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?

Answer 1

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.

Answer 2

1. function name onchange event of select is different! you have written onchange="chosenadmin(23)" and for AJAX call the name is function changeadmin(verkochtid){...}

2. 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;
 };