In changename.php I have this DIV:
<div id="resultDiv"></div>
In the same file I have the PHP code:
<?php
include_once ('connect.php');
if(isset($_POST['name']))
{
$postedname = $_POST['name'];
$safename = mysqli_real_escape_string($con, $postedname);
$checkname = "SELECT * from accounts where name='$safename' LIMIT 1";
$query = mysqli_query($con, $checkname);
$row = mysqli_num_rows($query);
if($row) {
echo 'Name is already taken!';
} else {
$changename = "UPDATE accounts SET name='$safename' WHERE Name='$_SESSION[username]'";
$query = mysqli_query($con, $changename);
if($query)
{
echo 'Name is changed!';
$_SESSION['username'] = $safename;
}
}
}
?>
And this is my Jquery script (in a seperate file):
$(function(){
$("form").submit(function() {
event.preventDefault();
var name = $('#inputName').val();
$.post("changename.php",
{
name: name
},
function(data)
{
$("#resultDiv").text(data);
});
});
});
The code is just working fine, it successfully changes my name. If I want my name to be 'John', it successfully changes it in the DB.
The problem is, it's supposed to change the div with ID resultDiv to Name is already taken!. Instead of that, it takes the whole changename.php HTML code and puts it into that DIV. So instead of 1 line, I have 100+ lines of my page in there. So it looks like this:
https://gyazo.com/9320a5f15ed67a8ad9298d6172ab6909
Any idea why it's not just the message I want?
Ajax will fetch all data from file given as url for it.
If you want to avoid that, you have 2 options.
Write condition in changename.php in such a way that only update to db part will be executed.
On changename.php keep only required code to update to db.
Related
I realize that there are several similar questions that have been asked, but none of those have been able to get me over the top. Maybe what I wnat to do is just not possible?
I have a page on which there is an order form. The admin can create an order for any user in the database by selecting them in the dropdown menu and then fill out the form. But each user may have a PriceLevel that will give them a discount. So I need to be able to make a database call based on the username selected in the dropdown and display their price level and be able to use the username and pricelevel variables in my PHP.
I have the an add_order.php page on which the form resides, and an ajax.php which makes a quick DB call and returns the results in a json format.
The problem I am running into is actually getting the information from jQuery into the PHP. I have tried using the isset method, but it always comes back as false.
Here's what I have:
add_order.php
<?php
// $username = $_POST['orderUser']['Username'];
$username = isset($_POST['orderUser']) ? $_POST['orderUser']['Username'] : 'not here';
echo 'hello, ' . $username;
?>
...
$('#frm_Username').change(function() {
orderUser = $(this).val();
$.post('/admin/orders/ajax.php', {
action: 'fetchUser',
orderUser: orderUser
}
).success(function(data) {
if(data == 'error') {
alert('error');
} else {
console.log(data);
}
})
})
ajax.php
<?php
$action = $_POST['action'];
if($action == "fetchUser"):
$un = $_POST['orderUser'];
/*if($un):
echo $un;
exit;
endif;*/
// SET THE REST UP WITH MYSQL
if($un):
$qid = $DB->query("SELECT u.Username, u.PriceLevel FROM users as u WHERE u.Username = '" . $un . "'");
$row = $DB->fetchObject($qid);
// $row = jason_decode($row);
echo json_encode($row);
exit;
endif;
echo "error";
endif;
?>
I am logging to the console right now and getting this:
{"Username":"dev2","PriceLevel":"Tier 2"}
Any help would be appreciated. Thanks.
After calling $.post('/admin/orders/ajax.php', ...), the PHP code which sees your POSTed variable is ajax.php.
You need to check in there (inside ajax.php), whereas currently your isset check is in add_order.php, which does not see the POST request you send.
You do seem to have some logic in ajax.php, but whatever you've got in add_order.php is not going to see the data in question.
I am building a website and I can not figure out one thing. I need a script that checks if 10 seconds have past since load time, then it would run another PHP script. But I am not sure if it is possible. I have attached my attempt at this problem. Any ideas? Thanks in advance!
if (
<script type="text/javascript">
function viewplus(){
}
setTimeout(viewplus, 10000);
</script>
)
$query = "SELECT * FROM users WHERE id=" . $rws2['id_user'];
$result2 = mysqli_query($db, $query);
$rws2 = mysqli_fetch_array($result2);
$views_total = $rws2['views_total'] + 1;
$views_week = $rws2['views_week'] + 1;
$views_today = $rws2['views_today'] + 1;
$id = $rws2['id'];
$query = "UPDATE users SET views_total='$views_total',views_week='$views_week',views_today='$views_today' WHERE id='$id'";
mysqli_query($db, $query);
It is possible using jQuery like this:
Put your code in a php file and call it after 10 seconds after page load with
<script type="text/javascript">
// Check if the page has loaded completely
$(document).ready( function() {
setTimeout( function() {
$('#some_id').load('index.php');
}, 10000);
});
</script>
Example of updating a table and receiving a status message:
if (isset($_POST['id'])&&
isset($_POST['var'])){
$con = new mysqli("localhost", "my_user", "my_password", "world");
$id = $con->real_escape_string($_POST['id1']); //In our example id is INT
$var = $con->real_escape_string($_POST['var']);
$result = $con->query("UPDATE table SET value = '$var' WHERE id = $id);
(!$result) ? echo "Update failed!" : "Update succeeded";
}
This will load the output of your php file(Update filed or succeeded) in a element (this case one with id some_id).
