my json script is:
// add button .click
$('a.add').click(function(){
$('#loader').show();
var url = "/?"+$("form[name='jsms_add']").serialize();
ajx = $.ajax({
url: url,
type: 'post',
data:{ajax:1},
dataType: 'json',
success: function(r) {
$('#loader').hide();
if(r.r != 0){
alert("ok");
jsmsalert($('#alert_add'),'success',r.m);
apendtable(r.r)
$("tr").removeClass("odd");
$("tr.viewrow:odd").addClass("odd");
$("tr.editrow:odd").addClass("odd");
$('td[colspan="7"]').remove();
}
else{
jsmsalert($('#alert_add'),'error',r.m,0);
}
},
error: function(request, status, err) {
$('#loader').hide();
jsmsalert($('#alert_add'),'error','error...');
alert( "ERROR: " + err + " - " );
}
});
and in my php file i have
$data = array(
'r' => '1',
'm' => '1',
);
json_encode($data);
now i want to know that how can i send a value to json that r.r != 0 be true and my success code execute?
In Firefox the error is "SyntaxError: JSON.parse: unexpected character"
this code used by another site by i don't know what happend in php file
my problem is that how can i send some data (for example r=1) to my json, because i want to check that if r=1 (means success) do something and show text from m (message) in my page
please help me to correct php file
thank
In your php file make sure you are echoing the encoded data,
like so:
$data = array(
'r' => '1',
'm' => '1',
);
echo json_encode($data);
If you don't echo the data it will not make it back to your JavaScript.
If your JavaScript is what's throwing the errors try commenting out the dataType: 'json', and console logging the return. It might be php is throwing errors that jquery is refusing to parse.
$.ajax({'url': url,type: 'post',data:{ajax:1}, success: function(returnData){
console.log(returnData);
}});
Related
I'm trying to pass a simple string variable with an onclick event, the request is successful but I get an empty response on the console, and the xvar variable is no coming through so I get the "NO DATA" p tag. This is my first time using Ajax so I tried to make it very simple to start. Here's my code:
JS
var xvar = 'D';
$('.test').click( () => {
$.ajax({
data: { xvar : xvar },
type: "POST",
url: 'test.php',
success: function(data, textStatus, XMLHttpRequest)
{
console.log('Success: '+data)
},
error: function(XMLHttpRequest, textStatus, errorThrown){
console.log("The request failed."+errorThrown);
}
});
});
test.PHP
$xvar = (isset($_POST['xvar'])) ? $_POST['xvar'] : '<p>NO DATA</p>';
echo $xvar;
I'm using JQuery 3.5.1 that is included with Wordpress. I'll appreciate any feedback.
EDIT
I managed to get the response on test.php as seen
here:
But I when I try to render the value with print_r is not shown.
$res = $_POST;
print_r($res);
On the data of your Ajax, try to add the quotes in the key of the data. Like this:
data: { "xvar" : xvar },
I'm trying to create a forum and jquery throws an 'illegal invocation' error.
Here is my jquery code:
$('#formSumbit').on('submit', function(e) {
e.preventDefault();
$.ajax({
url: 'data-get.php',
type: 'POST',
data: new FormData(this),
contentType: false,
dataType: 'json',
success: function(value) {
var serialize = $.parseJSON(value);
if (serialize.success == 'false') {
$('.alert').fadeIn().delay(3000).fadeOut();
$('.alert-msgText').html(serialize.datamsg);
}
}
});
});
And here is my PHP code:
<?php
$user = $_POST['user'];
$msg = $_POST['message'];
if(empty($user)&&empty($message)) {
$data = array(
'success' => 'false',
'datamsg' => 'Please fill the textboxes'
);
echo json_encode($data);
} else {
mysqli_query($con,"INSERT INTO forums(name,message) VALUES ('$user','$msg')");
$data = array(
'success' => 'true',
'datamsg' => 'Done!'
);
echo json_encode($data);
}
exit();
?>
When the textboxes are empty and i click the submit button, nothing seems to work and jquery throws an illegal invocation error. I don't understand what the problem is. Can you please help?
