empty($_FILES) is true when no files are transmited
!empty($_FILES) is false because of !. So no file is attach.
Check Developer Tools->Network and check Preserve log. Check what header is sent with what data (maybe nothing is sent etc, it is half success when you can inform if file is added to form data).
Screenshoot to show what we all need to help: developer tools

Related

Jquery & Ajax -> File upload not possible

I got totally lost.
Ive tried to make some Image Upload function in PHP and everything works fine. Because i dont want the whole Page to reload, when uploading a File i wanted to use AJAX with Jquery, to send the Form Content (Image) via POST to a file like upload.php with an hidden ajax request.
No matter what i try its impossible to send anything with formData(). I copied & pasted several Sample Codes, tried changing the Code, nothing happens when i use formData().
A normal request with Jquery / Ajax, using POST works fine.
Here ist the Sample of my last used Code..
Could my XamPP has been misconfigured, or what could cause that really not one of the Scripts from google, tutorial pages etc works?
<!DOCTYPE html>
<html>
<body>
<script type="text/javascript" src="jquery.min.js"></script>
<form id="Test" action="" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
</form>
<button id="Knopf">Knopf</button>
<div id="Disp">fghfgh</div>
</body>
<script>
$(document).ready(function(){
$("#Knopf").click(function(){
var formData = new FormData(Test);
$.ajax({
url : "uploadtest2.php",
type : "POST",
data : formData,
cache : false,
contentType : false,
processType : false,
success : function() {
$("#Disp").html(result);
}
});
});
});
</script>
</html>
<?php
$target_dir = "Media/uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "File is an image - " . $check["mime"] . ".";
$uploadOk = 1;
} else {
echo "File is not an image.";
$uploadOk = 0;
}
}
// Check if file already exists
if (file_exists($target_file)) {
echo "Sorry, file already exists.";
$uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 500000) {
echo "Sorry, your file is too large.";
$uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
echo "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";
$uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
echo "Sorry, your file was not uploaded.";
// if everything is ok, try to upload file
} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
echo "The file ". htmlspecialchars( basename( $_FILES["fileToUpload"]["name"])). " has been uploaded.";
} else {
echo "Sorry, there was an error uploading your file.";
}
}
?>
The problem is this:
if(isset($_POST["submit"])) {
There's no element named submit in the form, so this check fails. Even if you had a submit button in the form, it wouldn't be included in formData, because buttons are only automatically included in POST data when they trigger normal form submission.
You can add that to formData.
var formData = new FormData(Test);
formData.set("submit", "1");
Or you could change your test to
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
Please see: https://developer.mozilla.org/en-US/docs/Web/API/FormData/FormData
You must use a Form Element:
An HTML <form> element — when specified, the FormData object will be populated with the form's current keys/values using the name property of each element for the keys and their submitted value for the values. It will also encode file input content.
Consider the following example.
$(function() {
$("#Test").submit(function(e) {
e.preventDefault();
var formData = new FormData(this);
$.ajax({
url: "uploadtest2.php",
type: "POST",
data: formData,
cache: false,
contentType: false,
processType: false,
success: function(result) {
$("#Disp").html(result);
}
});
});
$("#Knopf").click(function() {
$("#Test").submit();
});
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<form id="Test" action="" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload" />
</form>
<button id="Knopf" type="submit">Knopf</button>
<div id="Disp">fghfgh</div>
It is better to bind to the submit callback. This way, if the User submits the form or clicks Submit, the callback is triggered. We need to .preventDefault() on the Event to ensure the Form doesn't post or submit the data. Now we can then perform the AJAX call without the page being refreshed.
In your success callback, you must pass in a variable to be used for the returned data. Otherwise result will be undefined.
With the proper FormData, there should be no issue uploading. this in the callback refers to the Form Element itself.
Consider updating your PHP as well:
if(isset($_POST["submit"])) {
Change this to:
if(isset($_POST["fileToUpload"])) {

