2012-03-28 76 views
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我目前正在为android开发并使用一些人脸识别服务。该服务返回一些复杂的JSON,如下所示;Android中复杂的JSON解析 - 多对象和数组

编辑 - 正确的JSON

{"photos":[{"url":"http:\/\/face.com\/images\/ph\/0b3ecef9a00b647570027e590b72c893.jpg","pid":"[email protected]_88f17d8482a3d3aac1d07b3062a3e29b","width":960,"height":637,"tags":[{"tid":"[email protected]_88f17d8482a3d3aac1d07b3062a3e29b_54.06_45.53_1_1","recognizable":true,"threshold":59,"uids":[{"uid":"[email protected]","confidence":17}],"gid":null,"label":"","confirmed":false,"manual":false,"tagger_id":null,"width":19.58,"height":29.51,"center":{"x":54.06,"y":45.53},"eye_left":{"x":49.55,"y":38.8},"eye_right":{"x":58.4,"y":38.92},"mouth_left":{"x":49.82,"y":52.1},"mouth_center":{"x":54.77,"y":53.04},"mouth_right":{"x":58.36,"y":52.16},"nose":{"x":54.98,"y":46.3},"ear_left":null,"ear_right":null,"chin":null,"yaw":-7.59,"roll":2.27,"pitch":0.01,"attributes":{"age_est":{"value":18,"confidence":30},"age_max":{"value":24,"confidence":30},"age_min":{"value":12,"confidence":30},"face":{"value":"true","confidence":99},"gender":{"value":"female","confidence":41},"glasses":{"value":"false","confidence":73},"lips":{"value":"parted","confidence":63},"mood":{"value":"happy","confidence":61},"smiling":{"value":"true","confidence":48}}}]}],"status":"success","usage":{"used":1,"remaining":4999,"limit":5000,"reset_time_text":"Wed, 28 Mar 2012 14:20:08 +0000","reset_time":1332944408}} 

我需要摆脱它的唯一的事情是UID的数组中的UID值。尝试过几种方法没有成功,有什么建议?

编辑 - 答案:)

String uid = new JSONObject(inputString).getJSONArray("photos").getJSONObject(0).getJSONArray("tags").getJSONObject(0).getJSONArray("uids").getJSONObject(0).getString("uid"); 
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那些' - '字符不对,或者它们是? – Ferdau 2012-03-28 13:32:32

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对不起,似乎更容易阅读,他们将它们添加到格式化的返回预览中,但它们实际上并未通过。 – user1298268 2012-03-28 13:35:23

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发布您尝试过的内容,以及您使用的是什么jsonparser – 2012-03-28 13:35:34

回答

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使用json.org库(http://www.json.org/java/index.html)试试这个例子:

String uid = new JSONObject(inputString).getJSONArray("photos").getJSONObject(0).getJSONArray("tags").getJSONObject(0).getJSONArray("uids").getJSONObject(0).getString("uid"); 

其中输入字符串为您的JSON字符串表示。

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感谢您的回复,我试过这个,但它引发了JSONException和dosnt的工作。 – user1298268 2012-03-28 13:49:29

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现在你已经发布了适当的JSON,我更新了我的代码。再试一次! – damaxxed 2012-03-28 13:58:45

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仍然抛出异常:( – user1298268 2012-03-28 14:05:30