ios - sqlite3_step(statement) == SQLITE_DONE 总是假的

Question

我正在尝试在我的应用程序中注册一个用户。我有一个带有 3 textfields (username, password and confirmpassword) 的 View Controller 和 submit按钮。

按下提交按钮时会调用以下方法:

-(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];
    }
}

此方法由 addUser 调用方法:

-(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， if (sqlite3_step(statement) == SQLITE_DONE)总是出来是 false , 声明之前有一些值(value)

sqlite3_prepare_v2(_mDb, insert_stmt, -1, &statement, NULL); 

但转向 nil上述语句执行后。我不明白为什么会这样。请帮忙。

Answer 1

这可能是您正在进行的检查的问题，请尝试 SQLITE_OK。它在 docs 中说如在旧接口(interface)(旧接口(interface))中，返回值将是 SQLITE_BUSY、SQLITE_DONE、SQLITE_ROW、SQLITE_ERROR 或 SQLITE_MISUSE .与 “v2”接口(interface) ，也可能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 语句后插入一个花括号，并在finalize 语句后关闭它。当你得到 SQLITE_MISUSE 时，可能是这个例程被不恰本地调用了。也许它是在已经完成的准备好的语句上调用的，或者是在之前返回 SQLITE_ERROR 或 SQLITE_DONE 的语句上调用的。或者可能是同一数据库连接同时被两个或多个线程使用。

希望这可以帮助 :)