Php 如何在for循环中为display获取json数据

php代码：

$query = mysqli_query($con,"INSERT INTO tlb_comments(user_id,comments) values ('$vw_id','$name')");
$comments = mysqli_query($con,"SELECT * FROM tlb_comments");
$avatar   = mysqli_query($con,"SELECT * FROM tlb_avatar WHERE `vw_id`='".$vw_id."'");
$avatar_select  =  mysqli_fetch_all($avatar);
$comment_select =  mysqli_fetch_all($comments);
array_push($avatar_select,$comment_select);
echo json_encode($avatar_select);

Jquery Ajax：

$(".comment").click(function() {
    var id = $(this).data("id");
    var name = $("#username_" + id).val();
    if (name == '') {
        alert("Please Fill All Fields");
    } else {

        $.ajax({
            type: "POST",
            url: "comments",
            data: {
                username: name
            },
            success: function(html) {

                $("#cmt_output").html(html);
                console.log(html);
            }
        });
    }
});

Json数组：

[
   [
      "11",
      "6",
      "1509947417_User_Avatar_2.png"
   ],
   [
      [
         "113",
         "0",
         "6",
         "all of us",
         "0"
      ],
      [
         "114",
         "0",
         "6",
         "all of us",
         "0"
      ],
      [
         "115",
         "0",
         "6",
         "welcome....",
         "0"
      ]
   ]
]

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我认为您必须从php返回关联数组，这样您就可以轻松地读取带有每个数组元素键的json数组…

尝试添加一些描述，解释我的前端json数组数据类型中显示的确切问题。如何将数据类型Now display转换为前端json数组数据类型。如何转换数据文本类型你的意思是你想将json转换成javascript，使用Hmmm，好的，我会试试。。。。。。