使用PHP從下拉列表中對SQL查詢進行排序

Question

我有一個使用JOIN連接在一起的表，我想創建一個帶有提交按鈕的表單，允許最終用戶更改他們希望表格被過濾的方式。 排序欄位於我的網頁上，但實際上並沒有做任何事情。 所以我的問題是我需要做什么才能讓排序按鈕工作。

<?php


 $sort = @$_POST['order']; 
if (($sort)) { // If you Sort it with value of your select options
 $query = "SELECT n.firstname, n.middlename, n.lastname, m.title, f.format, t.price, e.series, t.inventory 
                FROM book_main AS m
                JOIN book_author AS n ON n.author_id=m.author_id
                JOIN book_title AS t ON t.title_id=m.title_id
                JOIN book_format AS f ON f.format_id=m.format_id 
                LEFT JOIN book_series AS e ON e.series_id=m.series_id 
                ORDER BY '".$sort."' ASC";



} 
else { // else if you do not pass any value from select option will return this
    $query  =  "SELECT n.firstname, n.middlename, n.lastname, m.title, f.format, t.price, e.series, t.inventory 
                FROM book_main AS m
                JOIN book_author AS n ON n.author_id=m.author_id
                JOIN book_title AS t ON t.title_id=m.title_id
                JOIN book_format AS f ON f.format_id=m.format_id 
                LEFT JOIN book_series AS e ON e.series_id=m.series_id
                  ORDER BY n.lastname ASC";


$results = mysql_query($query);
echo $results['order'];

//can I have a couple points for trying so hard haha...
}

?>

<form name="sort" action="" method="post">
<select name="order">
   <option value="choose">Make A Selection</option>
   <option value='lastname'>Authors Last name</option>
   <option value='price'>Price</option>
   <option value='inventory'>Inventory</option>
</select>
<input type="submit" value="Sort" />
</form>

Answer 1

要確保$ sort具有select選項中列出的值，您需要進行如下檢查。 沒有回應的原因是，order不是$ results數組的一部分。

訪問結果的唯一方法是使用以下任何一個http://php.net/manual/en/function.mysql-result.php

if(in_array($sort, ['lastname', 'price', 'inventory'])) {
    $query = "SELECT n.firstname, n.middlename, n.lastname, m.title, f.format, t.price, e.series, t.inventory 
                FROM book_main AS m
                JOIN book_author AS n ON n.author_id=m.author_id
                JOIN book_title AS t ON t.title_id=m.title_id
                JOIN book_format AS f ON f.format_id=m.format_id 
                LEFT JOIN book_series AS e ON e.series_id=m.series_id 
                ORDER BY $sort ASC";
} else {
    $query  =  "SELECT n.firstname, n.middlename, n.lastname, m.title, f.format, t.price, e.series, t.inventory 
                FROM book_main AS m
                JOIN book_author AS n ON n.author_id=m.author_id
                JOIN book_title AS t ON t.title_id=m.title_id
                JOIN book_format AS f ON f.format_id=m.format_id 
                LEFT JOIN book_series AS e ON e.series_id=m.series_id
                  ORDER BY n.lastname ASC";
}
$results = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
    echo $row['firstname'];
    echo $row['middlename'];
    echo $row['lastname'];
    echo $row['title'];
    echo $row['format'];
}