URL 处的 FileNotFoundException

Question

我正在尝试从 php url 获取请求。运行时发生未找到文件异常。我在下面的代码中提到了错误行。Stacktrace 似乎没有从 url 获得响应。

堆栈跟踪：

10-30 05:57:23.000: D/URL(2735): Connectionhttp://example.php

10-30 05:13:37.600: W/System.err(2577): java.io.FileNotFoundException: http://myapi.phpBa
10-30 05:13:37.600: W/System.err(2577):     at com.android.okhttp.internal.http.HttpURLConnectionImpl.getInputStream(HttpURLConnectionImpl.java:186)
10-30 05:13:37.600: W/System.err(2577):     at com.manishkpr.autotextviewexample.JsonParse.getParseJsonWCF(JsonParse.java:32)
10-30 05:13:37.610: W/System.err(2577):     at com.manishkpr.autotextviewexample.SuggestionAdapter$1.performFiltering(SuggestionAdapter.java:39)
10-30 05:13:37.610: W/System.err(2577):     at android.widget.Filter$RequestHandler.handleMessage(Filter.java:234)
10-30 05:13:37.610: W/System.err(2577):     at android.os.Handler.dispatchMessage(Handler.java:102)
10-30 05:13:37.610: W/System.err(2577):     at android.os.Looper.loop(Looper.java:136)
10-30 05:13:37.610: W/System.err(2577):     at android.os.HandlerThread.run(HandlerThread.java:61)

JSONParse.java：

public List<SuggestGetSet> getParseJsonWCF(String sName)
        {
         List<SuggestGetSet> ListData = new ArrayList<SuggestGetSet>();
         try {
            String temp=sName.replace(" ","%20");
            URL js = new URL("example.php"+temp);

            Log.d("URL","Connection"+js);

            URLConnection urlConnection  = js.openConnection();
            BufferedReader reader = new BufferedReader(new InputStreamReader(urlConnection.getInputStream()));  ---32 nd line
            String line = reader.readLine();
            JSONObject jsonResponse = new JSONObject(line);
            JSONArray jsonArray = jsonResponse.getJSONArray("search");
            for(int i = 0; i < jsonArray.length(); i++){
                JSONObject r = jsonArray.getJSONObject(i);
                ListData.add(new SuggestGetSet(r.getString("id"), 
r.getString("name")));
            }
        } catch (Exception e1) {

            e1.printStackTrace();
        }
         return ListData;

        }

Answer 1

尝试替换此行

String temp=sName.replace(" ","%20");
URL js = new URL("http://192.11.my api.php"+temp);

与

String temp=URLEncoder.encode("http://192.11.my api.php/"+sName, "UTF-8");
URL js = new URL(temp);