MySQLi准备更新而不更新数据库

As the title states, my MySQLi prepared update is not actually updating the database. I have checked the MySQL logs - no errors. Here is the code in question:

正如标题所述,我的MySQLi准备更新实际上并没有更新数据库。我检查了MySQL日志 - 没有错误。这是有问题的代码:

public function update_tweet($tweet)
{
    $prepared_update = $this->connection->prepare("UPDATE Tweets 
                     SET `text` = ?, `algo_score` = ?, `has_algo_score` = ?, `baseline_score` = ?, `has_baseline_score` = ?, `is_sanitized` = ?
                     WHERE `twitter_id` = ?");
    mysqli_stmt_bind_param($prepared_update, "sssssss", $tweet['text'], $tweet['algo_score'], $tweet['has_algo_score'], $tweet['baseline_score'], $tweet['has_baseline_score'], $tweet['is_sanitized'], $tweet['tweet_id']);
    mysqli_execute($prepared_update) or die(mysqli_error($this->connection));
    $prepared_update->close();
}

And an example of the tweet array being passed in:

传递的推文数组的示例如下:

Array ( [id] => 2 
        [twitter_id] => 595463376026734592 
        [text] => History has a way of repeating itself 
        [algo_score] => 0 
        [has_algo_score] => 1 
        [baseline_score] => 0 
        [has_baseline_score] => 1 
        [is_sanitized] => 1 
)

And the table schema:

和表架构:

No PHP errors or MySQL errors. What am I doing wrong?

没有PHP错误或MySQL错误。我究竟做错了什么?

1 个解决方案

#1

Your condition:

WHERE `twitter_id` = ?

The variable that you're binding is:

您绑定的变量是:

$tweet['tweet_id']

While the array you're getting is:

你得到的阵列是:

Array ( [id] => 2 
        [twitter_id] => 595463376026734592

So you're using the wrong index and since the index tweet_id is undefined (it's just a notice, you're receiving no errors or warnings) your WHERE condition would never return true, therefore - no action is being taken.

所以你使用了错误的索引,因为索引tweet_id是未定义的(它只是一个通知,你没有收到任何错误或警告)你的WHERE条件永远不会返回true,因此 - 没有采取任何行动。