Question

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']."%'

");

Answer 1

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。

SELECT * FROM search_locations 
WHERE 'bs3 5qu' LIKE CONCAT(location_postcode,'%')

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！

Answer 2

大家。目标是拥有一个搜索框，用户可以搜索邮政编码或位置。

数据库被列出了：

location_id   location_name   location_postcode
1             bristol         bs
2             glasgow         g

有些邮政编码仅从一个字母开始，有些是从上面的两个字母开始的，例如上面的数据库示例。现在，下面的代码我将$ location_search变量更改为第一个字母，如果它是一个带有一个字母的邮政编码，则将其更改为第一个字母，如果它是带有两个字母的邮政编码，则将其更改为两个字母。现在，如果它的位置（例如布里斯托尔）。它会忽略邮政编码并将整个单词分配给变量$location_search.

现在我刚刚完成了此代码。可能会更好，但奏效。分配给变量后，您可以使用它来运行mysql查询。

<?php //Assign location ID 
    $location_word = mysql_real_escape_string($_POST["location_input"])

    if(!empty($location_word)){

        //Remove spaces
        $location_word = str_replace(" ", "", $location_word);

        //Work out if it's a postcode or a location.
        $location_3rd_character = substr($location_word, 2, 1);     //Get third character
        if(!ctype_alpha($location_3rd_character)){      //Check if it's a number or not
            $location_type = "postcode";
        }
        else {
            $location_type = "city";
        }

        if($location_type == "postcode"){       //Definding the postcode search, strip down to first or second letter.
            $location_2nd_character = substr($location_word, 1, 1);

            if(!ctype_alpha($location_2nd_character)){   //Trim postcode down to one letter, if only one letter in postcode.
                $location_search = substr($location_word, 0, 1);
            }
            else {      //Trim postcode down to two letters, if only two letter in postcode.
                $location_search = substr($location_word, 0, 2);
            }
        }
        else {
            $location_search = $location_word;  
        }
?>

Answer 3

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);