使用PHP對下拉表中的SQL查詢進行排序

Question

我有一個使用JOIN連接在一起的表格，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過濾表格的方式。 排序欄在我的網頁上，但它實際上沒有做任何事情。所以我的問題是我需要做什麼才能讓排序按鈕起作用。

<?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$那種在你需要如下檢查的選擇選項列出的值。同樣沒有任何迴應的原因是，order不是$ results數組的一部分。

訪問結果的唯一方法是使用其中任何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']; 
}