尝试在 Java 中将 JSON 解析为字符串
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Trying to parse JSON to String in Java
提问by Ravi Thapliyal
I'm trying to parse this stock info at:
我正在尝试在以下位置解析此股票信息：
http://www.google.com/finance/info?client=ig&q=csco
http://www.google.com/finance/info?client=ig&q=csco
that's in JSON format to a map, essentially following this tutorial I saw using the quick-json jar but it keeps giving me an exception and I can't figure out why. Here's the code, any help is greatly appreciated
这是地图的 JSON 格式，基本上按照本教程我看到使用 quick-json jar 但它一直给我一个例外，我不知道为什么。这是代码，非常感谢任何帮助
Tutorial link: https://code.google.com/p/quick-json/
教程链接：https: //code.google.com/p/quick-json/
public static void main(String args[]) throws IOException
{
String value="";
URL uri = new URL("http://www.google.com/finance/info?client=ig&q=csco");
BufferedReader input = new BufferedReader(new InputStreamReader(uri.openStream(), "UTF-8"));
while(input.readLine()!=null)
{
value+=input.readLine();
}
JsonParserFactory factory = JsonParserFactory.getInstance();
JSONParser parse = factory.newJsonParser();
Map jsonData =parse.parseJson(value);
System.out.println((String)jsonData.get("e"));
}
Here's the exception I get:
这是我得到的例外：
Exception in thread "main" com.json.exceptions.JSONParsingException: @Key-Heirarchy::root[0]/ @Key:: COMMA or ] is expected. but found :...@Position::5
at com.json.utils.JSONUtility.handleFailure(JSONUtility.java:124)
at com.json.parsers.JSONParser.stringLiteralTemplate(JSONParser.java:574)
at com.json.parsers.JSONParser.nonValidatingValueTemplate(JSONParser.java:698)
at com.json.parsers.JSONParser.jsonArrayTemplate(JSONParser.java:454)
at com.json.parsers.JSONParser.parseJson(JSONParser.java:170)
at parser.Scratch.main(Scratch.java:27)
EDIT: I also tried Map jsonData =parse.parseJson(value.substring(3) to start at [ but it still gives me an error
编辑：我也试过 Map jsonData =parse.parseJson(value.substring(3) 从 [ 开始，但它仍然给我一个错误
回答by Ravi Thapliyal
In addition to removing the leading //fix your loop as well. Change
除了删除前导//修复你的循环之外。改变
while(input.readLine()!=null) // skipping odd lines
{
value+=input.readLine(); // reading even lines
}
to
到
String line = null;
while((line = input.readLine()) !=null)
{
value +=line;
}
or, better use a StringBuilderlike
或者，更好地使用StringBuilder喜欢
String line = null;
StringBuilder json = new StringBuilder();
while((line = input.readLine()) !=null)
{
json.append(line);
}
value = json.substring(3); // removes the leading "// "
EDIT:
I'm not familiar with your JSON parser. With the org.json.Java parser you could do it this way.
编辑：
我不熟悉你的 JSON 解析器。使用org.json。Java解析器你可以这样做。
JSONArray jsonRoot = new JSONArray(value);
JSONObject quote = jsonRoot.get(0);
System.out.println ("e = " + quote.getString("e"));
But, as a workaround you could strip the []from StringBuilderas
但是，作为一种解决方法，您可以[]从StringBuilderas
// removes the leading "// [" and trailing "]"
value = json.substring(4, json.length() - 1);
回答by Wilker Iceri
This json is not a valid, have two "//".
