Send array with Ajax to PHP script

PhpJavascriptJqueryAjax

Php Problem Overview


I have array made by function .push. In array is very large data. How is the best way send this to PHP script?

   dataString = ??? ; // array?
   $.ajax({
        type: "POST",
        url: "script.php",
        data: dataString, 
        cache: false,

        success: function(){
            alert("OK");
        }
    });
    

script.php:

  $data = $_POST['data'];
  
  // here i would like use foreach:
  
  foreach($data as $d){
     echo $d;
  }

How is the best way for this?

Php Solutions


Solution 1 - Php

Encode your data string into JSON.

dataString = ??? ; // array?
var jsonString = JSON.stringify(dataString);
   $.ajax({
        type: "POST",
        url: "script.php",
        data: {data : jsonString}, 
        cache: false,

        success: function(){
            alert("OK");
        }
    });

In your PHP

$data = json_decode(stripslashes($_POST['data']));

  // here i would like use foreach:

  foreach($data as $d){
     echo $d;
  }

Note

When you send data via POST, it needs to be as a keyvalue pair.

Thus

data: dataString

is wrong. Instead do:

data: {data:dataString}

Solution 2 - Php

 dataString = [];
   $.ajax({
        type: "POST",
        url: "script.php",
        data:{data: $(dataString).serializeArray()}, 
        cache: false,

        success: function(){
            alert("OK");
        }
    });

http://api.jquery.com/serializeArray/

Solution 3 - Php

Data in jQuery ajax() function accepts anonymous objects as its input, see documentation. So example of what you're looking for is:

dataString = {key: 'val', key2: 'val2'};
$.ajax({
        type: "POST",
        url: "script.php",
        data: dataString, 
        cache: false,

        success: function(){
            alert("OK");
        }
    });

You may also write POST/GET query on your own, like key=val&key2=val2, but you'd have to handle escaping yourself which is impractical.

Solution 4 - Php

If you have been trying to send a one dimentional array and jquery was converting it to comma separated values >:( then follow the code below and an actual array will be submitted to php and not all the comma separated bull**it.

Say you have to attach a single dimentional array named myvals.

jQuery('#someform').on('submit', function (e) {
    e.preventDefault();
    var data = $(this).serializeArray();

    var myvals = [21, 52, 13, 24, 75]; // This array could come from anywhere you choose 
    for (i = 0; i < myvals.length; i++) {
        data.push({
            name: "myvals[]", // These blank empty brackets are imp!
            value: myvals[i]
        });
    }

jQuery.ajax({
    type: "post",
    url: jQuery(this).attr('action'),
    dataType: "json",
    data: data, // You have to just pass our data variable plain and simple no Rube Goldberg sh*t.
    success: function (r) {
...

Now inside php when you do this

print_r($_POST);

You will get ..

Array
(
    [someinputinsidetheform] => 023
    [anotherforminput] => 111
    [myvals] => Array
        (
            [0] => 21
            [1] => 52
            [2] => 13
            [3] => 24
            [4] => 75
        )
)

Pardon my language, but there are hell lot of Rube-Goldberg solutions scattered all over the web and specially on SO, but none of them are elegant or solve the problem of actually posting a one dimensional array to php via ajax post. Don't forget to spread this solution.

Solution 5 - Php

dataString suggests the data is formatted in a string (and maybe delimted by a character).

$data = explode(",", $_POST['data']);
foreach($data as $d){
     echo $d;
}

if dataString is not a string but infact an array (what your question indicates) use JSON.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionPaul AttuckView Question on Stackoverflow
Solution 1 - PhpxbonezView Answer on Stackoverflow
Solution 2 - PhpJohn xView Answer on Stackoverflow
Solution 3 - PhpVyktorView Answer on Stackoverflow
Solution 4 - PhpMohd Abdul MujibView Answer on Stackoverflow
Solution 5 - PhpRvdKView Answer on Stackoverflow