使用JQuery / PHP解析嵌套的JSON String對象？
[英]parsing nested JSON String objects with JQuery/PHP?
問題描述
這是我正在努力的頁面.... http://fremontchurch.net/json_test/
這是json http://fremontchurch.net/json_test/posts.php
我正在嘗試通過簡單的html鏈接<a href="URL_GOES_HERE">TRACK NAME GOES HERE</a>列出並鏈接到其網址的列表列表
我得到了其他一切，只是嵌套的部分不斷出現“[對象對象]，[對象對象]，[對象對象]”......等等......每個嵌套有兩個項目，一個軌道名稱和網址。 ..這樣做的正確方法是什么？
來自我的json的一個對象......
{
"id":"All Things Are Possible",
"key":"All Things Are Possible",
"doc":"All Things Are Possible",
"album":"All Things Are Possible",
"artwork":"http://godsgypsychristianchurch.net/music_artwork/DEFAULT_COVER2.png",
"baseurl":"http://www.godsgypsychristianchurch.net/music",
"church":"Atlanta GA",
"des":"All Things Are Possible from the Atlanta GA Church, Pastor Nick White",
"tracks":[
{"name":"1 Intro",
"url":"/Atlanta%20GA/All%20things%20are%20possible/01%20Intro.mp3"},
{"name":"2 Wo si O Drom",
"url":"/Atlanta%20GA/All%20things%20are%20possible/02%20Wo%20si%20O%20drom.mp3"},
{"name":"3 Nas Murrgo Shov",
"url":"/Atlanta%20GA/All%20things%20are%20possible/03%20Nas%20murrgo%20shov.mp3"},
{"name":"4 To Cho Vos",
"url":"/Atlanta%20GA/All%20things%20are%20possible/04%20To%20cho%20vos.mp3"},
{"name":"5 Tu Son Kai Sastridas",
"url":"/Atlanta%20GA/All%20things%20are%20possible/05%20Tu%20son%20kai%20sastridas.mp3"},
{"name":"6 Now I Am Strong",
"url":"/Atlanta%20GA/All%20things%20are%20possible/06%20Now%20I%20am%20strong.mp3"},
{"name":"7 Sorr Nevee",
"url":"/Atlanta%20GA/All%20things%20are%20possible/07%20Zorr%20nevee.mp3"},
{"name":"8 Preaching",
"url":"/Atlanta%20GA/All%20things%20are%20possible/08%20Preaching.mp3"},
{"name":"9 Arkadyan Amey",
"url":"/Atlanta%20GA/All%20things%20are%20possible/09%20Arkadyan%20amey.mp3"},
{"name":"10 O Christo Ka Wudarr",
"url":"/Atlanta%20GA/All%20things%20are%20possible/10%20O%20Christo%20ka%20wudarr.mp3"},
{"name":"11 Eloi, Eloi",
"url":"/Atlanta%20GA/All%20things%20are%20possible/11%20Eloi%2C%20Eloi.mp3"},
{"name":"12 Amadow Dell",
"url":"/Atlanta%20GA/All%20things%20are%20possible/12%20Amadow%20Dell.mp3"},
{"name":"13 Sastiar Amey Devla",
"url":"/Atlanta%20GA/All%20things%20are%20possible/13%20Sastiar%20amey%20Devla.mp3"},
{"name":"14 Tu Skepeese",
"url":"/Atlanta%20GA/All%20things%20are%20possible/14%20Tu%20skepeese.mp3"},
{"name":"15 Dov Ma Godgee",
"url":"/Atlanta%20GA/All%20things%20are%20possible/15%20Dov%20ma%20godgee.mp3"},
{"name":"16 The Lord is my strength",
"url":"/Atlanta%20GA/All%20things%20are%20possible/16%20The%20Lors%20is%20my%20strength.mp3"}
]
}
繼承了html / jquery / php的所有代碼
<html>
<body>
<div id="content">
</div>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
var url="posts.php";
$.getJSON(url,function(json){
$.each(json.posts,function(i,post){
$("#content").append(
'<div class="post">'+
'<h1>'+post.album+'</h1>'+
'<p><img src="'+post.artwork+'"width="250"></img></p>'+
'<p><strong>'+post.church+'</strong></p>'+
'<p>Description: <strong>'+post.des+'</strong></p>'+
'<p>Base url: <em>'+post.baseurl+'</em></p>'+
'<p>Tracks: <li>'+post.tracks+'</li></p>'+
'<hr>'+
'</div>'
);
});
});
});
</script>
</body>
</html>
請回復完整的代碼，因為我對這種編碼有點新鮮，有時我知道在哪里放...任何幫助將非常感謝，謝謝...
2 個解決方案
解決方案1
3 2011-09-29 06:39:36
您需要遍歷post.tracks中的每個項目並在包含它之前將該信息收集到字符串中，如下所示：
$.each(json.posts,function(i,post){
var content,
trackInfo = '',
tracks = post.tracks;
// loop over the tracks and collect info
$.each(tracks, function (i) {
trackInfo += '<a href="' + tracks[i].url + '">' + tracks[i].name + '</a> ';
});
content = '<div class="post">'+
'<h1>'+post.album+'</h1>'+
'<p><img src="'+post.artwork+'"width="250"></img></p>'+
'<p><strong>'+post.church+'</strong></p>'+
'<p>Description: <strong>'+post.des+'</strong></p>'+
'<p>Base url: <em>'+post.baseurl+'</em></p>'+
'<p>Tracks: <li>'+ trackInfo +'</li></p>';
'<hr>'+
'</div>'
$("#content").append(content);
});
這里的工作示例： http ： //jsfiddle.net/3SFE4/
解決方案2
1 已采納 2011-09-29 06:29:27
以同樣的方式使用each迭代帖子，您可以使用each來迭代軌道。 我在下面提供了完整的代碼。
<html>
<body>
<div id="content">
</div>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
var url="posts.php";
$.getJSON(url,function(json){
$.each(json.posts,function(i,post){
// Here we generate a fragment for the tracks.
var tracks = '';
$.each(post.tracks, function(j, track) {
tracks += '<a href="' + track.url + '">' + track.name + '</a>';
})
$("#content").append(
'<div class="post">'+
'<h1>'+post.album+'</h1>'+
'<p><img src="'+post.artwork+'"width="250"></img></p>'+
'<p><strong>'+post.church+'</strong></p>'+
'<p>Description: <strong>'+post.des+'</strong></p>'+
'<p>Base url: <em>'+post.baseurl+'</em></p>'+
'tracks' +
'<hr>'+
'</div>'
);
});
});
});
</script>
用jQuery解析PHP腳本返回的JSON字符串
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