send data by Jquery ajax to same php

Question

i need help because i'm stuck and don't know what's wrong ,i try to send user clicked button "id" to php to get related data from database in the same page

$(".button_class").on("click", function() {
        ToEditId = $(this).attr('id');

        console.log(ToEditId ); //to check clicked id is Ok

      $.ajax({
            type: "POST",
            url: same/php/page/path,
            data: { 
                ToEditId: ToEditId 
            },
            success: function(res, data) {
                console.log(res, data);
            },
            error: function(err) {
                alert(err);
            }

        });
    });

the ajax print success in console log ,here is php code to get the value if clicked id

<?php
if(isset($_POST['ToEditId'])){
    $to_edit_id=$_POST['ToEditId'];
    var_dump($to_edit_id);
}

but nothing happen in php file !!

You mean nothing happens when you do a POST request manually? — Julian
– Julian, Commented Feb 21, 2018 at 22:05
try echo $to_edit_id; in php file, so in your console you can see data — Krish
– Krish, Commented Feb 21, 2018 at 22:09
i mean , it suppose when jQuery ajax send the id by POST to the server same php file it should do what inside if (isset($_POST['ToEditId'])... — IBRAHIM EZZAT
– IBRAHIM EZZAT, Commented Feb 21, 2018 at 22:10

NoobishPro · Accepted Answer · 2018-02-21 22:17:52Z

1

Which is the expected behaviour.

PHP is not dynamic. It doesn't "update". PHP only runs once. This means that once your page is rendered, you cannot use PHP to change it again. You actually would have to use javascript to change the page, like so;

PHP side:

<?php
  if(isset($_POST['ToEditId'])){
    echo $_POST['ToEditId'];
    $to_edit_id=$_POST['ToEditId'];
    var_dump($to_edit_id);
    die(); // prevent entire page from re-rendering again.
  }

JS side:

$(".button_class").on("click", function() {
    ToEditId = $(this).attr('id');

    console.log(ToEditId ); //to check clicked id is Ok

  $.ajax({
        type: "POST",
        url: same/php/page/path,
        data: { 
            ToEditId: ToEditId 
        },
       success: function(res, data) {
          //Add your PHP file's response to the body through javascript.
          $('body').append(res);
        },
        error: function(err) {
            alert(err);
        }

    });
});

As @IncredibleHat mentioned, you should make sure your page doesn't render any of its usual HTML, so it won't return the entire page back to your ajax call. So put the PHP all the way above your html!

edited Feb 21, 2018 at 22:17

answered Feb 21, 2018 at 22:11

NoobishPro

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4 Comments

IncredibleHat Over a year ago

Unless he had his php block on the bottom of the script ;) As I should say, its important to note the php portion needs to be well ahead of any html output (if ajax is calling the same php).

NoobishPro Over a year ago

Agreed, I'll add that.

IBRAHIM EZZAT Over a year ago

thanks bro, it's clear now ,please modify your code and put body betwenn "" like $('body').append(res)

NoobishPro Over a year ago

@IBRAHIMEZZAT oops, my bad. I fixed it.

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