多语句查询中的PHP PDO错误

Question

我在我的一个实时网络应用程序中遇到了这个问题.看来如果你通过PHP PDO向MySQL发出一个多语句查询,并且第一个语句是一个insert语句,而第二个语句是一个update语句,那么PDO :: nextRowset()函数不会返回正确的数字结果集 (请注意,自PHP 5.3起,PDO应该支持每个MySQL查询的多个语句.)

这是一个例子:

SQL:

create database `test`character set utf8 collate utf8_general_ci;
create table `test`.`testtable`( `id` int ); 

PHP:

<?php
$link = new \PDO('mysql:host=localhost;dbname=test', 'username', 'password');

//Run one of the 4 $handle assignments at a time (comment out all but one). 
//Run #4 on an empty table to compare the results of #1 and #4.

//WORKS: INSERT, followed by SELECT, followed UPDATE
//Output: 
//Rowset 1
//Rowset 2
//Results detected
$handle = $link->prepare(' insert into testtable(id) values(1);
                           select * from testtable where id = ?;
                           update testtable set id = 2 where id = ?;');


//WORKS: SELECT, followed by UPDATE
//Output: 
//Rowset 1
//Results detected
$handle = $link->prepare('select * from testtable where id = ?; 
                          update testtable set id = 2 where id = ?;');

//WORKS: UPDATE, followed by SELECT
//Output: 
//Rowset 1
//Rowset 2
//Results detected
$handle = $link->prepare('select * from testtable where id = ?; 
                         update testtable set id = 2 where id = ?;');


//DOESN'T WORK: INSERT, followed by UPDATE, followed by SELECT
//Output: 
//Rowset 1
//Expected output: same as examples 1 and 3
$handle = $link->prepare('insert into testtable(id) values(1);
                          update testtable set id = 2 where id = ?;
                          select * from testtable where id = ?;');

$handle->bindValue('1', '1');
$handle->bindValue('2', '2');

$handle->execute();

$i = 1;
do{
    print('Rowset ' . $i++ . "\n");
    if($handle->columnCount() > 0)
     print("Results detected\n");
}while($handle->nextRowset());
?>

有没有人知道我做错了什么？为什么我不能把我的选择声明放在最后？

PHP 5.3.5

MySQL 5.1.54

Answer 1

我提交了有关此问题的PHP 错误报告，并且已提交了可能的补丁。所以看起来这是一个 PHP 错误。