Another way which I DO NOT recommend is using PHP's sleep() function PHP Documentation
I'm trying to make a query from my MySQL database here is the code
<?php
$link = new MySQLi('localhost','root','Rrtynt','copy');
if(isset($_POST['id'])){
$name = $_POST['id'];
$profile = 'profile';
$thestring = $name.$profile;
//echo $thestring;
$result = $link->query("SELECT Email,Name,idauth FROM user WHERE Email = '$name'");
echo $result;
}
?>
the code for the query
$result = $link->query("SELECT Email,Name,idauth FROM user WHERE Email = '$name'");
works in a different php script the same exact code but it keeps giving me http 500 error, I'm using this to post from a javascript file
$.post("/getfirstfolder.php", { id: value1 }, function (data) {
cop = data;
console.log("Data: " + data);
});
if I take out
$result = $link->query("SELECT Email,Name,idauth FROM user WHERE Email = '$name'");
and just echo $thestring it works fine, I cant figure out the problem so thank you for your time and your help is greatly appreciated
The fact that you mentioned $thestring, makes me think that perhaps you should be using $thestring as the parameter to the query like this:
$result = $link->query("SELECT Email,Name,idauth FROM user WHERE Email = '$thestring'");
I have two php files that handle a commenting system I have created for my website. On the index.php I have my form and an echo statement that prints out the user input from my database. I have another file called insert.php that actually takes in the user input and inserts that into my database before it is printed out.
My index.php basically looks like this
<form id="comment_form" action="insertCSAir.php" method="GET">
Comments:
<input type="text" class="text_cmt" name="field1_name" id="field1_name"/>
<input type="submit" name="submit" value="submit"/>
<input type='hidden' name='parent_id' id='parent_id' value='0'/>
</form>
<!--connects to database and queries to print out on site-->
<?php
$link = mysqli_connect('localhost', 'name', '', 'comment_schema');
$query="SELECT COMMENTS FROM csAirComment";
$results = mysqli_query($link,$query);
while ($row = mysqli_fetch_assoc($results)) {
echo '<div class="comment" >';
$output= $row["COMMENTS"];
//protects against cross site scripting
echo htmlspecialchars($output ,ENT_QUOTES,'UTF-8');
echo '</div>';
}
?>
I want users to be able to write comments and have it updated without reloading the page (which is why I will be using AJAX). This is the code I have added to the head tag
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
<script>
// this is the id of the form
$("#comment_form").submit(function(e) {
var url = "insert.php"; // the script where you handle the form input.
$.ajax({
type: "GET",
url: url,
data: $("#comment_form").serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
e.preventDefault(); // avoid to execute the actual submit of the form.
});
</script>
However, nothing is happening. The alert() doesn't actually do anything and I'm not exactly sure how to make it so that when the user comments, it gets added to my comments in order (it should be appending down the page). I think that the code I added is the basic of what needs to happen, but not even the alert is working. Any suggestions would be appreciated.
This is basically insert.php
if(!empty($_GET["field1_name"])) {
//protects against SQL injection
$field1_name = mysqli_real_escape_string($link, $_GET["field1_name"]);
$field1_name_array = explode(" ",$field1_name);
foreach($field1_name_array as $element){
$query = "SELECT replaceWord FROM changeWord WHERE badWord = '" . $element . "' ";
$query_link = mysqli_query($link,$query);
if(mysqli_num_rows($query_link)>0){
$row = mysqli_fetch_assoc($query_link);
$goodWord = $row['replaceWord'];
$element= $goodWord;
}
$newComment = $newComment." ".$element;
}
//Escape user inputs for security
$sql = "INSERT INTO parentComment (COMMENTS) VALUES ('$newComment')";
$result = mysqli_query($link, $sql);
//attempt insert query execution
header("Location:index.php");
die();
mysqli_close($link);
}
else{
die('comment is not set or not containing valid value');
it also filters out bad words which is why there's an if statement check for that.
