【发布时间】:2012-12-27 13:03:00
【问题描述】:
我正在制作一个 android 应用程序,这是它抛出的第一个错误:
Error parsing data org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject
这是我的类函数:
package com.ernest.httppost;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.util.List;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;
import android.util.Log;
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
// function get json from url
// by making HTTP POST or GET method
public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {
// Making HTTP request
try {
// check for request method
if(method == "POST"){
// request method is POST
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}else if(method == "GET"){
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
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();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
我认为它会在最后一个 try-catch 块中的 jObj = new JSONObject(json); 返回该错误。任何想法如何解决这个问题?
【问题讨论】:
-
Java 是一种编程语言。 Eclipse 是一个 IDE(集成开发环境)。
-
我认为你用来构造 JSONObject 的字符串应该是一个有效的 json 字符串。由于错误状态值
<br无法转换,我猜您正在尝试从 HTML 字符串构造一个 json 对象。确保您传递的是格式正确的 json 字符串。例如:"{\"name\":\"John Ernest Guadalupe\", \"reputation\":181, \"messages\":[\"msg 1\",\"msg 2\",\"msg 3\"]}" -
@Jules 你应该写一个答案
-
@Ondra 我不认为
android标签是相关的,即使它是一个 Android 应用程序。 -
对……我把这个习惯自动化的太多了。
标签: java json xmlhttprequest