在PHP中的mysql查询中的内联变量无法正常工作

问题描述：

I have following php code,

<?php
    $subject_set = mysqli_query($db_link, 'SELECT * FROM subjects');

    if(!$subject_set){
        die("Database subjects query failed: ".mysql_error());
    }

    while ($subject=mysqli_fetch_array($subject_set)) {
        echo "<li>{$subject["menu_name"]}</li>";
            $page_set = mysqli_query($db_link, 'SELECT * FROM pages WHERE subject_id= {$subject["id"]}');
            // $page_set = mysqli_query($db_link, 'SELECT * FROM pages WHERE subject_id='.$subject["id"]);

            if(!$page_set){
                die("Database pages query failed: ".mysql_error());
            }

            echo "<ul class=\"pages\">";

            while ($page=mysqli_fetch_array($page_set)) {
                echo "<li>{$page["menu_name"]}</li>";
            }

            echo "</ul>";

        }
    ?>

when I run this code it gives me an error when I getting 'page_set' values.In line $page_set = mysqli_query($db_link, 'SELECT * FROM pages WHERE subject_id= {$subject["id"]}'); But I used commented line instead of the above line code works fine. Also mysql_error() doesn't show me the error.

我有以下php代码， p>

 ＆lt;？php  
 $ subject_set = mysqli_query（$ db_link，'SELECT * FROM subject'）; 
 
 if if（！$ subject_set）{
 die（“数据库主题查询失败：”。mysql_error（））; 
} \  n 
 while（$ subject = mysqli_fetch_array（$ subject_set））{
 echo“＆lt; li＆gt; {$ subject [”menu_name“]}＆lt; / li＆gt;”; 
 $ page_set = mysqli_query（$ db_link，'  SELECT * FROM pages WHERE subject_id = {$ subject [“id”]}'）; 
 // $ page_set = mysqli_query（$ db_link，'SELECT * FROM pages WHERE subject_id ='。$ subject [“id”]）;  
 
 if if（！$ page_set）{
 die（“数据库页面查询失败：”。mysql_error（））; 
} 
 
 echo“＆lt; ul class = \”pages \“＆gt;”  ; 
 
 while（$ page = mysqli_fetch_array（$ page_set））{
 echo“＆lt; li＆gt; {$ page [”menu_name“]}＆lt; / li＆gt;”; 
} 
 
 echo“  ＆lt; / ul＆gt;“; 
 
} 
？＆gt; 
  code>  pre> 
 
  我运行此代码，当我获得'page_set'值时，它给我一个错误。在行 $ page_set = mysqli_query（$ db_link，'SELECT * FROM pages WHERE subject_id = {$ subject [“id”]}'）;   code> 
但是我使用注释行而不是上面的行代码工作正常。   mysql_error（） code>也没有显示错误。 p> 
  div>

答

Change

$page_set = mysqli_query($db_link, 'SELECT * FROM pages WHERE subject_id= {$subject["id"]}');

$page_set = mysqli_query($db_link, "SELECT * FROM pages WHERE subject_id= {$subject["id"]}");

because PHP variables won't be processed unless they are in double-quoted strings.

答

http://docs.php.net/language.types.string says

Note: Unlike the double-quoted and heredoc syntaxes, variables and escape sequences for special characters will not be expanded when they occur in single quoted strings.

答

Because $subject not treated as variable it is used as string in your code. Use

 $page_set = mysqli_query($db_link, "SELECT * FROM pages WHERE subject_id={$subject['id']}");

在PHP中的mysql查询中的内联变量无法正常工作

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