Develop Reference

PHP script status bar

I have created a php script that submits a query and downloads the results in a CSV. I have created a bootstrap button where users can click to download the file. Some of the reports take longer to create and I wanted to show the user some sort of status that the script is running in the background.
I tried something simple like BlockUI from inside the PHP script. Using echo '<script type="text/javascript">$.blockUI();</script>'; in the beginning of the script and echo '<script type="text/javascript">$.unblockUI();</script>'; at the end but it didn't work.
Can someone help? I don't need a progress bar or anything fancy. I just need to show some type of status while the php script is running.
HTML:
...
<td class="pull-right"><a type="button" href="report1_download_csv.php" class="btn">Download</a></td>
...
PHP:
<?php
/* Set up and execute the query. */
$sql = "SELECT
FROM TABLE ";
$stmt = sqlsrv_query( $conn, $sql);
while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {
foreach($row AS $key => $value){
$pos = strpos($value, '"');
if ($pos !== false) {
$value = str_replace('"', '\"', $value);
}
$out .= '"'.$value.'",';
}
$out .= "\n";
}
sqlsrv_free_stmt($results);
sqlsrv_close($conn);
// Output to browser with the CSV mime type
header("Content-type: text/x-csv");
$date = date('m-d-Y-His');
header("Content-Disposition: attachment; filename=Report1_{$date}_UTC.csv");
echo "Column1, Column2\n";
echo $out;
?>

use ajax to send request to server to do operation then show a spinner or any thing else untill server give you response.

Header not working in php

I have a program that brings an image from the database and displays it inside an image div in my website. The below code was working successfully on my local wamp server but when I moved it to an online server it did not work anymore.
<?php
session_start();
include 'dbConnector.php';
$uID = $_SESSION['loggedUserID'];
$sql = "SELECT photo FROM hostmeuser WHERE userID = '$uID'";
$result = $conn->query($sql);
if(!$row = mysqli_fetch_assoc($result))
{
$imgData = "Assets/man.jpg";
}
else
{
$imgData = $row['photo'];
}
header("content-type: image/jpg");
echo $imgData;
?>
I have noticed that all (header) functions are not working on the new server and I have no control over this server so I replaced every:
header("Location: example.php")
with:
?>
<script type="text/javascript">
window.location.replace("example.php");
</script>
<?php
it is working fine now on most cases but not this one!
header("content-type: image/jpg");
Can you suggest any solution for this? or at least do you know how to represent this command in javascript?

onClick echo PHP code to div (AJAX)

So what I'm trying to do is the following:
first: When the button add-location is clicked this loads a php file using AJAX.
jQuery
$('.add-location').click(function() {
var locationName = $(this).data('location');
$.post('http://www.example.com/reload-location-2.php?addparam',
{locationName: locationName}, function(data) {
$("textarea#location").html(data);
})
});
PHP FILE
<?php start_session();
$locationName = array_key_exists('locationName',
$_POST) ? (string) $_POST['locationName'] : '';
if(isset($_GET['delparam'])){
unset($_SESSION['locations'][$locationName]);
}
if(isset($_GET['addparam'])){
$_SESSION['locations'][$locationName] = $locationName;
}
?>
<?php foreach ($_SESSION['locations'] as $location): ?>
<?php echo htmlspecialchars($location); echo "\n"; ?>
<?php endforeach;?>
This all works like a charm! No worries here. No what I'm trying to do is when the button add-location is clicked this echo's the following code to a div called textarea#thema.
<?php foreach ($_SESSION['products'] as $product): ?>
<?php echo htmlspecialchars($product); echo "\n"; ?>
<?php endforeach;?>
OR, if that's easier when the page is loaded -> echo that code to the textarea#thema div.
I do not understand AJAX that much but I'm getting there. But I think the last option maybe the easiest solution?
Could anyone help me figure this one out?
What I tried
Right now, When the button add-location is clicked I reload a previous jQuery script:
$('.add-location').click(function() {
var productName = $(this).data('product');
$.post('http://example.com/reload-2.php?addparam',
{productName: productName}, function(data) {
$("textarea#thema").html(data);
})
});
This works, but it adds an extra <p>-tag leaving me with a line break in the code.
I also changed the .html(data); to .val(data); for the textareas. Also I made a new PHP file with just this code:
<?php foreach ($_SESSION['products'] as $product): ?>
<?php echo htmlspecialchars($product); echo "\n"; ?>
<?php endforeach;?>
And this jQuery code:
$('.add-location').click(function() {
var productName = $(this).data('product');
$.post('http://www.example.com/reload-3.php',
{productName: productName}, function(data) {
$("textarea#thema").val(data);
})
});
But still no go... And I don't think this is the way to go?? I hope I gave enough information.

You should use .val() function to fill textarea, not .html().

how to make the result appear in popup widow?

