我正在尝试在我的应用程序中注册用户。我有一个带有3 textfields (username, password and confirmpassword)
和submit
按钮的视图控制器。
按下提交按钮时,将调用以下方法:
-(IBAction)addUser
{
NSString *tempUser,*tempPass, *tempConfPass;
tempUser = [[NSString alloc]init];
tempPass = [[NSString alloc]init];
tempConfPass = [[NSString alloc]init];
tempUser = [NSString stringWithFormat:@"%@",_mUserName.text];
tempPass = [NSString stringWithFormat:@"%@",_mPassword.text];
tempConfPass = [NSString stringWithFormat:@"%@",_mConfPassword.text];
signupUser = [[UseDb alloc]init];
flagUser = [signupUser addNewUser:_mUserName.text:_mPassword.text:_mConfPassword.text];
if(flagUser)
{
myAlertViewUser = [[UIAlertView alloc] initWithTitle:@"Error" message:@"User Added"
delegate:nil cancelButtonTitle:@"OK" otherButtonTitles: nil];
[myAlertViewUser show];
}
else {
_mStatus.text = @"failed to add user";
myAlertViewUser = [[UIAlertView alloc] initWithTitle:@"Error" message:@"passwords don't match"
delegate:nil cancelButtonTitle:@"OK" otherButtonTitles: nil];
[myAlertViewUser show];
}
}
并且此方法由
addUser
方法调用:-(BOOL)addNewUser:(NSString *)newUser :(NSString *)newPassword :(NSString *)confirmPass
{
NSLog(@"%@....%@...%@",newUser, newPassword, confirmPass);
sqlite3_stmt *statement;
const char *dbpath = [_mDatabasePathDb UTF8String];
if (sqlite3_open(dbpath, &_mDb) == SQLITE_OK && [newPassword isEqualToString:confirmPass] && ![newUser isEqualToString:@""] && ![newPassword isEqualToString:@""])
{
self.userName = [NSString stringWithFormat:@"%@",newUser];
self.password = [NSString stringWithFormat:@"%@",newPassword];
NSString *insertSQL = [NSString stringWithFormat:
@"INSERT INTO USERDETAIL VALUES (\"%@\",\"%@\",\"%@\", \"%@\",\"%@\", \"%@\",\"%@\", \"%@\",\"%@\", \"%@\",\"%@\")",self.userName,self.password,@"",@"",@"",@"",@"",@"",@"",@"",@"" ];
NSLog(@"%@",insertSQL);
const char *insert_stmt = [insertSQL UTF8String];
sqlite3_prepare_v2(_mDb, insert_stmt, -1, &statement, NULL);
if (sqlite3_step(statement) == SQLITE_DONE)
{
return YES;
/* mUserName.text = @"";
mPassword.text = @"";
mConfPassword.text = @""; */
}
else {
NSLog(@"failed to add user");
}
sqlite3_finalize(statement);
sqlite3_close(_mDb);
}
}
在
addNewUser
方法中,if (sqlite3_step(statement) == SQLITE_DONE)
始终是false
,语句之前有一些值sqlite3_prepare_v2(_mDb, insert_stmt, -1, &statement, NULL);
但是在执行了以上语句后变成
nil
。我不明白为什么会这样。请帮忙。 最佳答案
这可能与您正在执行的检查有关,请尝试使用SQLITE_OK。它在docs中表示为在旧版接口(旧版接口)中,返回值将为 SQLITE_BUSY,SQLITE_DONE,SQLITE_ROW,SQLITE_ERROR或SQLITE_MISUSE 。使用“v2”接口,也可能会返回任何其他结果代码或扩展结果代码。
if (sqlite3_prepare_v2(_mDb, insert_stmt, -1, &statement, nil) == SQLITE_OK)
{
if (sqlite3_step(statement) == SQLITE_DONE)
{
return YES;
}
else
{
NSLog(@"failed to add user");
}
sqlite3_finalize(statement);
}
您还可以找到类似的问题及其答案here
在prepare语句后插入花括号,并在finalize语句后将其关闭。当您获得SQLITE_MISUSE时,可能是该例程被不适当地调用。可能是在已经完成的预准备语句或先前返回SQLITE_ERROR或SQLITE_DONE的语句上调用了它。也可能是两个或多个线程同时使用同一数据库连接。
希望这可以帮助 :)
关于ios - sqlite3_step(statement)== SQLITE_DONE始终为false,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/16891190/