php - 从不同的数据库表获取搜索结果并显示它们？

Question

我正在尝试创建一个搜索功能，该功能将检索我们在网站上创建的不同海报的搜索结果。如果一个人正在搜索“狗”，那么它会显示与狗相关的海报。该网站将以海报的形式发布不同的事件。

目前代码如下:

<?php

class Search
{
    public static $con;

    private $search = '';

    function __construct()
    { 
        self::$con = mysqli_connect('localhost', 'guest', 'guestpw', 'db_users');

        $this->search = mysqli_real_escape_string(self::$con, $_POST['search']);
    }

    if(isset($_POST['submit_search']))
    {

    $sql = mysqli_query($con, "SELECT * FROM Event WHERE eventNamn LIKE '%" . $search);

    $sql = mysqli_query($con, "SELECT * FROM Användare WHERE userName LIKE '%" . $search);

    $sql = mysqli_query($con, "SELECT * FROM Poster WHERE Kategori LIKE '%" . $search);

    $sql = mysqli_query($con, "SELECT * FROM EventTyp WHERE EventTyp LIKE '%" . $search);

    $result = mysqli_query($sql);

    }
}

我现在想要发生的是使用用户正在搜索的搜索词，然后显示与该词关联的事件。非常感谢所有帮助!谢谢。

Answer 1

您可能需要使用UNION 或UNION ALL 运算符。 SQL UNION 运算符组合两个或多个 SELECT 语句的结果。

 SELECT col FROM Event WHERE ...
 UNION ALL
 SELECT col FROM User WHERE ...

文档在这里:
MYSQL UNION 运算符:http://dev.mysql.com/doc/refman/5.0/en/union.html

<小时/>

你的代码可以像这样:

$sql  = "SELECT [your column] AS event FROM Event WHERE eventNamn LIKE '%" . $search . "'".
    "UNION ALL ".
    "SELECT [your column] AS event FROM Användare WHERE userName LIKE '%" . $search . "'".
    "UNION ALL".
    "SELECT [your column] AS event FROM Poster WHERE Kategori LIKE '%" . $search . "'".
    "UNION ALL".
    "SELECT [your column] AS event FROM EventTyp WHERE EventTyp LIKE'%" . $search . "'";

$result = mysqli_query($con,$sql);

while($row = mysqli_fetch_array($result)) {
  echo $row['event '];
  echo "<br>";
}
mysqli_close($con);

希望这有帮助。