<?php
if(!empty($_GET["field1_name"])) {
//protects against SQL injection
$field1_name = mysqli_real_escape_string($link, $_GET["field1_name"]);
$field1_name_array = explode(" ",$field1_name);
foreach($field1_name_array as $element)
{
$query = "SELECT replaceWord FROM changeWord WHERE badWord = '" . $element . "' ";
$query_link = mysqli_query($link,$query);
if(mysqli_num_rows($query_link)>0)
{
$row = mysqli_fetch_assoc($query_link);
$goodWord = $row['replaceWord'];
$element= $goodWord;
}
$newComment = $newComment." ".$element;
}
//Escape user inputs for security
$sql = "INSERT INTO parentComment (COMMENTS) VALUES ('$newComment')";
$result = mysqli_query($link, $sql);
//attempt insert query execution
if ($result)
{
http_response_code(200); //OK
//you may want to send it in json-format. its up to you
$json = [
'commment' => $newComment
];
print_r( json_encode($json) );
exit();
}
//header("Location:chess.php"); don't know why you would do that in an ajax-accessed file
//die();
mysqli_close($link);
}
else{
die('comment is not set or not containing valid value');
}
?>
<script>
// this is the id of the form
$("#comment_form").submit(function(e) {
var url = "insert.php"; // the script where you handle the form input.
$.ajax({
type: "GET", //Id recommend "post"
url: url,
dataType: json,
data: $("#comment_form").serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
$('#myElement').append( data.comment );
}
});
e.preventDefault(); // avoid to execute the actual submit of the form.
});
</script>
To get a response from "insert.php" you actually need to print/echo the content you want to handle in the "success()" from the ajax-request.
Also you want to set the response-code to 200 to make sure "success: function(data)" will be called. Otherwise you might end up in "error: function(data)".
I'd like to find a way of having a single page in the root of each of my web sections to hold all of the databae queries I'm calling.
I'm using a little script .....
<script type="text/javascript">
$(function() {
var availableTags = <?php include('fn-search-em.php'); ?>;
$("#quick-add").autocomplete({
source: availableTags,
autoFocus:true
});
});
</script>
.... to do SQL searches that appear as the user is typing. Similar to this ....
$sql = "SELECT * FROM stock_c_colours WHERE current_c_status = 'current' AND deleted = 'no'";
$result = mysqli_query($conn, $sql);
$results_list = array();
while($row = mysqli_fetch_array($result))
{
$colour_id = $row['id'];
$range_name = $row['range_name'];
$range_colour = $row['colour'];
$colour_code = $row['code'];
$p1 = $row['piece_size_1'];
$p2 = $row['piece_size_2'];
if($p1 > 1){
$p_mark = 'x';
}
else {
$p_mark = '';
}
$results_list[] = $range_name.' ('.$range_colour.' '.$colour_code.' '.$p1.$p_mark.$p2.') ID:'.$colour_id;
}
echo json_encode($results_list);
Echos a list in the form of a JSON array back to the text box and voila, a list. However, the site I'm working on at the moment has about 20 search boxes for various reasons scattered around (user request), does this mean I have to have 20 separate php function pages, each with their own query on, or can a single page be used?
I suspect the java needs modifying a little to call a specific function on a page of multiple queries, but I'm not good with Java, so some help would be greatly appreciated.
I did initially try adding ?action= to the end of the PHP address in the Java script, hoping a GET on the other end would be able to separate the PHP end into sections, but had no luck.
You need to change <?php include('fn-search-em.php'); ?>; to <?php $action = 'mode1'; include('fn-search-em.php'); ?>;.
Then in your fn-search-em.php file, use the $action variable to determine what kind of MySQL query you make.
For example:
if ($action == 'mode1')
$sql = "SELECT * FROM stock_c_colours WHERE current_c_status = 'current' AND deleted = 'no'";
else
$sql = "SELECT * FROM stock_c_colours WHERE current_c_status = 'mode1' AND deleted = 'no'";
You can do this with by creating a php file with a switch statement to control what code is executed during your Ajax call:
JS:
$.ajax({url: 'ajax.php', method: 'POST', async:true, data: 'ari=1&'+formData,complete: function(xhr){ var availableTags = JSON.parse(xhr.responseText);}});
PHP:
<?php
switch($_REQUEST['ari']){
case 1:
$sql = "SELECT * FROM stock_c_colours WHERE current_c_status = 'current' AND deleted = 'no'";
$result = mysqli_query($conn, $sql);
$results_list = array();
while($row = mysqli_fetch_array($result)){
$colour_id = $row['id'];
$range_name = $row['range_name'];
$range_colour = $row['colour'];
$colour_code = $row['code'];
$p1 = $row['piece_size_1'];
$p2 = $row['piece_size_2'];
if($p1 > 1){$p_mark = 'x';}
else { $p_mark = ''; }
$results_list[] = $range_name.' ('.$range_colour.' '.$colour_code.' '.$p1.$p_mark.$p2.') ID:'.$colour_id;
}
echo json_encode($results_list);
break;
case 2:
// another SQL Query can go here and will only get run if ARI == 2
break;
}
?>
This allows you to keep multiple AJAX handlers in the same file, you just need to pass the index for the desired handler when you make calls to the PHP file or nothing will happen.