【发布时间】:2018-10-07 06:07:35
【问题描述】:
我在插入时关注了这个人的answer。
这是我的AsyncTask 的代码
private class sendRequest extends AsyncTask<String, String, String> {
String z = "";
boolean isSuccess = false;
@Override
protected void onPreExecute() {
progressDialog.setMessage("Sending request...");
progressDialog.show();
super.onPreExecute();
}
@Override
protected String doInBackground(String... params) {
try {
Connection con = connectionClass.CONN();
if (con == null) {
z = "Please check your internet connection";
} else {
myArrayList.clear();
String query = "insert into notifications (id, suggestion, type, isIgnored) values ('', '"+ sv.getQuery().toString() +"', 'medicine', 'false')";
Statement stmt1 = con.createStatement();
stmt1.executeUpdate(query);
}
} catch (Exception ex) {
isSuccess = false;
z = "Exceptions" + ex;
}
return z;
}
@Override
protected void onPostExecute(String s) {
if(!isSuccess && !z.equals("")) {
Toast.makeText(getBaseContext(), z, Toast.LENGTH_LONG).show();
}
progressDialog.hide();
}
}
但是它在下面给了我这个错误消息,我想知道为什么或如何这是一个错误,因为我的查询是插入而不是截断。请帮忙
【问题讨论】: