CODEWITHSUNDEEP
phpmysqlajaxmysqli
Lamberto Fredrick Nababan
Lamberto Fredrick Nababan

Reputation: 101

PHP / MySQL / AJAX - Update multiple data

How to update multiple data using AJAX ?

Example :

TableA

id : 1, 2

name : Jack, John

It's only working with id 1, when I am trying to edit name for id 2 it's not working.

I have try with this code but failed.

HTML/PHP :

...

    while($row=mysqli_fetch_array($query)){
    echo'    
    <form class="btn-group">
        <input type="text" class="form-control" name="id_user" id="id_user" data-user="'.$row['id'].'" value="'.$row['id'].'">
        <input type="text" class="form-control" name="id_status" id="id_status" data-status="'.$row['id'].'" value="'.$row['id'].'">
        <button type="submit" id="likestatus" class="btn btn-primary btn-outline btn-xs"><i class="fas fa-thumbs-up"></i></button>
        </form>
    ';
    }

AJAX :

$(document).ready(function(){
    $("#likestatus").click(function(){
    var id_user=$("#id_user").data("user");
    var id_status=$("#id_status").data("status");
        $.ajax({
        url:'status/like-status.php',
        method:'POST',
        data:{
            id_user:id_user,
            id_status:id_status
            },
            success:function(response){
            alert(response);
            }
        });
    });
});

Upvotes: 0

Views: 1716

Answers (4)

Capricorn
Capricorn

Reputation: 2089

You're using the id's multiple times. Thus your query for var id_user=$("#id_user").data("user"); always finds the first input field on the page. You should avoid using the same id multiple times on one page (see this Question).

You may subscribe to the jQuery submit event of the form and then search for the input fields within that form, to properly extract the id_user and status_user values. For that you have to add an appropriate event listener to the <form> element. To find the form I would recommend adding a css-class like like-status-form.

$(document).ready(function(){
    // We're attaching a submit-event listener to every element with the css class "like-status-form"
    $(".like-status-form").submit(function(event){
       // Form get's submitted
       // Prevent that the Browser reloads the page
       event.preventDefault();

       // Extract the user id and status from the form element (=== $(this))
       var id_user = $(this).find('[name="id_user"]').data('user');
       var id_status = $(this).find('[name="id_status"]').data('status');

       // TODO Perform AJAX Call here
    });
});

To detect the form elements one can use the jQuery Attribute Equals Selector.

Find a working example at https://jsfiddle.net/07yzf8k1/

Upvotes: 1

Masivuye Cokile
Masivuye Cokile

Reputation: 4772

Make your ID unique so make them dynamic

 <?php

 $counter = 0;

 while($row=mysqli_fetch_array($query)){
    $counter++;

    echo'    
    <form class="btn-group">
        <input type="text" class="form-control" id="userid_$counter" data-user="'.$ruser['id'].'" value="'.$ruser['id'].'">
        <input type="text" class="form-control" name="id_status" id="status_$counter" data-status="'.$rtimeline['id'].'" value="'.$rtimeline['id'].'">
        <button type="submit" id="likestatus_$counter" class="btn btn-primary btn-outline btn-xs"><i class="fas fa-thumbs-up"></i></button>
        </form>
    ';
    }
?>

Then

<script type="text/javascript">
$(document).ready(function(){
    $('[id^="likestatus_"]').on('click',function(){
         var index = $(this).attr('id').split("_")[1];
    var id_user=$("#user_"+index).data("user");
    var id_status=$("#status_"+index).data("status");
        $.ajax({
        url:'status/like-status.php',
        method:'POST',
        data:{
            id_user:id_user,
            id_status:id_status
            },
            success:function(response){
            alert(response);
            }
        });
    });
});

Upvotes: 1

Adder
Adder

Reputation: 5868

The problem with your code is that ids should be unique, but in the loop you create elements with same id.

Use this in the event handler to find the siblings of the button that has been clicked - closest returns the parent of type form.

$(document).ready(function(){
    $(".btn-primary").click(function(){
        var $form = $(this).closest('form');
        var id_user=$form.find('[name="id_user"]').data("user");
        var id_status=$form.find('[name="id_status"]').data("status");
            $.ajax({
            url:'status/like-status.php',
            method:'POST',
            data:{
                id_user:id_user,
                id_status:id_status
                },
                success:function(response){
                alert(response);
                }
            });
    });
});

You might want to use your own class instead of .btn-primary because this affects all buttons on the page.

Upvotes: 1

Wouter H
Wouter H

Reputation: 51

Judging from the incomplete PHP, it appears as if you're not assigning to $ruser within your loop. This would mean you're always posting the same id to like-status.php.

PS: Would've posted as comment, but I can't.

Upvotes: 1

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