And thanks in advance!
1) You have a typo mismatch between your form and your JavaScript:
<form id="formSubmit" and $('#formSumbit') - it should be $('#formSubmit') to match the spellings.
2) Unless you are trying to upload files via this AJAX request, then you can simplify things by replacing data: new FormData(this), contentType: false, with just data: $(this).serialize(). This will get rid of the illegal invocation error.
3) Writing dataType: 'json' means that jQuery will automatically try to parse the data coming from the server as JSON, and convert it. Therefore, in your "success" function, value will already be parsed and converted to an object. In turn therefore, using $.parseJSON is not necessary. You can just access value.success directly, for instance.
Here's a fixed version:
$('#formSubmit').on('submit', function(e) {
e.preventDefault();
$.ajax({
url: 'data-get.php',
type: 'POST',
data: $(this).serialize(),
dataType: 'json',
success: function(value) {
if (value.success == 'false') {
$('.alert').fadeIn().delay(3000).fadeOut();
$('.alert-msgText').html(value.datamsg);
}
}
});
});
Working demo: https://jsfiddle.net/khp5rs9m/2/ (In the demo I changed your URL for a fake one, just so it would get a response, but you can see where I have altered it and left your settings in the commented-out part).
i have problem with ajax post submit
here is my script
function postad(actionurl) {
if (requestRunning) return false ;
if (! $("#editadform").valid()) {
validator.focusInvalid();
return false ;
}
$('#ajxsave').show() ;
requestRunning = true ;
var postData = $('#editadform').serializeArray();
$.ajax(
{
url : actionurl,
type: "POST",
data : postData,
success:function(data, textStatus, jqXHR)
{
$('#diverrors').html(data.errors) ;
$('#divalerts').html(data.alerts) ;
if (data.status=='success') { alert(data.status);
$('#siteid').val(data.siteid) ;
if ($('#adimager').val())
$('#divlmsg').html(data.alertimage) ;
$('#editadform').submit() ;
} else {
$('#ajxsave').hide() ;
}
},
error: function(jqXHR, textStatus, errorThrown)
{
$('#ajxsave').hide() ;
},
complete: function() {
requestRunning = false;
}
});
$('.btn').blur() // remove focus
return false ;
}
This works on if (textStatus=='success') {
but when the action fails, alert(data.status) shows Undefined.
Using FireBug, I can see that the data is correctly returned. Why is data.status "Undefined" then?
If you don't specify the dataType field of an $.ajax() call in jQuery, it formats the response as plain text. A workaround to your code would be to either include dataType: "JSON" into your $.ajax() parameters, or alternatively, in your success function, parse the plain text response as a JSON object, by using the folllowing:
data = JSON.parse(data); // You can now access the different fields of your JSON object
UPDATE:
yes i have not status field in action url, how to add data status field in php code?
When creating your PHP script that is intended to return the JSON data, you first need to build an array and then encode it as JSON.
So, suppose you have a PHP script that either succeeds and produces some data that you put into a $data variable, or fails, then the following style could be adopted:
<?php
// ^^ Do your PHP processing here
if($success) { // Your PHP script was successful?
$response = array("status" => "success!", "response" => $data);
echo json_encode($response);
}
else {
$reponse = array("status" => "fail", "message" => "something went wrong...");
echo json_encode($response);
}
?>
I've a such js-function, which sends data to file, which contains a function, which sends back JSON-response:
function deleteCategory(id){
console.log(id);
$.ajax({
url: '<?php echo $cfg['options']['siteurl']; ?>/gears/ajax.deletePkgCatAsideMenu.php',
type: 'POST',
dataType: 'JSON',
data: {idItem:id},
success: function(data) {
console.log(data);
if (data.type=='error') {
notify(data.type, data.type, data.text);
} else{
document.location.reload();
}
},
error: function(v1,v2,v3) {
alert('Ошибка!\nПопробуйте позже.');//in english it will be: alert('Error!\nTry again later.');
console.log(v1,v2,v3);
}
});
}
In the end of that file these actions take place:
$q = 'DELETE FROM `pkg_cat_aside_menu` WHERE pkg_cat_ddlist_id='.$idItem;
$db->query($q);
exit(json_encode(array('type'=>'ok','text'=>'Удаление произведено!')));
//or in english: exit(json_encode(array('type'=>'ok','text'=>'Deleted!')));
This is a response text:
"int(13) {"type":"ok","text":"\u0423\u0434\u0430\u043b\u0435\u043d\u0438\u0435 \u043f\u0440\u043e\u0438\u0437\u0432\u0435\u0434\u0435\u043d\u043e!"}"
And error syntax error: unexpected token i(...)
