PHP和JSON评论帮助

我的问题是，我似乎无法让jquery显示我生成的数据。

这里是我的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"}])

这里是我的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>' 
); });  
}); 
    });

来源

2010-06-16 jay

这是什么都用PHP做什么？ – 2010-06-16 08:22:25

json是由php生成的。实际问题与此无关，但涉及到。 – 2010-06-17 12:24:43

我刚才想验证使用http://www.jsonlint.com/

您的JSON，但失败：

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

整体（外）括号需要从(和)改为{和}这将验证您的JSON和脚本应该可以正常

来源

2010-06-16 08:24:04 Lizard

我没有JSON检查语法适合你，但如果它是正确的，那么尝试这在发送输出之前（在PHP文件中）

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

来源

2010-06-16 08:25:03

您的Json对象缺少前导和尾随大括号，因此无效。尝试添加他们：

{"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"}]}

来源

2010-06-16 08:25:37 mamoo

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 ..

来源

2010-06-16 08:25:42 Reigel

PHP和JSON评论帮助

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