此 json 无效，有两个“//”。
Use http://jsonlint.com/to validate this
使用http://jsonlint.com/来验证这一点
回答by Brendan Long
The response from that URL starts with //, which isn't valid JSON:
来自该 URL 的响应以 开头//，这不是有效的 JSON：
// [ { "id": "99624" ,"t" : "CSCO" ,"e" : "NASDAQ" ,"l" : "24.00" ,"l_cur" : "24.00" ,"s": "2" ,"ltt":"4:00PM EDT" ,"lt" : "Jun 25, 4:00PM EDT" ,"c" : "-0.05" ,"cp" : "-0.21" ,"ccol" : "chr" ,"el": "24.00" ,"el_cur": "24.00" ,"elt" : "Jun 25, 5:54PM EDT" ,"ec" : "0.00" ,"ecp" : "0.00" ,"eccol" : "chb" ,"div" : "0.17" ,"yld" : "2.83" } ]
// [ { "id": "99624" ,"t" : "CSCO" ,"e" : "NASDAQ" ,"l" : "24.00" ,"l_cur" : "24.00" ,"s": "2 ","ltt":"4:00PM EDT","lt":"6 月 25 日下午 4:00 EDT","c":"-0.05","cp":"-0.21","ccol":" chr","el":"24.00","el_cur":"24.00","elt":"美国东部时间 6 月 25 日下午 5:54","ec":"0.00","ecp":"0.00"," eccol”：“chb”，“div”：“0.17”，“yld”：“2.83”}]
According to thisand this, the Google Finance API is deprecated anyway, so you may want to find something else.
回答by Java Guru
Following blog has enough number of very good examples on quick-json parser
下面的博客有足够多的关于 quick-json 解析器的很好的例子
It has got other competitive parsers examples as well
它还有其他有竞争力的解析器示例
http://codesnippets4all.com/html/parsers/json/quick-json.htm
http://codesnippets4all.com/html/parsers/json/quick-json.htm
回答by wazy
Add this to your code:
将此添加到您的代码中：
String line = null;
while((line = input.readLine()) !=null)
{
value += line;
}
value = value.replace("// ", "");
You need to replace the //at the beginning to "clean" the JSON before you can parse it.
您需要先替换//开头的 以“清理”JSON，然后才能解析它。
回答by swetha
It seems you are using old quick-json parser version. Use the latest version for parsing
看来您使用的是旧的 quick-json 解析器版本。使用最新版本进行解析
quick-json-1.0.2.3.jar
快速json-1.0.2.3.jar
I could see that the json is coming as follows,
我可以看到 json 如下，
// [
{
"id": "99624"
,"t" : "CSCO"
,"e" : "NASDAQ"
,"l" : "25.41"
,"l_cur" : "25.41"
,"s": "2"
,"ltt":"3:59PM EDT"
,"lt" : "Jul 10, 3:59PM EDT"
,"c" : "+0.25"
,"cp" : "1.01"
,"ccol" : "chg"
,"el": "25.55"
,"el_cur": "25.55"
,"elt" : "Jul 10, 7:07PM EDT"
,"ec" : "+0.14"
,"ecp" : "0.55"
,"eccol" : "chg"
,"div" : "0.17"
,"yld" : "2.68"
}
]
This is not valid JSON, it should not be preceded by //
这不是有效的 JSON，它不应以 //
// [
remove //and just use from [till end of the json string
删除//并从[json 字符串的末尾开始使用
i was able to parse successfully the below json string without //
我能够成功解析下面的json字符串而没有 //
[
{
"id": "99624"
,"t" : "CSCO"
,"e" : "NASDAQ"
,"l" : "25.41"
,"l_cur" : "25.41"
,"s": "2"
,"ltt":"3:59PM EDT"
,"lt" : "Jul 10, 3:59PM EDT"
,"c" : "+0.25"
,"cp" : "1.01"
,"ccol" : "chg"
,"el": "25.55"
,"el_cur": "25.55"
,"elt" : "Jul 10, 7:07PM EDT"
,"ec" : "+0.14"
,"ecp" : "0.55"
,"eccol" : "chg"
,"div" : "0.17"
,"yld" : "2.68"
}
]
Below is the output i've got with version quick-json-1.0.2.3.jar
以下是我使用版本quick-json-1.0.2.3.jar获得的输出
{root=[{e=NASDAQ, c=+0.25, div=0.17, l=25.41, lt=Jul 10, 3:59PM EDT, ec=+0.14, ltt=3:59PM EDT, elt=Jul 10, 7:07PM EDT, id=99624, yld=2.68, el_cur=25.55, t=CSCO, cp=1.01, s=2, el=25.55, l_cur=25.41, eccol=chg, ccol=chg, ecp=0.55}]}
