有趣的错误与我的数据库 [英] funny error with my database
问题描述
我的应用是联系人的书,终于一切都工作正常,但是当我,用户,添加新的联系人,我得到强制关闭..但应用程序没有关闭,而是刚刚重启!当我检查联系人列表中我看到了我所要补充的是真正加入!
logcat中说,错误是在这里:
返回cursorToContact(光标);
这是这个附加功能的一部分:
/ **添加新的联系人到数据库中。 * /
公开联系方式的createContact(联系方式联系我们){
//用来存储数据,如:COLUMN_NAME,VALUE。
ContentValues值=新ContentValues();
values.put(MySQLiteHelper.COLUMN_FIRST_NAME,contact.getFirstName());
values.put(MySQLiteHelper.COLUMN_LAST_NAME,contact.getLastName());
//插入新的接触,并获得它的ID。
长insertId = database.insert(MySQLiteHelper.TABLE_NAME_FTS,空,
值);
//为了展示如何查询(要获取它的id接触)。
光标光标= database.query(MySQLiteHelper.TABLE_NAME_FTS,
allColumns,MySQLiteHelper.COLUMN_ID +=+ insertId,
空值,
NULL,NULL,NULL);
//将光标移动到第一行。
cursor.moveToFirst();
返回cursorToContact(光标);
}
那么,为什么这种情况发生> plz帮助我
请注意,我期运用的SQLite数据库FTS3
在cursorToContact:
/ **用于获取联系人数据从光标到备忘录对象。 * /
私人联系cursorToContact(光标光标){
联系方式联系我们=新的联系人();
contact.setId(cursor.getLong(0));
contact.setFirstName(cursor.getString(1));
contact.setLastName(cursor.getString(2));
返回接触;
}
这行实际上将数据添加到您的数据库:
长insertId = database.insert(MySQLiteHelper.TABLE_NAME_FTS,空,
值);
任何之后发生的意外(大部分)。
如果您的查询搞砸了,你的其他部分将失败。您的疑问:
光标光标= database.query(MySQLiteHelper.TABLE_NAME_FTS,
allColumns,MySQLiteHelper.COLUMN_ID +=+ insertId,
空值,
NULL,NULL,NULL);
看起来合适。该allColumns可以搞砸了,虽然。
另外,我不知道为什么你去设立在CursorToContact新联系人的所有的麻烦,只返回一些其他的价值。
/ **用于获取联系人数据从光标到备忘录对象。 * /
私人联系cursorToContact(光标光标){
联系方式联系我们=新的联系人();
contact.setId(cursor.getLong(0));
contact.setFirstName(cursor.getString(1));
contact.setLastName(cursor.getString(2));
返回接触;
}
您从来不用接触。我本来期望你回来吧。
my app is contacts book , finally everything's working fine but when I, "the user", add new contact I get force close .. but the app didn't close, but rather just restarted ! and when i check the contact list i see that what i tried to add is really been added !!! logcat say the error is here :
return cursorToContact(cursor);
it is part of this add function:
/** Add new Contact to Database. */
public Contact createContact(Contact contact) {
//Used to store data like : COLUMN_NAME , VALUE.
ContentValues values = new ContentValues();
values.put(MySQLiteHelper.COLUMN_FIRST_NAME, contact.getFirstName());
values.put(MySQLiteHelper.COLUMN_LAST_NAME, contact.getLastName());
//Insert new contact and get the id of it.
long insertId = database.insert(MySQLiteHelper.TABLE_NAME_FTS, null,
values);
//To show how to query (To get contact by it id).
Cursor cursor = database.query(MySQLiteHelper.TABLE_NAME_FTS,
allColumns, MySQLiteHelper.COLUMN_ID + " = " + insertId,
null,
null, null, null);
//Move Cursor to the first row.
cursor.moveToFirst();
return cursorToContact(cursor);
}
so why is that happen > plz help me note that i'm useing sqlite fts3 database
the cursorToContact :
/** Used to get Contact data from Cursor to Memo Object. */
private Contact cursorToContact(Cursor cursor) {
Contact contact = new Contact();
contact.setId(cursor.getLong(0));
contact.setFirstName(cursor.getString(1));
contact.setLastName(cursor.getString(2));
return contact;
}
This line actually adds the data to your database:
long insertId = database.insert(MySQLiteHelper.TABLE_NAME_FTS, null,
values);
Anything that happens after that is incidental (mostly).
If your query is messed up, your other parts will fail. Your query:
Cursor cursor = database.query(MySQLiteHelper.TABLE_NAME_FTS,
allColumns, MySQLiteHelper.COLUMN_ID + " = " + insertId,
null,
null, null, null);
looks appropriate. The allColumns could be messed up, though.
Also, I'm not sure why you go to all the trouble of setting up a new contact in CursorToContact, only to return some other value.
/** Used to get Contact data from Cursor to Memo Object. */
private Contact cursorToContact(Cursor cursor) {
Contact contact = new Contact();
contact.setId(cursor.getLong(0));
contact.setFirstName(cursor.getString(1));
contact.setLastName(cursor.getString(2));
return contact;
}
You never use contact. I would have expected for you to return it.
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