PHP 事务和 mysqli_insert_id

Question

我有以下表格:

thread: id, title, content, created
thread_tags: tag_id, thread_id
tag: id, name
author_threads: thread_id, author_id

这是我通常将值插入所有这些字段的方法(为简单起见，省略了一些步骤):

  $sql_thread = "INSERT INTO thread (title, content)
           VALUES ('some title', 'some content')";

   #  this is normally a loop, as there are more than one tags:   
   $sql_tags = "INSERT INTO tag (name) 
               VALUES ('onetag')";


   #  normally I would check the return value
   mysqli_query($link, $sql_thread);

   #  get the thread id:
   $thread_id = mysqli_insert_id($link);

   mysqli_query($link, $sql_tags);

   #  get the tag id:
   $tag_id = mysqli_insert_id($link);

   #  insert into thread_tags:
   mysqli_query($link, "INSERT INTO thread_tags (thread_id, tag_id)  VALUES ($thread_id, $tag_id)");

   #  insert into author_threads, I already know author_id:
   mysqli_query($link, "INSERT INTO author_threads (author_id, thread_id)  VALUES ($author_id, $thread_id)")

我想确保所有这一切都发生或没有发生(即创建线程的过程)。我如何编写代码以使用交易？或者任何其他方式来确保所有这一切发生？

Answer 1

1. 将您的表转换为 innodb(如果尚未转换)
2. 在所有查询之前执行 mysqli_autocommit($link, FALSE);
3. 在所有查询之后执行 mysqli_commit($link);，如果它们已成功执行
4. 如果任何查询失败 - 执行 mysqli_rollback($link); 并停止执行

更多详情:

• > http://ru.php.net/manual/en/mysqli.commit.php
• > http://ru.php.net/manual/en/mysqli.rollback.php