how to request a single file as object

I just start to learn javascirpt, php about 2 days. The problem I face is I already have a x.dcm file under server root, and I already known that path(e.g. http://localhost:8888/....)
My question is how can I simply grab that file from server to use, maybe something like:
var file= 'http://localhost:8888/....'; ////file is not an object
I ask this question because I already known how to use input method:
<input type="file" name="file" id="file">
<script>
$('#file').on('change',function(e){
var file = e.target.file; ///file is an object
});
</script>
but that is not what I want, what I want is to use an existed file rather than input.
So the whole thing is that:
<form id="input" method="post" enctype="multipart/form-data">
<input type="file" id="fileToUpload" name="fileToUpload">
</form>
I firstly make a input to upload some file,then in script
<script>
$("form#input").submit(function(){
var formData = new FormData($(this)[0]);
$.ajax({
url: 'segmentation.php',
type: 'POST',
data: formData,
async: false,
success: function (html) {
$('#segbound').html(html);
},
cache: false,
contentType: false,
processData: false
});
return false;
});
</script>
I sent this file(e.g image.dcm) to do something( run a exec) on the server side, then it generates another image(imgproc.dcm) in an expected path(http://localhost:8888/....), and then the next thing is that I what that processed image display on the screen. To do that I need to use a js called cornerstone, and the function in it imageLoader.fileManager.get(file)
which file is that one I what to display.
When I select from input using var file = e.target.file; as I mentioned above, it works perfect, then I check the file type it is a [file object].
But when I want to simply display that 'imgproc.dcm' by using var file= 'http://localhost:8888/....'; the file type is not an object which comes out my question, how can I simply grab that known path image to use as an object.
Or, to improve that, it is possible to get the return (generated imgproc.dcm) directly after it process on server side, and then to use that return(maybe give it an id...do not know) to display (call cornerstone function imageLoader.fileManager.get(file))
On server side, it looks like:
<?php
$target_dir = "/Applications/MAMP/htdocs/dicomread/temp/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
if (file_exists($target_file)) {
echo "file has already been uploaded.";
$uploadOk = 0;
}
if ($uploadOk == 0) {
echo "Sorry, your file was not uploaded.";
} else {
if (move_uploaded_file($_FILES['fileToUpload']['tmp_name'], $target_file)) {
echo "The file ". basename( $_FILES['fileToUpload']['name']). " has been uploaded.";
} else {
echo "Sorry, there was an error uploading your file.";
}
}
$cmd = "/Applications/MAMP/htdocs/dicomread/abc 2>&1";
$Output_fileName = "imgproc.dcm";//$_FILES['fileToUpload']['name'];
exec("$cmd $target_file $Output_fileName);
echo "<br/>done";
?>
Any help would be appreciated.
Use fopen with URL to the file:
$file = fopen("http://localhost:8888/x.dcm", "r");
Refer to this for fopen: http://php.net/manual/en/function.fopen.php