I want to make the result comes out in a popup window.
Code:
<?php
$con=mysqli_connect("localhost","root","1234","fyp");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT password FROM admin WHERE email = '$_POST[email]' AND Admin = '$_POST[admin]'");
while($row = mysqli_fetch_array($result))
echo "your password is : " . " $row['password']" ;
mysqli_close($con);
?>
Is it possible to make it echoed in popup window like javascript alert messages ??
I have tried this but still not working
echo "<script> alert ("<?php echo 'your password is: ' . '$row['password']'?>")</script>";

I found a maybe strange solution for this a while back.
echo "<style onload=\"jsfunc($row['password']);\"></style>";
//in your html or javascript add this function
<script type='text/javascript'>
function jsfunc(data){
alert(data);
}
</script>
the style tag is invisible and will run the function onload, so when its gets echoed. Also I use the style tag because its one of the few tags where the onload works when you echo it like this.
Its a strange solution but it works.

There are some serious flaws in your code, including it being open to SQL Injection. I'd recommend changing it to look more like this:
$con = new MySQLi("localhost","root","1234","fyp");
if($con->connect_errorno) {
echo "Failed to connect to MySQL: ".$sql->connect_error;
}
$email = $sql->real_escape_string($_POST['email']);
$admin = $sql->real_escape_string($_POST['admin']);
$query = "SELECT password FROM admin WHERE email = '$email' AND Admin = '$admin'";
$result = $sql->query($query);
if($result) {
$row = mysqli_fetch_assoc($result);
$pass = $row['password'];
echo '<script> alert("Your password is '.$pass.'");</script>';
} else {
//do some error handling
}
//the closing of a mysqli connection is not required but
mysqli_close($con);
Real escape string is not 100% proof against injection but is a good place to start with sanitising your inputs. Additionally I would strongly advise against storing your passwords in plain text. Please take a look at the PHP sha1() function or the SQL equivalent when storing the passwords initially.

Use this code
<?php
$alert='your password is: '.$row['password'];
?>
<script>
alert("<?php echo $alert; ?>");
</script>
It will work for sure.

Javascript and PHP scan for nudity

I am trying to not allow the uploading of files that have nudity to my server. I found javascript online that will scan a photo for nudity. It comes with demo pics and an html file and js files. I am using PHP to upload the file and I am having trouble not allowing if the scan find that the pic has nudity.
Here is my code sample:
$q= "insert into $table values('', '$email', '$aim', '$icq', '$yahoo', '$homepage', '0', '0', '0', '0', '0', '0', '', now(),'$myip','$email2','$password','$title','$download','$approved','$allowdelete','$author','$facebook','$piclink','$domain','$option3','$secret')";
$result = mysql_query($q) or die("Failed: $sql - ".mysql_error());
$q = "select max(id) from $table";
$result = mysql_query($q);
$resrow = mysql_fetch_row($result);
$id = $resrow[0];
$file = $_FILES['file']['name'];
move_uploaded_file($_FILES['file']['tmp_name'], "pics/".$id.".".$picext);
$picfile=$id.".".$picext;
echo '<script type="text/javascript" <src="nude.js">';
echo 'nude.load("pics/".<? echo $picfile; ?>);nude.scan(function(result){if(!result){ <? $nude = false; ?>;}else{ $nude = true;}})';
echo '</script>';
if ($nude === false) {
$q = "update $table set picfile = '".$id.".".$picext."' where id='$id'";
$result = mysql_query($q);
Header("Location: index.php?id=$id");
} else{
echo '<script type="text/javascript">';
echo 'alert("Nudity found. Please try again.")';
echo '</script>';
$q = "delete from $table where id='$id'";
$result = mysql_query($q);
unlink("pics/".$picfile);
Header("Location: new2.php");
}
The code uploads the file and then it's supposed to check the file for nudity and delete it and tell the user to try again if nudity is found. If nudity is not found the user is brought to the main page of the site.(This is the add new photo page). All of the PHP is working fine, but since the javascript doesn't seem to be running the file i uploaded and then since $nude isn't set it goes into the else of the if statement and again the js doesnt run(no alert box), and then the file is deleted. How can I make the javascript run to scan my uploaded pic for nudity? What am I doing wrong here?
Any help is greatly appreciated!
P.S.
For those that would like to see the js file that is doing the scanning: http://pastebin.com/MpG7HntQ

The problem is that this line:
echo 'nude.load("pics/".<? echo $picfile; ?>);nude.scan(function(result){if(!result){ <? $nude = false; ?>;}else{ $nude = true;}})';
Doesn't do what you think it does.
When you output JavaScript via echo(), that code runs on the browser or client side and doesn't run until after the PHP script has finished.
You'll either need to port the code to PHP or use an AJAX call to report the validity of the images.