Question

-(void)setFlagToZero:(BOOL)flg id:(NSString *)qid
{
    NSLog(@"flag %hhd and ID:%@ ",flg,qid);
    sqlite3 *database;
    quiz = [[NSMutableArray alloc] init];
    if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK)
    {
        const char *sqlStatement = &"UPDATE  Question set Question_Flag="+flg+"where question_ID="+qid;
        sqlite3_stmt *compiledStatement;
        if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK)
        {
            while(sqlite3_step(compiledStatement) == SQLITE_ROW)
            {
                NSString *aQuestion = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 2)];
            }
        }
        else
        {
            NSLog(@"Error occured in connection");
        }
        sqlite3_finalize(compiledStatement);
    }
    sqlite3_close(database);
}
@end

显示如下错误

二进制表达式的操作数无效（＆＃39; char（*）[36]＆＃39;＆＃39; char *＆＃39;）

请帮我修正更正的代码。

Answer 1

问题在于以下声明（您已在该字符串中添加了&和+）：

const char *sqlStatement = &"UPDATE  Question set Question_Flag="+flg+"where question_ID="+qid;

将其更改为：

NSString *query          = [NSString stringWithFormat:@"UPDATE  Question set Question_Flag=%d where question_ID=%d",flg,qid];
const char *sqlStatement = [query UTF8String];

Answer 2

试试这个更新，它的工作

 -(void)updateDataBaseRecords:(NSString *)databaseQuery
{
    NSString *replaceStatement = databaseQuery;
    // Execute the replace
    char *error;
    if (sqlite3_exec(database, [replaceStatement UTF8String], NULL, NULL, &error) == SQLITE_OK)
    {
        UIAlertView *alert = [[UIAlertView alloc]initWithTitle:@"Message" message:@"Updated successfully" delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil, nil ];
        [alert show];
    }
    else
    {
        NSLog(@"Error: %s", error);
    }

}

在此功能中直接传递您的查询

二进制表达式的操作数无效（＆＃39; char（）[36]＆＃39;＆＃39; char ＆＃39;）

2 个答案: