ajax post information into php $_POST

Question

I am trying to get the information from my ajax into a PHP $_POST so that i can update my database.

HTML

<form id="23" method="post">
   <select name="admin" onchange="chosenadmin(23)">
      <option value="rick">rick</option>
      <option value="john">john</option>
      <option value="dick">dick</option>
    </select>
</form>

AJAX

function changeadmin(verkochtid){
        id = verkochtid;
        console.log(id);
    $.ajax({

            url: 'winkels.php',
            id: id,
            type: 'POST',
            data: $('#'+id).serialize(),
            success: function(data, id){
                console.log(this.data);
                console.log(this.id);  
               }  
         });            
 };

PHP

if (isset($_POST["serialize"])) {
                $data = $_POST["serialize"];
                $medewerker = $data["chosen_admmin"];
                $verkoopid = $data["id"];
                $sql = "UPDATE verkocht SET medewerker_verwerkt = '$medewerker' WHERE verkocht_id='$verkoopid'";

                echo $sql;
 };

the PHP will never be executed but in the console log i get to see the id of the form and admin=rick.

I can tell from this that AJAX gets the information and procces it, but how do i set it in my PHP?

Put a `print_r($_POST);` in front of that PHP code and see what and how the data is being tranfered. There is no `$_POST["serialize"]` occurance. But there shoud be a `$_POST['admin']` — RiggsFolly, May 27 '17 at 10:07
Your script is at risk of [SQL Injection Attack](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) Have a look at what happened to [Little Bobby Tables](http://bobby-tables.com/) Even [if you are escaping inputs, its not safe!](http://stackoverflow.com/questions/5741187/sql-injection-that-gets-around-mysql-real-escape-string) Use [prepared parameterized statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) — RiggsFolly, May 27 '17 at 10:10
@RiggsFolly i get a array() from the print_r and i took the prepare away for the form — LeondenBoer, May 27 '17 at 10:12
change in ajax request data : { 'serialize' : $('#'+id).serialize() } now user print_r($_POST["serialize"]); and check that. — Ranjeet Singh, May 27 '17 at 10:15
@RanjeetSingh it is giving me Undefined index: serialize back, also if i change it. — LeondenBoer, May 27 '17 at 10:39
it is like that my winkels.php doenst get the POST type back, the console.log does it fine — LeondenBoer, May 27 '17 at 10:40
You are getting these output in your success function as server response right? — Parantap Parashar, May 27 '17 at 10:49
yes because there i am asking for console.log and that is on the page — LeondenBoer, May 27 '17 at 10:50
Okay. Try `console.log($('#'+id).serialize())` before `$.ajax` — Parantap Parashar, May 27 '17 at 10:52

Jeppsen · Answer 1 · 2017-05-27T10:23:04.093

0

The data in the $_POST will have the index of the input name fields in your html when you serialize it.

You can use a hidden input field to get the ID serialized

<form id="23" method="post">
    <select name="admin" onchange="chosenadmin(23)">
        <option value="rick">rick</option>
        <option value="john">john</option>
        <option value="dick">dick</option>
    </select>
    <input type="hidden" name="id" value="23" />
</form>

Then you can do the following in you php:

if (isset($_POST["admin"]) && isset($_POST["id"])) {
    $admin = $_POST["admin"];
    $id = $data["id"];

    $sql = "UPDATE verkocht SET medewerker_verwerkt = '$admin' WHERE verkocht_id='$id'";

    echo $sql;
}

You can do the following in ajax.php to check the structure of the $_POST.

echo '<pre>';
print_r($_POST);
echo '</pre>';
die;

You should also pay attention to SQL injections.

edited May 27 '17 at 10:23

answered May 27 '17 at 10:14

Jeppsen

487
3
10

If i use print_r($_POST); it says : array() – LeondenBoer May 27 '17 at 10:33
If i use print_r($_POST["id"]); it says: Undefined index: id. doesnt matter if i make a change. the consol log will give me the good values back tho – LeondenBoer May 27 '17 at 10:35
sorry guys, still nothing is happening in the php if – LeondenBoer May 27 '17 at 11:02
Have you tried using the debugger in your browser to check the Request Headers and Payload for the Ajax call? – Jeppsen May 27 '17 at 12:32

score 0 · Answer 2 · edited May 27 '17 at 10:54

function name onchange event of select is different! you have written onchange="chosenadmin(23)" and for AJAX call the name is function changeadmin(verkochtid){...}
To get the value in post you must use name in PHP

HTML

<form id="23" name="frm_admin" method="post">
    <select name="admin" onchange="chosenadmin(23)">
        <option value="rick">rick</option>
        <option value="john">john</option>
        <option value="dick">dick</option>
    </select>
    <input type="hidden" name="id" value="23" />
</form>

AJAX

function chosenadmin(verkochtid){
        id = verkochtid;
        console.log(id);
        $.ajax({
            url: 'winkels.php',
            id: id,
            type: 'POST',
            data: $('#'+id).serialize(),
            success: function(data, id){
                console.log(this.data);
                console.log(this.id);  
               }  
       });            
 };

PHP

if (isset($_POST["frm_admin"])) {
                $medewerker = $_POST["admin"];
                $verkoopid = $data["id"];
                $sql = "UPDATE verkocht SET medewerker_verwerkt = '$medewerker' WHERE verkocht_id='$verkoopid'";  
                echo $sql;
 };

still just array(), i tried document.getElementById("test").innerHTML = (this.data); and that worked also — LeondenBoer, May 27 '17 at 11:08
Make sure you are posting page through AJAX and try to print log on console for $('#'+id).serialize() — Mansi Raja, May 27 '17 at 11:16

ajax post information into php $_POST

2 Answers2