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I have editText's (Username, Firstname, Lastname and Email Address) in my activity for user registration. The user has the privilege to search if the username that he input is already existing or not, by clicking the Search Button. If the username is existing, all the information with regards to that username like the name of the user will be showed. However, if the user click the button if it is not existing, my app crashes and I am getting CursorIndexOutOfBoundsException error. How can I debug that one?

MainActivity.java

btn_Search.setOnClickListener(new OnClickListener() {
    @Override
    public void onClick(View arg0) {
        String searchableUser = txt_User.getText().toString();

        ConsUserRegistration consUserRegistration = db.searchUser(searchableUser);

        String searchUser = consUserRegistration.getUser().toString();
        String searchFirst = consUserRegistration.getFirstName().toString();
        String searchLast = consUserRegistration.getLastName().toString();
        String searchEmail = consUserRegistration.getEmail().toString();

        txt_User.setText(searchUser);
        txt_First.setText(searchFirst);
        txt_Last.setText(searchLast);
        txt_Email.setText(searchEmail);             
    }
});

DatabaseHandler.java

public ConsUserRegistration searchUser(String username){

    SQLiteDatabase db = this.getReadableDatabase();
    Cursor cursor = db.query(Constants.TABLE_USER, new String[] {Constants.KEY_USER, Constants.KEY_FIRST, 
            Constants.KEY_LAST, Constants.KEY_EMAIL}, Constants.KEY_USER + " =? ", 
            new String[] { String.valueOf(username) }, null, null, null);
    if (cursor != null)
        cursor.moveToFirst();

    ConsUserRegistration search = new ConsUserRegistration (cursor.getString(0), cursor.getString(1), cursor.getString(2), cursor.getString(3));

    return search;

}
4

2 回答 2

1

尝试这个：

MainActivity.java

btn_Search.setOnClickListener(new OnClickListener() {
    @Override
    public void onClick(View arg0) {
        String searchableUser = txt_User.getText().toString();

        ConsUserRegistration consUserRegistration = db.searchUser(searchableUser);
        if (consUserRegistration != null){

           String searchUser = consUserRegistration.getUser().toString();
           String searchFirst = consUserRegistration.getFirstName().toString();
           String searchLast = consUserRegistration.getLastName().toString();
           String searchEmail = consUserRegistration.getEmail().toString();

           txt_User.setText(searchUser);
           txt_First.setText(searchFirst);
           txt_Last.setText(searchLast);
           txt_Email.setText(searchEmail);
        }else{
         Toast.makeText(getApplicationContext(), "Username Not Found", Toast.LENGTH_LONG).show();
        }               
    }
});

数据库处理程序.java

public ConsUserRegistration searchUser(String username){
    ConsUserRegistration search;
    SQLiteDatabase db = this.getReadableDatabase();
    Cursor cursor = db.query(Constants.TABLE_USER, new String[] {Constants.KEY_USER, Constants.KEY_FIRST, 
            Constants.KEY_LAST, Constants.KEY_EMAIL}, Constants.KEY_USER + " =? ", 
            new String[] { String.valueOf(username) }, null, null, null);
    if (cursor != null && cursor.moveToFirst()){
            search = new ConsUserRegistration (cursor.getString(0), cursor.getString(1), cursor.getString(2), cursor.getString(3));
        }else{
            search = null;
        }

    return search;

}
于 2013-07-17T01:09:41.210 回答
1

所以...如果您的光标为空或没有元素会发生什么？您仍在尝试访问它...

尝试类似...

if (cursor != null && cursor.moveToFirst())  {    

...
于 2013-07-17T00:43:05.823 回答