I have the following Jquery code that sends a ajax request to add-to-cart.php.
var productData = {
"product_id": s_product_id,
"product_opties": s_product_opties,
"product_aantal": s_product_aantal
}
productData = JSON.stringify(productData);
$.ajax({
url: 'inc/add-to-cart.php',
dataType: "json",
contentType: "application/json; charset=utf-8",
data: productData,
type: 'POST',
success: function(response) {
alert(response.d);
},
error: function(e){
console.log(e);
}
});
Add-to-cart.php:
<?php
print $_POST['product_id'];
?>
I am having two problems, the first one is that the $_POST['product_id'] does not exists when i ask for it and secondly when the ajax response returns it goes directly to the error: function and not succes function
The console log says:
[object Object]{readyState: 4, responseText: "<br /> <b>N...", status: 200, statusText: "OK"}
If the status is OK why does it jump to the error: part?
Thanks!
Try with:
...
var productData = {
'product_id': s_product_id,
'product_opties': product_opties,
'product_aantal': product_aantal,
}
$.ajax({
url: 'inc/add-to-cart.php',
dataType: 'json',
data: productData,
type: 'POST',
success: function(response) {
console.log(response.d);
},
error: function(e){
console.log(e);
}
});
...
Omitting the AJAX contentType parameter and the part where you stringify your JSON, that's already ready to be sent to your url, so, isn't needed.
Remove the line
productData = JSON.stringify(productData);
Then do not return HTML <br> ... from add_to_cart.php, you need to return a JSON string created by the PHP function json_encode and NOTHING ELSE.
Like in:
<?php echo json_encode(array('d' => 'value of d'));
First: status Code from the Webserver is 200 cause he deliverd an existing site.
Second: You dont need to stringify the json object by urself
Related
I have this small but annoying problem. I really not usual with a web thing. I try to request to my php file using ajax jquery. When I want to retrieve the data I send from ajax, it return undefined index. I dunno what's the problem, it make me spend a lot of time to solve it. Thanks
Below is my ajax code
var at=this.name.substring(this.name.length,7);
var value_header = $("#key"+at).val();
var jsObj = { new_value:value_header, id:at, data:'header'};
console.log(JSON.stringify(jsObj));
$.ajax({
type: 'POST',
headers: 'application/urlformencoded',
url: 'admin_crud.php',
data: jsObj,
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function(data){
console.log("Sukses");
}
When I call the below code in my php file, the result is 'Undefined index: data'
echo $_POST['data'];
//Edit
So, when I try var_dump($_POST);, the result is array(0) {}. Where is my mistake? I thought I had send the right one
//Edit
As I mention above, I want it to run perfect without error. Thanks
Remove headers, change your datatype to text and catch errors in the ajax call
$.ajax({
type: "POST",
dataType: "text",
data: jsObj,
url: "admin_crud.php",
success: function (result) {
console.log("success", result);
},
error: function (e) {
console.log("Unsuccessful:", e);
}
});
I have another solution beside #Marco Sanchez too, I don't know it always work or not, but in my case, it work :
$.ajax({
type: 'POST',
url: 'admin_crud.php',
headers: "Content-type: application/x-www-form-urlencoded"
data: "new_value="+value_header+"&id="+at+"&data=header",
success: function(data){
console.log("Sukses");
console.log(data);
}
});
I Am trying to send value from ajax to php and retrieve it just to test that everything is work, when i click in a button to test i got error and alert('Failed') Appears , how can i fix it in order to get success? thanks
Ajax :
var a = "test";
$.ajax({
url: "search.php",
dataType: "json",
data: a ,
success: function(data) {
alert('Successfully');
},
error: function(data) {
alert('Failed');
}
})
PHP :
<?php
$pictures = "img1";
echo json_encode($pictures);
?>
I refined your code slightly and it works.
var a = "test";
$.ajax({
type: 'POST',
url: 'search.php',
data: 'a=' + a,
dataType: 'json',
cache: false,
success: function (result) {
alert('Successful');
},
error: function (result) {
alert('Failed');
}
});
If you're requesting a JSON, use the $.getJSON from jQuery, it's aready parse the JSON into a JSON object for you.
Seems that you're not return an actual JSON from server, maybe this is what is causing the error.
If you're seeing the 'Failed' message probably the problem is a 500 error which is a server error.
Try this code above.
Javascript:
var a = "test";
$.getJSON("search.php", {
a: a
}, function (json) {
console.log(json);
});
PHP:
<?php
$pictures = ["img1"];
echo json_encode($pictures);
The only way to this not work, is if you have a huge mistake on you webserver configuration.
Your ajax is wrong, it should be:
var a = "test";
$.ajax({
type: "POST",
url: "search.php",
dataType: "json",
data: {a:a},
success: function(data) {
alert('Successfully');
},
error: function(data) {
alert('Failed');
}
});
I am posting my form using AJAX:
$(function () {
$("#Compare").click(function (e) {
$.ajax({
url: '#Url.Action("_Compare","API")',
dataType: 'application/json',
data: {
model: $("#CompareForm").serialize()
},
type: "post",
success: function(response) {
alert (response);
}
});
e.preventDefault();
});
});
I am trying to deserialize my JSON result but I am getting an 'Invalid Json Primitive Exception'.