How to add a progress bar to script that downloads file from url to server

I have searched for a script that download files from url to server and I came across this script. It works fine but I want to add some things to it like multi urls and progress bar. As I am new in php I know a little .
Here is the script in which I want to implement:
<html>
<form method="post">
<input name="url" size="50" />
<input name="submit" type="submit" />
</form>
<?php
// maximum execution time in seconds
set_time_limit (24 * 60 * 60);
if (!isset($_POST['submit'])) die();
// folder to save downloaded files to. must end with slash
$destination_folder = 'downloads/';
$url = $_POST['url'];
$newfname = $destination_folder . basename($url);
$file = fopen ($url, "rb");
if ($file) {
$newf = fopen ($newfname, "wb");
if ($newf)
while(!feof($file)) {
fwrite($newf, fread($file, 1024 * 8 ), 1024 * 8 );
}
}
if ($file) {
fclose($file);
}
if ($newf) {
fclose($newf);
}
?>
</html>
For the multiple links you can tell the user to input all links seperated with commas. Then you could create an array with these links (using explode function) and loop over it until you download all files from the links. The code is below:
<html>
<form method="post">
<input name="url" size="50" />
<input name="submit" type="submit" />
</form>
<?php
// maximum execution time in seconds
set_time_limit (24 * 60 * 60);
if (!isset($_POST['submit'])) die();
// folder to save downloaded files to. must end with slash
$destination_folder = 'downloads/';
$urls = $_POST['url']; // I changed this variable's name to urls
$links = explode(",",$urls); // links is now an array with the links
foreach ($links as $url)
{
$newfname = $destination_folder . basename($url);
$file = fopen ($url, "rb");
if ($file) {
$newf = fopen ($newfname, "wb");
if ($newf)
while(!feof($file)) {
fwrite($newf, fread($file, 1024 * 8 ), 1024 * 8 );
}
}
if ($file) {
fclose($file);
}
if ($newf) {
fclose($newf);
}
}
?>
Useful links:
Explode: http://php.net/manual/en/function.explode.php
Foreach: http://php.net/manual/en/control-structures.foreach.php
If you want to have another input box appear you would have to use javascript and it is a bit complicated. But if you want, I can show you that way too.
For the progress bar, you will need javascript and css Bootstrap. I am trying to create a code that does what you want. When and if i manage to do so, I will upload it here, so you might want to come back later and check it out.
Since you want to have the ability to show new input boxes here is the code:
<html>
<form method="post">
<input type="text" id="input1" name="input1" size="50" />
<div id="new"></div>
<br><input type="button" value="+" onclick="writeBtn();"><br><br>
<input name="submit" value="Download" type="submit" />
</form>
<script type="text/javascript">
url_id=1;
function writeBtn()
{
if (document.getElementById("input"+url_id).value != "")
{
url_id++;
var input = document.createElement("input");
input.type="text";
input.id="input"+url_id;
input.name="input"+url_id;
input.size="50";
input.required=true;
var br1 = document.createElement("br");
var br2 = document.createElement("br");
setTimeout(function(){
$(input).click();
},200);
document.getElementById('new').appendChild(br1);
document.getElementById('new').appendChild(input);
document.getElementById('new').appendChild(br2);
}
else
{
alert("Please complete the last url first.");
}
}
</script>
<?php
// i have just added this maximum execution time in seconds
set_time_limit (0);
function downloadFromUrl($url, $destination)
{
$newfname = $destination . basename($url);
$file = #fopen ($url, "rb");
if ($file)
{
$newf = fopen ($newfname, "wb");
if ($newf)
while(!feof($file))
{
fwrite($newf, fread($file, 1024 * 8 ), 1024 * 8 );
}
// and replaced this with if command
if ($file) {
fclose($file);
}
if ($newf) {
fclose($newf);
}
return true;
}
else
{
return false;
}
}
if (!isset($_POST['submit'])) die();
// folder to save downloaded files to. must end with slash
$destination_folder = 'downloads/';
for ($i=1; isset($_POST["input".$i]); $i++)
{
$url = $_POST["input".$i];
if(downloadFromUrl($url, $destination_folder))
echo "<br>Succesfully Downloaded from: ".$url;
else
echo "<br>Error Downloading from: ".$url;
}
?>
Explanation:
In the html form there is a div with id="new". In this div is where an input box is dynamically added when the button with the + sign is clicked. If you want to learn more about how this is achieved you can visit this link: http://www.w3schools.com/jsref/met_document_createelement.asp. The values from input boxes are afetrwards accessed by their names, which are input1, input2 etc. For this dynamic name giving the variable url_id is used in javascript part and is increased for every new input box. Also there is a feature I added, which doesn't allow user to create new input if the last one is empty and alerts a message. The if statement for this is:
if (document.getElementById("input"+url_id).value != "")
so, if you don't want this just delete this line and the else block. I also included your code for downloading in the function downloadFromUrl for simplicity reasons, which returns true if file was succesfully downloaded and false otherwise. In the php part there is a for loop that parses all inputs downloads the file that is in the url and echos a message if it was or not succesfully downloaded by calling the function downloadFromUrl.
As for your other question about the progress bar, I spent like the whole day trying to find a solution. Long story short, I wasn't able to make it work and I think it can't possibly work. I will explain you why. From php we can't change the value of the progress bar because php is executed and then prints all the results on screen, so we will only have the last change of the progress bar which will always be 100%. The other way is to use javascript. In that case we have to download the files from javascript because if we download them from php, by the time javascript starts executing, the files will have already been downloaded from php. Even if we execute a php script from javascript that downloads the files , we won't be able to tell when a file is downloaded, so we can't update the progress bar at the right time. So what is left? Downloading the files from javascript using XMLHttpRequest, which can't be achieved for 2 reasons. First, browsers don't allow this for security reasons and second and most important you can't upload the file to the server from javascript because javascript is a client-side language and doesn't have access to the server.
As you can see I tried many ways. If someone finds a way to do this, please let me know. I hope I helped you with your project. For any questions, problems, errors on the first half of the answer and suggestions or corrections for the second part of the answer are very welcome.
For the progress bar, I have already spent some time. There are two ways. Storing the total size of the remote file in a txt/Database file then compare each time. Or you can use AJAX at realtime. Hope if anyone can apply these ideas would share.

