Question

我的桌子在插入时没有提升。也没有显示在tableview中，但点击后添加“新记录”消息附带“添加记录”，所以没有任何错误

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;
}

还试过这个：

-(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;
}

Answer 1

不是

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

应该是

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

要进一步调试，请使用check this tutorial

肯定会有帮助

iOS - 无法更新表格？

1 个答案: