java.lang.IllegalArgumentException：索引59处的查询中存在非法字符 [英] java.lang.IllegalArgumentException: Illegal character in query at index 59

问题描述

我在做什么：
我试图在android中进行反向地理编码

错误为::
java.lang.IllegalArgumentException：索引59处的查询中存在非法字符： http://maps.google.com/maps/api/geocode/json?address=Agram ，印度卡纳塔克邦，班加罗尔& sensor = false

注意：该请求在浏览器中获得json响应，但不在我的课程下面获得。

这个错误::

  HttpGet httpget = new HttpGet（url）; 
  

---

JSONfunctions.java

  public class JSONfunctions {
 
 $ public static JSONObject getJSONfromURL（String url）{
 InputStream is = null; 
 String result =; 
 JSONObject jArray = null; 
 
 //从URL下载JSON数据
尝试{
 HttpClient httpclient = new DefaultHttpClient（）; 
 HttpGet httpget = new HttpGet（url）; 
 
 HttpResponse响应= httpclient.execute（httpget）; 
 HttpEntity entity = response.getEntity（）; 
 is = entity.getContent（）; 
 $ b} catch（Exception e）{
 Log.e（log_tag，http connection中的错误+ e.toString（））; 
 
 
 //将响应转换为字符串
 try {
 BufferedReader reader = new BufferedReader（new InputStreamReader（
 is，iso-8859-1 ），8）; 
 StringBuilder sb = new StringBuilder（）; 
 String line = null; （（line = reader.readLine（））！= null）{
 sb.append（line +\\\
）; 
 while 
} 
 is.close（）; 
 result = sb.toString（）; 
} catch（Exception e）{
 Log.e（log_tag，Error conversion result+ e.toString（））; 
} 
 
尝试{
 
 jArray = new JSONObject（result）; 
 catch（JSONException e）{
 Log.e（log_tag，Error parsing data+ e.toString（））; 
} 
 
 return jArray; 
} 
} 
  

解决方案

URLEncoder.encode（） 来编码您的地址参数Agram，Bengaluru ，印度卡纳塔克邦，然后将其放入URL字符串中，以便它变成类似于

的

  http ：//maps.google.com/maps/api/geocode/json？address = Agram，+ Bengaluru，+ Karnataka，+ India& sensor = false 
  

即空格变为 + 和其他特殊的八位字节表示为％xx 。

浏览器会自动为在地址栏中输入的字符串进行智能网址编码，这就是为什么它可以在那里运行。

What i am doing: I am trying to make a reverse geocoding in android

I am getting error as:: java.lang.IllegalArgumentException: Illegal character in query at index 59: http://maps.google.com/maps/api/geocode/json?address=Agram, Bengaluru, Karnataka, India&sensor=false

NOte: that request gets a json response in browser but not from my class below

This line is giving this error::

HttpGet httpget = new HttpGet(url);

---

JSONfunctions.java

public class JSONfunctions {

    public static JSONObject getJSONfromURL(String url) {
        InputStream is = null;
        String result = "";
        JSONObject jArray = null;

        // Download JSON data from URL
        try {
            HttpClient httpclient = new DefaultHttpClient();
            HttpGet httpget = new HttpGet(url);

            HttpResponse response = httpclient.execute(httpget);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();

        } catch (Exception e) {
            Log.e("log_tag", "Error in http connection " + e.toString());
        }

        // Convert response to string
        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            result = sb.toString();
        } catch (Exception e) {
            Log.e("log_tag", "Error converting result " + e.toString());
        }

        try {

            jArray = new JSONObject(result);
        } catch (JSONException e) {
            Log.e("log_tag", "Error parsing data " + e.toString());
        }

        return jArray;
    }
}

解决方案

Use URLEncoder.encode() to encode the value of your address parameter "Agram, Bengaluru, Karnataka, India" before putting it in the URL string so that it becomes something like

http://maps.google.com/maps/api/geocode/json?address=Agram,+Bengaluru,+Karnataka,+India&sensor=false

i.e. spaces changed to + and other special octets represented as %xx.

Browsers do smart URL encoding for strings entered in the address bar automatically so that's why it works there.

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