Passing JSON object to PHP script

Question

I am trying to pass a JSON object that looks similar to this:

 {"service": "AAS1", "sizeTypes":[{"id":"20HU", "value":"1.0"},{"id":"40FB","2.5"}]}

Just a note: In the sizeTypes, there are a total of about 58 items in the array.

When the user clicks the submit button, I need to be able to send the object to a PHP script to run an UPDATE query. Here is the javascript that should be sending the JSON to the PHP script:

 $('#addNewSubmit').click(function()
 {
   var payload = {
     name: $('#addservice').val();
     sizeTypes: []
   };

   $('input.size_types[type=text]').each(function(){
     payload.sizeTypes.push({
       id: $(this).attr('id'),
       value: $(this).val()
      });
   });

   $.ajax({
     type: 'POST',
     url: 'api/editService.php',
     data: {service: payload},
     dataType: 'json',
     success: function(msh){
       console.log('success');
     },
     error: function(msg){
       console.log('fail');
     }
   });
 });

Using the above click function, I am trying to send the object over to php script below, which is in api/editService.php:

 <?php
 if(isset($_POST['service']))
 {
   $json = json_decode($_POST['service'], true);

   echo $json["service"]["name"] . "<br />";

   foreach ($json["service"]["sizeTypes"] as $key => $value){
     echo $value["value"] . "<br />";
   }
 }
 else
 {
   echo "Nooooooob";
 }
 ?>

I do not have the UPDATE query in place yet because I am not even sure if I am passing the JSON correctly. In the javascript click function, you see the SUCCESS and ERROR functions. All I am producing is the ERROR function in Chrome's console.

I am not sure where the error lies, in the JavaScript or the PHP.

Why can I only produce the error function in the AJAX post?

Edit

I removed the dataType in the ajax call, and added JSON.stringify to data:

 $.ajax({
   type: 'POST',
   url: 'api/editService.php',
   data: {servce: JSON.stringify(payload)},
   success: function(msg){
     console.log('success');
   },
   error: function(msg){
     console.log('fail'), msg);
   }
 });

In the PHP script, I tried this:

 if(isset($_POST['service'))
 {
   $json = json_decode($_POST['service'], true);

   foreach ($json["service"]["sizeTypes"] as $key => $value){
     $insert = mysqli_query($dbc, "INSERT INTO table (COLUMN, COLUMN, COLUMN) VALUES (".$json["service"] . ", " . "$value["id"] . ", " . $value["value"]")");
   }
 }
 else
 {
   echo "noooooob";
 }

With this update, I am able to get the success message to fire, but that's pretty much it. I cannot get the query to run.

And there is no reason to use json_decode() in your PHP, the $_POST variable will not contain json. — Gerald Schneider
– Gerald Schneider, Commented Apr 12, 2016 at 13:41
I posted an answer, the error might be that you are echoing html but ajax is expecting json (dataType: 'json') — imvain2
– imvain2, Commented Apr 12, 2016 at 13:42
@HoodCoderMan You're getting error may be because of this line. echo $json["service"]["name"]. You don't have any name key in the given example array string. — Alok Patel
– Alok Patel, Commented Apr 12, 2016 at 13:45

imvain2 · Accepted Answer · 2016-04-12 13:41:49Z

2

without seeing the error, I suspect the error is because ajax is expecting json (dataType: 'json',) but you are echoing html in your php

answered Apr 12, 2016 at 13:41

imvain2

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Comments

pie6k · Accepted Answer · 2016-04-12 13:41:46Z

1

Try to change

 error: function(msg){
   console.log('fail');
 }

to

 error: function(msg){
   console.log(msg);
 }

There might be some php error or syntax issue and you should be able to see it there.

Also try to debug your php script step by step by adding something like

echo "still works";die;

on the beginning of php script and moving it down till it'll cause error, then you'll know where the error is.

Also if you're expecting JSON (and you are - dataType: 'json' in js , don't echo any HTML in your php.

answered Apr 12, 2016 at 13:41

pie6k

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1 Comment

John Beasley Over a year ago

I changed the error function as suggested, and I got the below console message: Object {readyState: 4, responseText: "↵↵", status: 200, statusText: "OK"} I also removed the HTML tags from the PHP script

jeroen · Accepted Answer · 2016-04-12 13:44:05Z

1

As you are sending an object in your service key, you probably have a multi-dimensional array in $_POST['service'].

If you want to send a string, you should convert the object to json:

data: {service: JSON.stringify(payload)},

Now you can decode it like you are doing in php.

Also note that you can only send json back from php if you set the dataType to json. Anything other than valid json will have you end up in the error handler.

edited Apr 12, 2016 at 13:44

answered Apr 12, 2016 at 13:43

jeroen

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3 Comments

pie6k Over a year ago

You don't need to do that with jQuery if you're setting it to json. You can pass js object and have php array on back-end

Chay22 Over a year ago

^depend on jquery version I guess, no?

jeroen Over a year ago

@AdamPietrasiak I'm pretty sure you do, otherwise you end up with values like $_POST['service']['sizeTypes'][..], which is also perfectly valid and common when you use arrays in your name attributes of the form inputs. So the OP probably doesn't even have to decode the current value :-)

mferly · Accepted Answer · 2016-04-12 14:16:05Z

Example how to handle a JSON response from editService.php. Typically, the editService.php script will be the worker and will handle whatever it is you need done. It will (typically) send a simple response back to the success method (consider updating your $.ajax to use the latest methods, eg. $.done, etc). From there you handle the responses appropriately.

$.ajax({
    method: 'POST',
    url: '/api/editService.php',
    data: { service: payload },
    dataType: 'json'
})
 .done(function(msh) {
    if (msh.success) {
        console.log('success');
    }
    else {
        console.log('failed');
    }
})
 .fail(function(msg) {
    console.log('fail');
});

Example /editService.php and how to work with JSON via $.ajax

<?php
$response = [];
if ( isset($_POST['service']) ) {
    // do your stuff; DO NOT output (echo) anything here, this is simply logic
    // ... do some more stuff

    // if everything has satisfied, send response back
    $response['success'] = true;

    // else, if this logic fails, send that response back
    $response['success'] = false;
}
else {
    // initial condition failed
    $response['success'] = false;
}
echo json_encode($response);

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