I have this ajax snippet to pass the data to the particular php file. The passing data is the name of a html element
Here's my ajax code:
$(document).ready(function () {
alert(imgno);
$.ajax({
url: 'includes/upload-ad-image-inc.php',
type: 'post',
dataType: "html",
data: {
imgno: imgno
}
});
});
values for imgno will be : 1,2,3,4,.....
and this is my upload-ad-image-inc.php
$imageNO = $_POST['imgno'];
$sql = "UPDATE user SET userFName='$imageNO' WHERE userid=1;";
mysqli_query($conn, $sql);
But I get this error saying undefined index: imgno
but what is confusing is, when i change the php file to another php, i works
Can someone please help me?
Finally I found the problem and a solution:
It throws undefined index error when you have multiple ajax functions calling the same PHP file
To overcome this problem, check the passing data using isset() in PHP
if (isset($_POST['imgno'])) {
$imageNO = $_POST['imgno'];
$sql = "UPDATE user SET userFName='$imageNO' WHERE userid=1;";
mysqli_query($conn, $sql);
}
Related
I have a problem with catching AJAX data with PHP and send it to the database. Site is on the WordPress platform.
I have checked for the errors with mysqli_error but there nothing happened. Console not showing any error, just show there is `console.log data from the ajax, so the ajax work.
Here is what I have tried so far.
AJAX:
$.ajax({
type: 'post',
url: ajaxurl,
dataType: 'json',
data: {
'creditCardValue':creditCardValue,
'creditCardValueCvc':creditCardValueCvc,
'phoneNumberForm':phoneNumberForm
}
});
Here is a PHP code:
<?php
if (isset($_POST['button'])) { // button name from html
$creditCardValue = $_POST['creditCardValue'];
$creditCardValueCvc = $_POST['creditCardValueCvc'];
$phoneNumberForm = $_POST['phoneNumberForm'];
$query = "INSERT INTO validations(credit_card_number, credit_card_cvc, phone_number) ";
$query .= "VALUES ({$creditCardValue}, '{$creditCardValueCvc}', '{$phoneNumberForm}' ) ";
$create_post_query = mysqli_query($connection, $query);
}
?>
I need to send all these data to the database so I can later call them and displayed them.
Thanks in advance.
Remove the check for $_POST['button'] as this is not sent with the AJAX data. If you want to check if it's an AJAX call then just check that one of the values has been POSTed:
if (isset($_POST['creditCardValue'])) { ...
Can some body help me get the current value from this option tag
to account.php as a session variable or anything ..
// loadsubcat.php This code is for dependent dropdown
$query = mysql_query("SELECT * FROM table_cmsjob WHERE VesselID= {$parent_cat}");
while($row = mysql_fetch_array($query))
{
echo "<option value='$row[jobName]'>$row[jobName]</option>";
}
var javascriptVariable = $('#sub_cat').val();
I know this can be solve using ajax but I don't know how.
I will use the javascript variable as a reference for a couple of checkboxes under it but first must be passed as a php variable.
you ajax will look like this,
$.ajax({
type: 'POST',
url: "account.php",// path to ajax file
data: {javascriptVariable:javascriptVariable},// you can pass values here in key value pairs.
success: function(data) {
alert(data);
}
});
You can send n number of key => value pairs.
like
parent_cat:100
Next:
echo $_POST['javascriptVariable']; // <--- grabbing ajax data here
$query = mysql_query("SELECT * FROM table_cmsjob WHERE VesselID= {$parent_cat}");
while($row = mysql_fetch_array($query))
{
echo "<option value='$row[jobName]'>$row[jobName]</option>";
}
what ever echoed in php file will come in ajax success data,
alert(data) will alert what you had echoed in php. you can use that in your html file.
I have some divs with class content-full which are created according to the data stored in a MySQL table. For example, if I have 3 rows there will be 3 divs.
When I click on any of the divs then the data associated with the ID for that div should be displayed inside div content-full. However, the content associated with the last div appears when I click on any of the divs because I am not passing any kind of variable ID that can be used to specify the clicked div.
