Question

我有一个mysql表，其中用户有一个有优势游戏的列。他们像这样保存：

id     username     games
1      user1        ,game1, game2, game3
2      user2        ,game2, game4
3      user3        ,game4, game1, game3

现在我想选择至少有一个游戏和我一样的所有用户。

$mygames = ,game2, game5

$statement = $pdo->prepare("SELECT * FROM users WHERE games = ?????");
$statement->execute();
$users = $statement->fetchAll();

它应该给我user1和user2，因为它们都包含game2和我！我怎么能这样做？

感谢您的帮助：）

Answer 1

尝试使用Like语句，例如：

for

Answer 2

我找到了解决方案

$mygames = ,game2, game5
$ug = explode(",", $mygames); //each word is splitted at the ,
$ug = array_filter($ug); //every empty array is deleted

$statement = $pdo->prepare("SELECT * FROM users");
$statement->execute();
$users = $statement->fetchAll();

foreach ($users as $row) { 

$usg = explode(",", $row['games']); //each word is splitted at the ,
$usg = array_filter($usg);          //every empty array is deleted
$gsame= count(array_intersect($usg, $ug)) > 0; //if a user has at least one game as me it counts(onegame=1, twogames=2 etc...)

if($gsame > 0) { //if gsam is more than 0 it displays the array
<html></html>
}

希望这有助于你们中的一些人：）

Answer 3

你可以使用正则表达式吗？

select * FROM users where games REGEXP 'game4|game2';如果您想要包含game4或game2的所有行。

所以你的php需要构建正则表达式。

<?php

$games = ['game2', 'game4']; // an array of the required games

$regexp = implode('|', $games); // converts array to string with pipe as delimiter

echo ($regexp); // outputs 'game2|game4'

所以你的mysql现在变成了

$sql = 'select * FROM users where games REGEXP '. $regexp .';

Answer 4

Matt说，使用LIKE语句

更好的＆amp;可扩展的方式：

$mygames = ",game2, game5"; // or whatever

// split & trim the gamenames
$games = array_map('trim',split(trim($mygames, ','))); 

// for ANY number of games you have, set the condition that selects anyone who have ANY of them.
$statement = $pdo->prepare("SELECT * FROM users WHERE " . substr(str_repeat(" games LIKE ? OR ", count($games)), 0, -3));
$statement->execute();
$users = $statement->fetchAll();

如果pdo包含字符串

4 个答案: