download2565 2018-05-11 11:15
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PHP / MySQL / AJAX - 更新多个数据

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);
            }
        });
    });
});
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4条回答 默认 最新

  • doulu3399 2018-05-11 11:28
    关注

    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.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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