这是我得到的-
$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,如果我注释掉嵌套在 while() 中的 for(),页面将正常加载并且我将看到我的所有 h3,但是,每当我尝试嵌套 for() 或 while () 在原始 while() 中,页面在重新加载时变为空白。
我做错了什么?
最佳答案
您需要在嵌入字符串的数组变量周围加上大括号,否则会出现解析错误。
echo "<li><strong>{$categories['category_name']}</strong>";
http://php.net/manual/en/language.types.string.php#language.types.string.parsing
关于php - 循环嵌套,我得到一个空白页 (php),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/1208688/