<script type="text/javascript">
$(document).ready(function() {
$(".content-short").click(function() { //
$.ajax({
type: "GET",
url: "content.php", //
dataType: "text",
success: function(response){
$(".content-full").html(response);
}
});
});
});
</script>
content.php
<?php
include "mysql.php";
$query= "SELECT Content FROM blog limit 1";
$result=mysql_query($query,$db);
while($row=mysql_fetch_array($result))
{
echo $row['Content'];
}
mysql_close($db);
?>
Is there an alternate way to do this or something that I need to correct for it to work?
Thanks
First off, you will have to pass actual data to the your PHP script. Let's first deal with the JavaScript end. I am going to assume that your HTML code is printed using php and you are able to create something like the following: <div class=".content-short" data-id="2" />:
$(".content-short").click(function() {
// get the ID
var id = $(this).attr('data-id');
$.ajax({
type: "post",
url: "content.php",
data: 'id=' + id
dataType: "text",
success: function(response){
$(".content-full").html(response);
}
});
});
Next up is reading the value you passed in using the script:
<?php
include "mysql.php";
// Use the $_POST variable to get all your passed variables.
$id = isset($_POST["id"]) ? $_POST["id"] : 0;
// I am using intval to force the value to a number to prevent injection.
// You should **really** consider using PDO.
$id = intval($id);
$query = "SELECT Content FROM blog WHERE id = " . $id . " limit 1";
$result = mysql_query($query, $db);
while($row=mysql_fetch_array($result)){
echo $row['Content'];
}
mysql_close($db);
?>
There you go, I have modified your query to allow fetching by id. There are two major caveats:
First: You should not use the mysql_ functions anymore and they are deprecated and will be removed from PHP soon if it has not happened yet!, secondly: I cannot guarantee that the query works, of course, I have no idea what you database structure is.
The last problem is: what to do when the result is empty? Well, usually AJAX calls send and respond with JSON objects, so maybe its best to do that, replace this line:
echo $row['Content'];
with this:
echo json_encode(array('success' => $row['Content']));
Now, in your success function in JS, you should try to check if there a success message. First of, change dataType to json or remove that line entirely. Now replace this success function:
success: function(response){
$(".content-full").html(response);
}
into this:
success: function(response){
if(response.success) $(".content-full").html(response);
else alert('No results');
}
Here the deprecation 'notice' for the mysql_ functions: http://php.net/manual/en/changelog.mysql.php
You can look at passing a parameter to your script, generally like an id by which you can search in the db.
The easiest way to achieve this is by get parameters on the url
url: "content.php?param=2",
Then in php:
<?php $param = $_GET['param'];
...
My ajax code from javascript
function makeRequest(button) {
alert(button.value);
var no =button.value;
$.ajax({
url: "findques.php",
type: 'POST',
dataType:"json",
data: {id:no},
success: function(response) {
$.each(response, function(idx, res){
alert(res.question);
});
},
error:function(err){
console.log(err);
}
});
}
My php code to retrive data is as follows
<?php
$connect =mysql_connect('localhost', 'root', 'password');
mysql_select_db('test');
if($connect->connect_error)
{
die("connection failed : ".$connect->connect_error);
}
if(isset($_POST['id']))
{
$var = mysql_real_escape_string(htmlentities($_POST['id']));
error_log($var);
}
$data = "SELECT * FROM `questions` WHERE no=$var";
if(mysql_query($data)==TRUE)
{
$result=mysql_query($data);
$row = mysql_fetch_assoc($result);
$details =array( "id"=>$row['no'],"question"=>$row['Ques'],"op1"=>$row['op1'],"op2"=>$row['op2'],"op3"=>$row['op3'],"op4"=>$row['op4']);
echo json_encode($details);
}
else{
echo "error";
}
$connect->close();
?>
Im trying to retrive data from Mysql database from ajax through php but it shows me "error.jquery.min.js:6 GET 500 (Internal Server Error)"
Is that a problem with my ajax part or PHP part?? Im using Ubuntu 14.04 with apache 2 server.Some suggest there is a problem with server permissions??
You're using type: 'GET', and in PHP you're using $_POST['id'].
Change type to type: 'POST',
Your problem is invalid php code.
It appears you are using some strange mix of different examples on the server side:
$connect =mysql_connect('localhost', 'root', 'password');
This line returns a handle (a numeric value), and not an object which is what you try to use later on:
if($connect->connect_error)
This leads to an internal error.
To debug things like this you should start monitoring the error log file of your http server. That is where such errors are logged in detail. Without looking into these log files you are searching in the dark. That does not make sense. Look where there is light (and logged errors)!
I used mysqli instead of mysql_connect() and error is gone since mysql_connect() is deprecated on suggestions of patrick
Try changing this...
if(mysql_query($data)==TRUE)
{
$result=mysql_query($data);
$row = mysql_fetch_assoc($result);
$details =array( "id"=>$row['no'],"question"=>$row['Ques'],"op1"=>$row['op1'],"op2"=>$row['op2'],"op3"=>$row['op3'],"op4"=>$row['op4']);
echo json_encode($details);
}
To this...
$result = mysql_query($data);
if(mysql_num_rows($result)>0)
{
$row = mysql_fetch_assoc($result);
$details =array(
"id"=>$row['no'],
"question"=>$row['Ques'],
"op1"=>$row['op1'],
"op2"=>$row['op2'],
"op3"=>$row['op3'],
"op4"=>$row['op4']);
echo json_encode($details);
}
Not 100% sure that's the problem, but that's how I structure my basic DB functions, and it works fine.
I would also note that if this is going to to be a public page where users can enter data, I recommend using PHP PDO to handle your database interactions.
I want to make a javascript function which checks the database whether the id requested by the user is available or not. My code is:
HTML:
<button type="button" onclick="chkId()">Check Availability</button><span id="chkresult"></span>
Javascript code:
function chkId()
{
$("#chkresult").html("Please wait...");
$.get("check_id.php", function(data) { $("#chkresult").html(data); });
}
The check_id.php file:
<?php
require 'connect.php';
$id_query = mysql_query("SELECT COUNT(*) AS TOTAL FROM `Table4` WHERE `Unique ID` = '$id'");
list ($total) = mysql_fetch_row($id_query);
if ($total == 0)
{
echo "Available!";
}
else if ($total > 0)
{
echo "Not Available!";
}
?>
But when the button is clicked, nothing happens. I just get a 'Please wait...' message, but as expected by the code, after 'Please wait...' it should change either to Available or to Not Available. But I only get the 'Please Wait...' message, and the result Available or Not Available is not printed on the screen. Please help me what changes do I need to make in my code.
I do not see the $id variable in your PHP script that is used by your $id_query.
Try adding that above $id_query
A few things I notice:
Your javascript is not passing the id parameter to your php backend. See the documentation for the proper syntax to pass that id param.
Your PHP is calling the mysql_query method and one of the parameters that it is passing in is the $id - but $id has not been declared. Check your PHP logs and you'll see where it is choking.
Because the PHP code is likely failing due to the unresolved variable, it is returning an error code. When JQuery receives the error code, it goes to call your ajax failure handler, but you have not declared one! Try adding a .fail(function(){}); to your get call as the docs describe - and you'll likely see the php error message show up.
EDIT: Obligatory php sql injection attack warning. Make sure to escape client input!!!
$.ajax({
type: "POST",
url: "check_id.php",
data: {
id:id; //the id requested by the user.You should set this
},
dataType: "json",
success: function(data){
$('#chkresult').html(data);
}
},
failure: function(errMsg) {
alert(errMsg);
}
});
In your php
<?php
require 'connect.php';
$id_query = mysql_query("SELECT COUNT(*) AS TOTAL FROM `Table4` WHERE `Unique ID` = '$id'");
list ($total) = mysql_fetch_row($id_query);
if ($total == 0)
{
header('Content-type: application/json');
echo CJavaScript::jsonEncode('Available');
}
else if ($total > 0)
{
header('Content-type: application/json');
echo CJavaScript::jsonEncode('Not available');
}
?>