嘿,我的应用程序遇到无法解决的问题,我尝试了有关异步任务的所有知识,但我并不擅长。
错误:
03-17 19:43:38.794 1662-1681/com.cwpsiproject E/AndroidRuntime﹕ FATAL EXCEPTION: AsyncTask #5
java.lang.RuntimeException: An error occured while executing doInBackground()
at android.os.AsyncTask$3.done(AsyncTask.java:299)
at java.util.concurrent.FutureTask.finishCompletion(FutureTask.java:352)
at java.util.concurrent.FutureTask.setException(FutureTask.java:219)
at java.util.concurrent.FutureTask.run(FutureTask.java:239)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1080)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:573)
at java.lang.Thread.run(Thread.java:856)
Caused by: java.lang.IllegalArgumentException: Illegal character in query at index 72: http://sqlphptry.co.nf/createUser.php?teamName=vcxvxcvcxvxcvxc&game=COD: AW&consola=Xbox One
at java.net.URI.create(URI.java:727)
at org.apache.http.client.methods.HttpGet.<init>(HttpGet.java:75)
at com.cwpsiproject.ApiConnector.createTeam(ApiConnector.java:196)
at com.cwpsiproject.CreateTeamActivity$createTeamTask.doInBackground(CreateTeamActivity.java:131)
at com.cwpsiproject.CreateTeamActivity$createTeamTask.doInBackground(CreateTeamActivity.java:124)
at android.os.AsyncTask$2.call(AsyncTask.java:287)
at java.util.concurrent.FutureTask.run(FutureTask.java:234)
at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1080)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:573)
at java.lang.Thread.run(Thread.java:856)
代码CreateTeamTask:
private class createTeamTask extends AsyncTask<ApiConnector,Long,JSONArray> {
@Override
protected JSONArray doInBackground(ApiConnector... params) {
// it is executed on Background thread
String teamName = f.getTeamNameC(); // returns string with teamName
String game = f.getGameC(); // returns string with game
String consola = f.getConsoleC(); // returns string with console
return params[0].createTeam(teamName, game, consola);
}
@Override
protected void onPostExecute(JSONArray jsonArray) {
Toast.makeText(CreateTeamActivity.this, "Team Successfully Created!", Toast.LENGTH_SHORT).show();
}
}
ApiConnector代码:
public JSONArray createTeam(String teamName, String game, String consola) {
String url = "http://sqlphptry.co.nf/createUser.php?teamName="+teamName+"&game="+game+"&consola="+consola;
// Get HttpResponse Object from url.
// Get HttpEntity from Http Response Object
HttpEntity httpEntity = null;
try {
DefaultHttpClient httpClient = new DefaultHttpClient(); // Default HttpClient
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
httpEntity = httpResponse.getEntity();
} catch (ClientProtocolException e) {
// Signals error in http protocol
e.printStackTrace();
//Log Errors Here
} catch (IOException e) {
e.printStackTrace();
}
// Convert HttpEntity into JSON Array
JSONArray jsonArray = null;
if (httpEntity != null) {
try {
String entityResponse = EntityUtils.toString(httpEntity);
Log.e("Entity Response : ", entityResponse);
jsonArray = new JSONArray(entityResponse);
} catch (JSONException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
return jsonArray;
}
PHP代码:
include 'connection.php';
if (isset($_GET['teamName']) && isset($_GET['game']) && isset($_GET['consola'])) {
$teamName = $_GET['teamName'];
$game = $_GET['game'];
$consola = $_GET['consola'];
$teamName = mysql_real_escape_string($teamName);
$game = mysql_real_escape_string($game);
$consola = mysql_real_escape_string($consola);
mysql_query("INSERT INTO `team_data` (`teamName`, `game`, `console`) VALUES('$teamName','$game','$consola')");
print('[{"created":"'.$teamName.'"}]');
}
mysql_close($dbhandle);
有人可以帮我吗?我被这个错误困住了,有趣的是,我有几乎相同的代码来注册用户,并且没有给出这个错误。
感谢您的时间。
参考方案
您的网址中有非法字符(主要是空格和冒号),您需要确保使用正确的url编码。一种简单的方法是使用Uri.Builder()来构建您的URL。
编辑:
在您的异常中,它显示以下URL引发错误:“ http://sqlphptry.co.nf/createUser.php?teamName=vcxvxcvcxvxcvxc&game=COD:AW&consola = Xbox One”。请注意,即使显示为此处,第一个非法字符也会破坏超链接。
要清理此网址,您需要正确地附加查询参数,例如使用上面的url:
Uri.Builder builder = Uri.parse("http://sqlphptry.co.nf/createUser.php").buildUpon();
builder.appendQueryParameter("teamName", "vcxvxcvcxvxcvxc");
builder.appendQueryParameter("game", "COD: AW");
builder.appendQueryParameter("consola", "Xbox One");
String url = builder.build().toString();
结果:http://sqlphptry.co.nf/createUser.php?teamName=vcxvxcvcxvxcvxc&game=COD%3A%20AW&consola=Xbox%20One
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