如何减少多个依赖查询

Question

I was wondering if i can clean this up a bit, i wanted to make a nested sql statement but am unsure how.

$serverfile = mysqli_fetch_array(mysqli_query($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']."'"));

i wanted something like this?

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

Answer 1

This query should do it; since you're only querying one item, you don't really need a join:

SELECT `ExternalDomain`,`AccessFile` 
FROM `FileServers` 
WHERE `ID` IN (SELECT `Server_ID` 
    FROM `FileServerFiles` 
    WHERE `File_ID` ='$fileid' 
    limit 1
);

Btw, I've assumed that $fileid has been properly sanitized; you could also consider prepared statements (and/or).

Answer 2

你可以试试IN，

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

Answer 3

This is not nested, but is an equivalent using the join syntax, which is sometimes considered more concise (and modern, for what that's worth). May be not what you want, but here for your reference (as you're likely to come up against this syntax if you're working with databases).

The INNER JOIN means that rows from the left side of the join are discarded from the result if there's no row matched on the right side of the join.

I've added heaps of white space so it's clear, feel free to strip that out. But no need to.

$server = mysqli_fetch_array(mysqli_query($link,
   "SELECT 
       fs.ExternalDomain,
       fs.AccessFile

    FROM 
       FileServers fs

       INNER JOIN FileServerFiles fsf
       ON fs.fileid = fsf.server_id

     WHERE 
       fsf.File_ID ='$fileid' 

     LIMIT 1
   "));

As mentioned by @Jack before me, you should ensure $fileid is sanitized. Refer http://en.wikipedia.org/wiki/SQL_injection

Answer 4

Syntax may be just abit off, but a left join should do the trick. Shorter is in the eye of the beholder, but one query is better than 2.

$serverfile = mysqli_fetch_array(mysqli_query($link,"
    SELECT f.`Server_ID`, s.`ExternalDomain`, s.`AccessFile`
    FROM `FileServerFiles` f 
    LEFT JOIN `FileServers` AS s ON s.`ID` = f.`Server_ID` 
    WHERE f.`File_ID` ='$fileid'
    LIMIT 1
  "));