我需要使用一組標簽中的LIKE，使用MySQL表中的$ _GET下拉菜單獲得幫助

Question

所以我有一個附有多個類別的項目表。 我一直遇到問題。 這是我的代碼片段：

    <form class="forms" action="form.php" method="get">
    <fieldset>
    <legend>Sort by Category</legend>
    <select name="tags">
    <option value="business" title="Business">Business</option>
    <option value="environmental" title="Environmental">Environmental</option>
    <option value="hospitality" title="Hospitality">Hospitality</option>
    </select><input type="submit" class="inputButton" value="Select Category" />
    </fieldset>
    </form>

    <?php
    $tags = $_GET['tags'];


    $connection = mysql_connect($local,$user,$pass) or die (mysql_errno().' : '.mysql_error().'<br />');
    mysql_select_db($db);
    $query = 'SELECT * FROM awards WHERE active=1 ORDER BY name';
    $result = mysql_query($query);
    $num = mysql_num_rows($result);

    while ($row = mysql_fetch_array($result)) 
    {
        if ($tags == 'business')
        {
            $where = "SELECT * FROM awards WHERE active=1 AND
tags LIKE '%business%' ORDER BY name";
}
    else if ($tags == 'environmental')
    {
        $where = "SELECT * FROM awards WHERE active=1 AND 
tags LIKE '%environmental%' ORDER BY name";
}
    else if ($tags == 'hospitality')
    {
        $where = "SELECT * FROM awards WHERE active=1 AND 
tags LIKE '%hospitality%' ORDER BY name";
    }
        else
        {
            $where = 'SELECT * FROM awards WHERE active=1 ORDER BY name';
            $title = 'Awards in Alphabetical Order';
        }
    }
    include 'conn.php';
    $connection = mysql_connect($local,$user,$pass) or die (mysql_errno().' : '.mysql_error().'<br />');
    mysql_select_db($db);
    $query = 'SELECT * FROM awards $where';
    $result = mysql_query($query);
    $num = mysql_num_rows($result);
?>

然后，回聲等，我沒有問題。 我似乎無法讓它僅提取我想要的記錄。 我會以錯誤的方式處理嗎？ 我表中的行“標簽”具有多個逗號分隔的值。 因此，如果有人選擇“業務”，則不應拉出任何沒有“業務”作為標記之一的記錄。 我應該將我的“標簽”作為單獨的行嗎？

Answer 1

<?php
    $tags = $_GET['tags'];

if ($tags == 'business' || $tags == 'environmental' || $tags == 'hospitality')
{
            $where = "SELECT * FROM awards WHERE active=1 AND
tags LIKE '%$tags%' ORDER BY name";
}
else
{
     $where = 'SELECT * FROM awards WHERE active=1 ORDER BY name';
            $title = 'Awards in Alphabetical Order';
}

include 'conn.php';
$connection = mysql_connect($local,$user,$pass) or die (mysql_errno().' : '.mysql_error().'<br />');
mysql_select_db($db);
$result = mysql_query($where);
$num = mysql_num_rows($result);
while ($row = mysql_fetch_array($result)){
    //echo here whatever u want from the rows fetched with matching tags
}

並且您應該開始使用PHP PDO進行數據庫訪問，因為此處使用的方法即將貶值。

PDO手動鏈接