I think, it take place, because I did actions with data base (delete), which response a deleted record number (13), and it has been included to json-respose. How to fix it?
int(13) makes me think that you have a var_dump( $idItem ); somewhere earlier in your PHP code, which is causing the response to be invalid.
I have my AJAX code here
$("#add-student").click(function(e) {
e.preventDefault();
formData = $("#student-form").serialize();
if (cleanFormInput()) {
sendTheInfo(formData);
} else {
shakeForm();
}
});
function sendTheInfo(formData) {
$.ajax({
type: "POST",
url: "../classes/ajax/postNewStudent.php",
data: formData,
statusCode: {
404: function() {
alert( "page not found" );
}
},
success: function(formData) {
console.log("New student submitted:\n" + formData);
//clearForms();
},
error: function(result, sts, err) {
console.warn("Connection error:\n" + err + " : " + sts);
console.log(result);
shakeForm();
},
complete: function() {
console.log("Everything complete");
}
});
}
Always without fail outputs this error:
Connection error:
SyntaxError: Unexpected end of input : parsererror
But still gives the complete message: Everything complete
Update, PHP code here:
require '../../core/init.php';
require '../../classes/Config.php';
header('Content-Type: application/json');
if (!empty($_POST)) {
$id = $_POST["sid"];
$first = $_POST["first"];
$last = $_POST["last"];
$fav = "0";
$sql = "INSERT INTO `students` (`id`, `first`, `last`, `active`) VALUES ('{$id}', '{$first}', '{$last}', '{$fav}')";
$link = mysql_connect(Config::get('mysql/host'),Config::get('mysql/username'),Config::get('mysql/password')) or die("could not connect");;
mysql_select_db(Config::get('mysql/db'), $link);
$result = mysql_query($sql, $link);
if ($result) {
header('Content-Type: application/json');
$student_data = $id . $first . $last . $fav;
echo json_encode($student_data);
}
}
I'm a bit confused, am I doing my ajax set up wrong? Or is it something else in by backend code wrong? I'm using MySQL and jQuery 2.0.3
Updated code here: here
I have had a look at your code. I saw that from the PHP side you are sending a JSON object. but you didn't specified the return dataType for the response. Try to add the dataType in the ajax call. Maybe that will solve the problem
function sendTheInfo(formData) {
$.ajax({
type: "POST",
url: "../classes/ajax/postNewStudent.php",
data: formData,
dataType : 'json',
statusCode: {
404: function() {
alert( "page not found" );
}
},
success: function(formData) {
console.log("New student submitted:\n" + formData);
//clearForms();
},
error: function(result, sts, err) {
console.warn("Connection error:\n" + err + " : " + sts);
console.log(result);
shakeForm();
},
complete: function() {
console.log("Everything complete");
}
});
}
It should be noted that the Ajax COMPLETE method will fire even if the back end does not return a result.
complete: function() {
console.log("Everything complete");
}
will thus show the log'ed entry every time an ajax call is 'finished executing', even if the submit failed.
I would also suggest NOT having 2 headers or the same declaration (you have one up top, and one in the if(result) call.
In a comment thread, you pointed out that you're working on the server but not locally, And thus that implies you have some pathing issues. Check your
../
relative path style urls and make sure that you have the same basepoints.
removed my old answer. I don't think it is an ajax/javascript error. it's definitely a PHP error. It's this lines:
$student_data = $id . $first . $last . $fav;
echo json_encode($student_data);
You $student_data is not an array, it's just a string. You need to pass an array into the json_encode function