Question

感谢您寻找

我的问题是我似乎无法通过jquery来显示我生成的数据。

这是我的JSON输出

("posts":[{"id":"1-2","time":"0","name":"dash","avatar":"http:\/\/www.gravatar.com\/avatar\/9ff30cc2646099e31a4ee4c0376091b0?s=182&d=identicon&r=PG","comment":"rtetretrete tet rt uh utert"},{"id":"2-2","time":"0","name":"james","avatar":"http:\/\/www.gravatar.com\/avatar\/d41d8cd98f00b204e9800998ecf8427e?s=182&d=identicon&r=PG","comment":"fsdfdfsdf\r\n"}])

这是我的jquery

    $(document).ready(function(){
        var url="comments.php";
        $.getJSON(url,function(json){
            $.each(json.posts,function(i,post){
$("#content").append(
    '<div class="post">'+
        '<h1>'+post.name+'</h1>'+
        '<p>'+post.comment+'</p>'+
        '<p>added: <em>'+post.time+'</em></p>'+
        '<p>posted by: <strong>'+post.name+'</strong></p>'+
        '<p>avatar: <strong>'+post.avatar+'</strong></p>'+
    '</div>'
); });      
});
    });

Answer 1

我刚尝试使用http://www.jsonlint.com/

验证您的JSON

它失败了：

syntax error, unexpected TINVALID, expecting '{' or '[' at line 1
Parsing failed

整体（外部）括号需要从(和)更改为{和}这将验证您的JSON并且脚本应该可以正常工作

Answer 2

我没有检查你的json的语法，但如果它是正确的，那么在发送输出之前尝试这个（在PHP文件中）

header ('Content-type: application/json');

Answer 3

您的Json对象缺少前导和尾随花括号，因此无效。尝试添加它们：

{"posts":[{"id":"1-2","time":"0","name":"dash","avatar":"http:\/\/www.gravatar.com\/avatar\/9ff30cc2646099e31a4ee4c0376091b0?s=182&d=identicon&r=PG","comment":"rtetretrete tet rt uh utert"},{"id":"2-2","time":"0","name":"james","avatar":"http:\/\/www.gravatar.com\/avatar\/d41d8cd98f00b204e9800998ecf8427e?s=182&d=identicon&r=PG","comment":"fsdfdfsdf\r\n"}]}

Answer 4

json应该是

{
    "posts": [
        {
            "id": "1-2",
            "time": "0",
            "name": "dash",
            "avatar": "http:\/\/www.gravatar.com\/avatar\/9ff30cc2646099e31a4ee4c0376091b0?s=182&d=identicon&r=PG",
            "comment": "rtetretrete tet rt uh utert" 
        },
        {
            "id": "2-2",
            "time": "0",
            "name": "james",
            "avatar": "http:\/\/www.gravatar.com\/avatar\/d41d8cd98f00b204e9800998ecf8427e?s=182&d=identicon&r=PG",
            "comment": "fsdfdfsdf\r\n" 
        }
    ]
}

使用jsonLint验证json ..

PHP和JSON评论帮助

4 个答案: