【发布时间】:2018-04-22 04:37:01
【问题描述】:
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"
]
]
]
【问题讨论】:
-
尝试添加一些描述来解释确切的问题
-
现在显示到我的前端 json 数组数据类型。如何转换数据类型
-
现在显示到我的前端 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 转换为 javascript abject use jQuery.parseJSON()
-
嗯,好的,我会试试......