删除之前的内容

Question

问题内容已被删除。

Answer 1

mysql_query("

SELECT
    location_postcode,
    CASE SUBSTRING('".$_POST['user_postal_search']."', 1, 2)
        WHEN 'bs' THEN 'Bristol'
        WHEN 'ba' THEN 'Bath'
        ELSE 'Unknown City'
    END CASE as City
FROM
    search_locations
WHERE
    location_postcode LIKE '".$_POST['user_postal_search']."%'

");

也是，如果您不希望SQL Server处理子字符串，那么只需让PHP做到这一点：

mysql_query("

SELECT
    location_postcode,
    CASE '".substr($_POST['user_postal_search'], 0, 2)."'
        WHEN 'bs' THEN 'Bristol'
        WHEN 'ba' THEN 'Bath'
        ELSE 'Unknown City'
    END CASE as City
FROM
    search_locations
WHERE
    location_postcode LIKE '".$_POST['user_postal_search']."%'

");

如果我理解您的问题，我不会解决，但我认为此查询可能会对您有所帮助。

Answer 2

它用作“对面的通配符”，将通配符与位置_postCode相连而不是用户的输入。 其他选项，如果我们假设邮政编码长2个字符，则是使用MySQL左功能来获得一个子字符串，如下所示：

SELECT * FROM search_locations
WHERE location_postcode LIKE LEFT('bs3 5qu', 2)

希望它有帮助！

这个问题更多地在您的Web服务器代码上，而不是DB。如果您使用PHP检索用户的输入，则是您如何做的示例：

<?php

// Retrieve the user's input.
$postCode = $_POST["postCode"];

// Sanitize, making sure it may only contain alpha-numerical characters for security and make lower-case.
$postCode = strtolower(preg_replace("/[^a-z]/i", "", $postCode));

// Retrieve the first two characters from the user's input.
$postCode = substr($postCode, 0, 2);

// Now query the database like you did before.
$location_query = mysql_query("SELECT * FROM search_locations WHERE location_postcode LIKE '" . $postCode . "%'",$db_connect);

Answer 3