php – 如何减少多个依赖查询

发布时间：2020-12-13 22:48:23 所属栏目：PHP教程来源：网络整理

导读：我想知道我是否可以清理一下这个,我想做一个嵌套的sql语句,但我不确定如何. $serverfile = mysqli_fetch_array(mysqli_query($link,"SELECT `Server_ID` FROM `FileServerFiles` WHERE `File_ID` ='$fileid' limit 1"));$server = mysqli_fetch_array(mysqli

我想知道我是否可以清理一下这个,我想做一个嵌套的sql语句,但我不确定如何.

$serverfile = mysqli_fetch_array(mysqli_query($link,"SELECT `Server_ID` 
    FROM `FileServerFiles` 
    WHERE `File_ID` ='$fileid' 
    limit 1"));
$server = mysqli_fetch_array(mysqli_query($link,"SELECT `ExternalDomain`,`AccessFile` 
    FROM `FileServers` 
    WHERE `ID` ='".$serverfile['Server_ID']."'"));

我想要这样的东西？

$server = mysqli_fetch_array(mysqli_query($link,`AccessFile` 
        FROM `FileServers` 
        WHERE `ID` ='(SELECT `Server_ID` 
            FROM `FileServerFiles` 
            WHERE `File_ID` ='$fileid' 
            limit 1)'"
    ));

解决方法

你可以试试IN,

$server = mysqli_fetch_array(mysqli_query($link,`AccessFile` FROM `FileServers` WHERE `ID` IN (SELECT `Server_ID` AS ID FROM `FileServerFiles` WHERE `File_ID` ='$fileid' limit 1)"));

（编辑：李大同）

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