ios - Xcode : sqlite3_step returning 101, 但不在 block 内执行代码

Question

如果 sqlite3_step 语句返回 101，我有一段代码要执行，但即使是，代码块中的代码也会被跳过。为什么会这样？

sqlite3_stmt *statement;
if (sqlite3_open([dbPathString UTF8String], &database) == SQLITE_OK) {
    NSLog(@"Value: %li",(long)sender.tag);
    NSString *delete_feed = [NSString stringWithFormat:@"DELETE FROM user_data WHERE id = %li", (long)sender.tag];
    if (sqlite3_prepare_v2(database, [delete_feed UTF8String], -1, &statement, NULL) == SQLITE_OK) {
        NSLog(@"%d", sqlite3_step(statement));
        if (sqlite3_step(statement) == SQLITE_DONE) { <== MY ISSUE
            NSLog(@"Got here"); <== NEVER GETS HERE
            for (UIView *subview in [self.view subviews]) {
                if (subview.tag == 2) {
                    [subview removeFromSuperview];
                    [self createListPage];
                }
            }
            sqlite3_finalize(statement);
            sqlite3_close(database);
        }
    }
}

Answer 1

您正在调用 sqlite3_step 两次。我认为第一次是 SQLITE_DONE (101)，但第二次(当你进入你的 block 时)，它返回了一些其他错误代码。

可以尝试类似(伪代码)

    int ret_code = sqlite3_step(statement);
    NSLog(@"ret_code=[%d]", ret_code);
    if(ret_code == SQLITE_DONE) {
       // do some stuff here...
    }