循环嵌套，我得到一个空白页面(php)

Question

这就是我得到的-

    $awards_sql_1 = mysql_query('SELECT * FROM categories WHERE section_id = 1') or die(mysql_error());
    $awards_rows_1 = mysql_num_rows($awards_sql_1);
    $awards_sql_2 = mysql_query('SELECT * FROM categories WHERE section_id = 2') or die(mysql_error());
    $awards_sql_3 = mysql_query('SELECT * FROM categories WHERE section_id = 3') or die(mysql_error());
    $awards_sql_4 = mysql_query('SELECT * FROM categories WHERE section_id = 4') or die(mysql_error());

    $i = 0;
    $records = mysql_num_rows($sections_query);
    while($row_sections = mysql_fetch_array($sections_query)) {
        echo "<h3>" . $row_sections['section_name'] . "</h3>";
        echo "<ul>";
        //while($categories = mysql_fetch_array($awards_sql_1)) {
        for ($i = 0; $i < $awards_rows_1; $i++) {
            echo "<li><strong>$categories['category_name']</strong>";

        }
        echo "</ul>";
    }

出于某种原因，如果我注释掉了嵌套在while()中的for()，页面将正常加载，并且我将看到所有的h3，然而，每当我试图在原始while()中嵌套for()或while()时，页面在重新加载时就会变成空白。

我做错了什么？

Answer 1

你需要用大括号将嵌入在字符串中的数组变量括起来，否则你会得到一个解析错误。

 echo "<li><strong>{$categories['category_name']}</strong>";

http://php.net/manual/en/language.types.string.php#language.types.string.parsing

Answer 2

尝试使用，您将得到以下错误：

<?php
error_reporting(E_ALL);
...