mysqli多个查询-设置变量产生布尔错误/如何跳过呢?

问题描述

得到以下简单查询，该查询通过phpmyadmin可以正常工作，但是当我将其添加到php网站时，没有返回结果，也没有错误/警告消息.如果我删除"SET @ N = -1;"然后就可以了.

Got the following simple query which works fine through phpmyadmin but when I add it to my php website no results are returned and no error/warning messages either. If I remove "SET @N=-1;" then it works fine.

<?php 
$db_connect = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD, true);
mysql_select_db(DB_NAME, $db_connect);

$test_query = mysql_query("SET @N=-1;SELECT `id`, (@N:=@N+1) AS `mycount` FROM `mydb`;");

for ($i = 0; $i <= mysql_num_rows($test_query)-1; $i++) {
   echo mysql_result($db_directorymap, $i, 0) . " " . mysql_result($db_directorymap, $i, 1) . "<br />";
}
?>

更新:我刚搬到mysqli，但是当然我仍然对mysql语句和mysqli_multi_query有问题.似乎当它运行查询的第一部分时，返回的结果为空，因此给出了布尔错误.我猜我必须跳过第一组结果，但是我不知道该怎么做?

UPDATE: I just moved to mysqli but of course I'm still having a problem with the mysql statement and mysqli_multi_query. It seems when it runs the first part of the query the results returned are empty thus a boolean error is given. I'm guessing I have to skip the first set of results but I don't know how to do that?

推荐答案

这是因为mysql_query函数将仅接受一个查询，但您却给了两个查询，并用分号隔开.尝试之一:

It's because the mysql_query function will only accept one query, but you've given it two, separated by a semicolon. Try either:

1. 分别运行每个查询(不知道是否可行):

1. Running each query separately (don't know if this will work):

mysql_query( "SET @N=-1" );
mysql_query( "SELECT `id`, (@N:=@N+1) AS `mycount` FROM `mydb`" );

将 mysqli 与 multi_query 函数(或等同于PDO (如果有的话).

Using mysqli with the multi_query function (or a PDO equivalent if there is one).

要回答更新的问题:请检查PHP手册页中的multi_query.我认为您将要使用mysqli::next_result.使用程序样式，类似这样:

To answer your updated question: check the PHP manual page for multi_query. I think you'll want to use mysqli::next_result. Something like this, using procedural style:

mysqli_multi_query($link, $query);
mysqli_next_result($link);

if ($result = mysqli_store_result($link)) {
    while ($row = mysqli_fetch_row($result)) {
        printf("%s\n", $row[0]);
    }
    mysqli_free_result($result);
}

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