My Json Result:
"%5B0%5D.Id=1&%5B0%5D.Description=Sutherland+Silver+Plans+offers+you...&%5B0%5D.Price=30&%5B0%5D.Title=Silver+Plan&%5B0%5D.isSelected=true&%5B0%5D.isSelected=false&%5B1%5D.Id=2&%5B1%5D.Description=Sutherland+Gold+Plans+offers+you...&%5B1%5D.Price=50&%5B1%5D.Title=Gold+Plan&%5B1%5D.isSelected=true&%5B1%5D.isSelected=false&%5B2%5D.Id=3&%5B2%5D.Description=Sutherland+Platinum+Plans+offers+you...&%5B2%5D.Price=80&%5B2%5D.Title=Platinum+Plan&%5B2%5D.isSelected=false"
You seem confused about what JSON is, and whether the issue is with the request or the response.
The issue is with the request. You are attempting to put the querystring that serialize() creates in to the model parameter of an object, which itself will be serialised and encoded again. Instead, just pass the querystring that serialise generates to the action:
$("#Compare").click(function (e) {
$.ajax({
url: '#Url.Action("_Compare","API")',
dataType: 'application/json',
data: $("#CompareForm").serialize(),
type: "post",
success: function(response) {
console.log(response);
}
});
e.preventDefault();
});
You have specified the response will be JSON. If this is the case, use console.log to inspect it, otherwise the alert() will only show [object Object].
I have the 'unexpected character error' problem, my Jquery ajax code looks like that:
function test(){
if(true){
$.ajax({
type: 'POST',
url: 'test.php',
dataType: 'json',
data: {
godot: 'godot',
jobadze: 'jobadze'
},
success: function(data){
alert(data);
},
error: function(jqXHR, textStatus, errorThrown) { alert("Error Status: "+textStatus+"\nMessage: "+errorThrown);
}
});
and this is the php code:
<?php
echo 'test';
?>
it should alert "test", but it calls error. What is going on?
You're not returning any JSON. You returning text but you've specified in the AJAX that it will return json.
You have: dataType: 'json',
You could change the dataType: 'text', if you will always be returning text
or in your php change echo 'test'; to echo json_encode('test');
Hope this helps
You wrote dataType: 'json', so the PHP script is required to return valid JSON. Since you're not, the it gets an error when it tries to parse the response as JSON, and reports that error.
You should use json_encode:
<?php
echo json_encode('test');
?>
it should alert "test", but it calls error. What is going on?
Reason for this is your dataType : "json" in $.ajax() method which expects the response from serverside should be a json, which is not the case because that is just a simple text string nothing else, so what could you do:
Either remove the dataType or change the dataType: "text"
Or do a json_encode('string') at your serverside.
As you asked in your question
It should alert "test",
so you can skip the #2 and do this:
$.ajax({
type: 'POST',
url: 'test.php',
dataType: 'text',
data: {
godot: 'godot',
jobadze: 'jobadze'
},
success: function(data){
alert(data); // will alert "test".
},
error: function(jqXHR, textStatus, errorThrown) {
alert("Error Status: "+textStatus+"\nMessage: "+errorThrown);
}
});
but it calls error
$.ajax({
type: 'POST',
url: 'test.php',
dataType: 'json', //<----because of this
See json is a {key : value} pair js object and from your php you are just echoing a string not a object.
I use $.ajax to send some data to testajax.phpenter code here where data are processing. Result of this proces is link. How to get access to this? I think about sessions or cookies. Set in session $test and get access in another files.
$.ajax({
url: 'testajax.php',
type: 'post',
dataType: 'json',
success: console.log('success');
data: { json : jsonObj }
});
In testajax.php I have someting like this
echo "PDF: <a href=images/test.pdf>$dir.pdf</a>";
$test = "PDF: <a href=images/test.pdf>$dir.pdf</a>";
How to get the $test or output the echo after call $.ajax
You can use success' function to get request from PHP.
$.ajax({
url: 'testajax.php',
type: 'post',
dataType: 'html',
data: { json : jsonObj }
success: function(data) {
var request = data;
console.log(data);
$('#link_container').html(data);
}
});
Your console now holds PDF: <a href=images/test.pdf>($dir content).pdf</a>.
Also the variable request holds the same result that you can use anywhere in the script.
Result is appended to #link_container.
Use the success() method for display the result.
$.ajax({
url: 'testajax.php',
type: 'post',
dataType: 'json',
success: function (data){ // <= define handler for manupulate the response
$('#result').html(data);
}
data: { json : jsonObj }
});
Use below code It may help you.
$.ajax({
url: 'testajax.php',
type: 'post',
dataType: 'json',
data: { json : jsonObj },
onSuccess: successFunc,
onFailure: failureFunc
});
function successFunc(response){
alert(response);
}
function failureFunc(response){
alert("Call is failed" );
alert(response);
}