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iOS-更新表失败?_Ios_Objective C_Xcode_Sqlite - Fatal编程技术网

iOS-更新表失败?

iOS-更新表失败?,ios,objective-c,xcode,sqlite,Ios,Objective C,Xcode,Sqlite,我的桌子在插入过程中没有上升。在tableview中也不显示,但单击“添加新记录”后,消息会显示“已添加记录”,因此没有任何错误 在表函数中插入: -(int) insert:(NSString *)filePath withBrand:(NSString *)brand { sqlite3* db = NULL; int rc=0; rc = sqlite3_open_v2([filePath cStringUsingEncoding:NSUTF8StringEncod

我的桌子在插入过程中没有上升。在tableview中也不显示,但单击“添加新记录”后,消息会显示“已添加记录”,因此没有任何错误

在表函数中插入:

-(int) insert:(NSString *)filePath withBrand:(NSString *)brand
{
    sqlite3* db = NULL;
    int rc=0;
    rc = sqlite3_open_v2([filePath cStringUsingEncoding:NSUTF8StringEncoding], &db, SQLITE_OPEN_READWRITE , NULL);
    if (SQLITE_OK != rc)
    {
        sqlite3_close(db);
        NSLog(@"Failed to open db connection");
    }
    else
    {
        NSString * query  = [NSString
                             stringWithFormat:@"INSERT INTO brands (bikeBrandName) VALUES (?)"];
        char * errMsg;
        rc = sqlite3_exec(db, [query UTF8String] ,NULL,NULL,&errMsg);
        if(SQLITE_OK != rc)
        {
            NSLog(@"Failed to insert record  rc:%d, msg=%s",rc,errMsg);
        }
        sqlite3_close(db);
    }
    return rc;
}
我也试过这个:

-(int) insert:(NSString *)filePath withBrand:(NSString *)brand
{
    sqlite3* db = NULL;
    int rc=0;
    rc = sqlite3_open_v2([filePath cStringUsingEncoding:NSUTF8StringEncoding], &db, SQLITE_OPEN_READWRITE , NULL);
    if (SQLITE_OK != rc)
    {
        sqlite3_close(db);
        NSLog(@"Failed to open db connection");
    }
    else
    {
        NSString * query  = [NSString
                                 stringWithFormat:@"INSERT INTO brands (bikeBrandName) VALUES (\"%@\")",brand];
        char * errMsg;
        rc = sqlite3_exec(db, [query UTF8String] ,NULL,NULL,&errMsg);
        if(SQLITE_OK != rc)
        {
            NSLog(@"Failed to insert record  rc:%d, msg=%s",rc,errMsg);
        }
        sqlite3_close(db);
    }
    return rc;
}
不是吗

应该是

INSERT INTO brands (bikeBrandName) VALUES ('%@')",brand
                                           ^  ^

要进一步调试,请使用check


当然会有帮助的

谢谢你的回答。。但是不起作用仍然不更新直到不起作用…尝试了教程,但很难实现到我的代码中,因为使用其他方法插入到数据库中
INSERT INTO brands (bikeBrandName) VALUES ('%@')",brand
                                           ^  ^