Trying to load a response sheet from php in the same page

I'm trying to load a response from the php onto the same page. My Client side html looks like this.
<p>
<script type="text/javascript">// <![CDATA[
function sendForm() {
var dataSend = "?ClientName=" + $("#ClientName").val();
$.post("AddClient.php", dataSend, function(data) {
$("#responseDiv").html(data);
});
// ]]></script>
</p>
<div id="responseDiv"> </div>
<form action="AddClient.php" onsubmit="sendForm()">
<h1>Client Wizard <span>Please fill all the texts in the fields.</span></h1>
<label> <span>Client Name :</span> <input id="ClientName" type="text" name="ClientName" /> </label> <span> </span> <input class="button" type="Submit" value="Send" />
</form>
My Server side php looks like this:
<?php
$dbhost='127.0.0.1';
$dbuser='name';
$dbpass='password';
$dbname='dbname';
$conn=mysqli_connect($dbhost,$dbuser,$dbpass,$dbname);
if(!$conn)
{
die('Could not connect:'.mysqli_connect_error());
}
$client=$_REQUEST["ClientName"];
$retval=mysqli_query($conn,"INSERT into client (clientid,clientname) VALUES (NULL,'$client')");
if(!$retval)
{
die('Could not add client:'.mysql_error());
}
$display_string="<h1>Client Added Successfully</h1>";
echo $display_string;
mysqli_close($conn);
?>
Unfortunately not only is the response being shown in anew html page, Its not accepting any name typed in the form. When I check the sql table the Column has a blank entry under it. I have not been able to figure out where I'm going wrong. Any help would be really appreciated.
All right. Your code have some room for improvement, but it's not an endless thing.
I saw somebody mention sanitization and validation. Alright, we got that. We can go in details here
This is how I will restructure your code using some improvements made by Samuel Cook (thank you!) and added a lot more.
index.html
<p>
<script type="text/javascript">// <![CDATA[
function sendForm() {
var dataSend = {clientName: $('#clientName').val()}
$.post("AddClient.php", dataSend, function(data) {
$('#responseDiv').html(data);
});
return false;
}
//]]>
</script>
</p>
<div id="responseDiv"></div>
<form action="AddClient.php" onsubmit="sendForm(); return false;">
<h1>Client Wizard <span>Please fill all the texts in the fields.</span></h1>
<label><span>Client Name :</span><input id="clientName" type="text" name="clientName"/><span></span><input type="submit" class="button" value="Send"></label>
</form>
Notice change in an input id and input name - it's now start with a lower case and now clientName instead of ClientName. It's look a little bit polished to my aesthetic eye.
You should take note on onsubmit attribute, especially return false. Because you don't prevent default form behavior you get a redirect, and in my case and probably your too, I've got two entries in my table with a empty field for one.
Nice. Let's go to server-side.
addClient.php
<?php
$dbhost = '127.0.0.1';
$dbuser = 'root';
$dbpass = '123';
$dbname = 'dbname';
$conn = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
if (!$conn) {
die('Could not connect: ' . mysqli_connect_error());
}
$client=$_REQUEST["clientName"];
$client = filter_var($client, FILTER_SANITIZE_STRING);
if (isset($client)) {
$stmt = $conn->prepare("INSERT into client(clientid, clientname) VALUES (NULL, ?)");
$stmt->bind_param('s', $client);
$stmt->execute();
}
if (!$stmt) {
die('Could not add client:' . $conn->error);
}
$display_string = "<h1>Client $client Added Successfully</h1>";
echo $display_string;
mysqli_close($conn);
?>
That is going on here. We are using PHP filters to sanitize our incoming from somewhere string.
Next, we check if that variable $client even exist (you remember that twice sended form xhr? Double security check!)
Here comes a fun part - to protect our selves even more, we start using prepared mySQL statements. There is no way someone could SQL inject you somehow.
And just check for any errors and display it. Here you go. I've tested it on my machine, so it works.
Forms default behavior is to redirect to the page given in the action attribute (and if it's empty, it refreshes the current page). If you want it to make a request without redirecting to another page, you need to use Javascript to intercept the request.
Here's an example in jQuery:
$('form').on('submit', function(e) {
e.preventDefault(); // This stops the form from doing it's normal behavior
var formData = $(this).serializeArray(); // https://api.jquery.com/serializeArray/
// http://api.jquery.com/jquery.ajax/
$.ajax($(this).attr('action'), {
data: formData,
success: function() {
// Show something on success response (200)
}, error: function() {
// Show something on error response
}, complete: function() {
// success or error is done
}
});
}
Would recommend having a beforeSend state where the user can't hit the submit button more than once (spinner, disabled button, etc.).
First off, you have a syntax error on your sendForm function. It's missing the closing bracket:
function sendForm() {
//...
}
Next, You need to stop the form from submitting to a new page. Using your onsubmit function you can stop this. In order to do so, return false in your function:
function sendForm() {
//...
return false;
}
Next, you aren't actually sending any POST data to your PHP page. Your second argument of your .post method shouldn't be a query string, but rather an object (I've commented out your line of code):
function sendForm() {
var dataSend = {ClientName:$("#ClientName").val()}
//var dataSend = "?ClientName=" + $("#ClientName").val();
$.post("AddClient.php", dataSend, function(data) {
$("#responseDiv").html(data);
});
return false;
}
Lastly, you have got to sanitize your data before you insert it into a database. You're leaving yourself open to a lot of vulnerabilities by not properly escaping your data.
You're almost there, your code just need a few tweaks!

Executing php file from another php file

Can I trigger the execution of a php file from another php file when performing an action? More specific, I have an anchor generated with echo, that has href to a pdf file. In addition of downloading the pdf I want to insert some information into a table. Here's my code, that doesn't work:
require('./database_connection.php');
$query = "select author,title,link,book_id from book where category='".$_REQUEST['categorie']."'";
$result = mysql_query($query);
$result2 = mysql_query("select user_id from user where username='".$_REQUEST["username"]."'");
$row2 = mysql_fetch_row($result2);
while($row= mysql_fetch_row($result))
{
echo '<h4>'.$row[0].' - '.$row[1].'</h4>';
if(isset($_SESSION["username"]) && !empty($_SESSION["username"]))
{
echo '<input type="hidden" name="id_carte" value="'.$row[3].'">';
echo '<input type="hidden" name="id_user" value="'.$row2[0].'">';
echo ' <script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
<script language="javascript">
function insert_download() {
$.ajax({
type: "GET",
url: "insert_download.php" ,
success : function() {
location.reload();
}
});
}
</script>
<a onclick="insert_download()" href="'.$row[2].'" download> download </a>';
}
And here's the insert_download.php:
<?php
require('./database_connection.php');
$query = "insert into download(user_id,book_id,date)values(".
$_REQUEST["id_carte"].",".
$_REQUEST["id_user"].",".
date("Y-m-d h:i:s").")";
mysql_query($query,$con);
mysql_close($con);
?>
Can anyone help me with this? Thanks!
As I understand correctly, you want to display a link, and when the user clicks that link,
some data is inserted into a database or something;
the user sees a download dialog, allowing him to download a file?
If this is correct, you can use this code:
On your webpage:
download
result.php:
<?php
$file = isset($_GET['file']) ? $_GET['file'] : "";
?>
<!DOCTYPE html>
<html>
<head>
<title>Downloading...</title>
<script type="text/javascript">
function redirect(url) {
//window.location.replace(url);
window.location.href = url;
}
</script>
</head>
<body>
Download is starting...
<script type="text/javascript">
redirect("http://example.com/download.php?file=dummy.pdf");
</script>
</body>
</html>
download.php:
<?php
$file = isset($_GET['file']) ? $_GET['file'] : "nullfile";
$file_url = "download_dir_or_something/".$file;
// Put some line in a log file...
file_put_contents("logfile.txt", "successful on ".date("Y-m-d H:i:s")."\n", FILE_APPEND);
// ...or anything else to execute, for example, inserting data into a database.
header("Content-Type: application/octet-stream");
header("Content-Transfer-Encoding: Binary");
header("Content-disposition: attachment; filename=\"".basename($file_url)."\"");
readfile($file_url);
?>
Why not use a redirection instead of "complicated" AJAX?
<!-- in your first document -->
echo '<input type="hidden" name="id_carte" value="'.$row[3].'">';
echo '<input type="hidden" name="id_user" value="'.$row2[0].'">';
echo 'download';
and in download_pdf.php
<?php
require('./database_connection.php');
...
mysql_close($con);
header("location: " . $_GET['redirect']);
You're lacking basic skill of debugging. If I was you, I should:
Use a browser which supporting, ex: Chrome with "Network" inspecting tab ready
Try click on the link <a onclick="insert_download()" ... and see if the ajax request is performed properly (via Network inspecting tab from your chrome). If not, re-check the generated js, otherwise, something wrong with the download_pdf.php, follow next step
Inspecting download_pdf.php: turn on error reporting on the beginning (put error_reporting(E_ALL); and ini_set('display_errors', 1); on top of your file) try echoing something before and/or after any line you suspect that lead to bugs. Then you can see those ajax response from your Network inspecting tab... By doing so, you're going to narrow down which line/scope of code is causing the problem.
Note that the "echoing" trick can be avoid if you have a solid IDE which is supporting debugger.
Hope it can help

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