java.lang.IllegalArgumentException：Android 中索引 77 处查询中的非法字符

Question

我正在尝试使用 json 解析 Facebook 提要

显示这个错误

06-02 16:53:33.112: D/ee:(29180): java.lang.IllegalArgumentException: 索引 77 处查询中的非法字符： https://graph.facebook.com/331394590231184/feed?access_token=******&client_id=**&client_secret=*****?

我使用代码：

private static String url = "https://graph.facebook.com/331394590231184/feed?access_token=**|*****&client_id=***&client_secret=****";
JSONParser jParser = new JSONParser();
List<NameValuePair> params = new ArrayList<NameValuePair>();
JSONObject json = jParser.makeHttpRequest(url, "GET", params);
JSONArray data = json.getJSONArray("data");

JSONParser 代码：

static InputStream is = null;
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();

Answer 1

您正在构建一个包含多个 ? 的无效 URL，您应该只传递方案、主机和路径作为 url 变量，然后分别传递参数：

private static String url = "https://graph.facebook.com/331394590231184/feed";
JSONParser jParser = new JSONParser();
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("access_token", "**|*****"));
params.add(new BasicNameValuePair("client_id", "***"));
params.add(new BasicNameValuePair("client_secret", "****"));
JSONObject json = jParser.makeHttpRequest(url, "GET", params);
JSONArray data = json.getJSONArray("data");

解决它的另一种方法是，在您的 JSONParser 代码中处理它之前，确保 params 包含任何内容：

static InputStream is = null;
DefaultHttpClient httpClient = new DefaultHttpClient();
if (params != null && params.size() > 0){
    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();

Answer 2

https://www.google.com/search?q=valid+url+characters -- 你在字符 77 周围有什么不寻常的地方吗？另外，https://www.google.com/search?q=encode+url

顺便说一句，编辑问题不会删除以前的修订：https://stackoverflow.com/posts/30601529/revisions