解析android中PHP Webservice的响应
我正在尝试在android中使用带有JSON的PHP Web服务
我从服务器得到的响应数据是
string(170) "["14","Samsung","1","15","Nokia","1","16","Sony Ericson","1","18","LG","1","19","Iphone","1","21","HTC","1","22","Motorola","1","23","Micromax","1","41","BlackBerry","1"]"我很熟悉如何通过创建自定义适配器在ListView上显示数据，但我无法解析此output.Please帮助
我按照本教程访问了PHP Web服务
http://www.codeproject.com/Articles/267023/Send-and-receive-json-between-android-and-php
给我一些推荐信。任何帮助都是非常感谢的。
谢谢..。
我想知道在php端生成的JSONArray的正确格式是什么?？
回答 3
Stack Overflow用户
发布于 2012-04-23 21:33:07
这是有效但奇怪的json响应。当你解析这个json时，你会得到27个值。14，三星，1，15，诺基亚...但似乎每3个项目都是一个类似"14，三星，1“，"15，诺基亚，1”的对象。而json并没有反映出这一点。
使用像这样的在线json解析器，您可以看到您可以得到什么结果：
http://jsonviewer.stack.hu/
你可以这样解析这个json：
List<String> list = new ArrayList<String>();
int counter = 0;
boolean first = true;
String json = "[\"14\",\"Samsung\",\"1\",\"15\",\"Nokia\",\"1\",\"16\",\"Sony Ericson\",\"1\",\"18\",\"LG\",\"1\",\"19\",\"Iphone\",\"1\",\"21\",\"HTC\",\"1\",\"22\",\"Motorola\",\"1\",\"23\",\"Micromax\",\"1\",\"41\",\"BlackBerry\",\"1\"]";
JSONArray getJSONArray;
try {
getJSONArray = new JSONArray(json);
for (int i = 0; i < getJSONArray.length(); i++) {
Log.d("", getJSONArray.getString(i) + "-" + i % 1 + "-" + i % 2);
if ((first == true && counter == 1) || (first == false && counter == 2)) {
list.add(getJSONArray.getString(i));
counter = 0;
first = false;
}
else {
counter += 1;
}
}
}
catch (JSONException e) {
e.printStackTrace();
}记录您的阵列以查看结果：
for (String item : list) {
Log.d("list", item);
}结果：
D/list ( 1669): Samsung
D/list ( 1669): Nokia
D/list ( 1669): Sony Ericson
D/list ( 1669): LG
D/list ( 1669): Iphone
D/list ( 1669): HTC
D/list ( 1669): Motorola
D/list ( 1669): Micromax
D/list ( 1669): BlackBerry和服务器站点：
通常情况下，json必须是这样的。
[
{
id: "14",
brand: "Samsung",
status: "1"
},
{
id: "15",
brand: "Nokia",
status: "1"
},
{
id: "16",
brand: "Sony Ericson",
status: "1"
}
]Stack Overflow用户
发布于 2012-04-23 21:23:30
首先它不是一个有效的Json格式，Json必须有一个label并且是对应的value ex。"label"："value"，但是您仍然可以像这样手动解析这个字符串
String yourString= "["14","Samsung","1","15","Nokia","1","16","Sony Ericson","1","18","LG","1","19","Iphone","1","21","HTC","1","22","Motorola","1","23","Micromax","1","41","BlackBerry","1"]";
yourString= yourString.subString(1, yourString.length()-1);
String[] arrayValue= yourString.split(",");现在你有了arrayValue= "14“arrayValue1=”三星“等等。
Stack Overflow用户
发布于 2012-04-23 21:27:10
看起来像一个‘笨蛋-json’数组。这基本上是没有问题的，因为你把它看作是一个字符串列表。通常我会建议使用一些JSON类，但在这种特殊的简单情况下，以下内容就足够了：
String jsonString = "["14","Samsung","1","15","Nokia","1","16","Sony Ericson","1","18","LG","1","19","Iphone","1","21","HTC","1","22","Motorola","1","23","Micromax","1","41","BlackBerry","1"]"
jsonString = jsonString.subString(1, substring.length() -1);
String[] singleValues = jsonString.split(",");这应该会导致如下结果
new String[]{"14","Samsung","1",....}您可以在自定义ListAdapter中传递该字符串数组(但默认的也可以)
https://stackoverflow.com/questions/10281078
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