我正在尝试包含一个单独的代码片段来从MySQL表中获取数据。我的页面代码：
<?php session_start(); ?>
<html>
<?php include('head.php'); ?>
<body>
<?php include('navigation.php'); ?>
<div id="container">
<?php
$section = "Movies";
print "THIS IS A TEST";
//$_SESSION['sectiontemp'] = $section;
include('section-grabinfo.php');
include('footer.php');
?>
</div>
</body>
</html>
我的部分抓取信息.php页面：
<?php
print "THIS IS A TEST";
$sql = mysqli_query($conn, "SELECT * FROM article JOIN person ON article.author=person.id WHERE section='".$section."' AND status=1 ORDER BY aid DESC LIMIT 1;");
if ($sql == 'false') {
print "SQL doesn't work";
}
elseif ($sql == 'true') {
print "Works all fine";
}
else {
print "Wrong code.";
}
while ($row = mysqli_fetch_assoc($sql)) {
$title = $row['title'];
$preview = $row['preview'];
$author = $row['name'];
$person_id = $row['author'];
$id = $row['id'];
$username = $row['username'];
include('featured-article.php');
}
//LEAVE SPACE FOR ADS
?>
<div id="secondaryArticleSection">
<?php
$sql2 = mysqli_query($conn, "SELECT * FROM article JOIN person ON article.author=person.id WHERE aid<(SELECT max(aid) FROM article WHERE section='."$section."' AND status=1) AND section='".$section."' AND status=1 ORDER BY aid DESC;");
while ($row = mysqli_fetch_assoc($sql2)) {
$title = $row['title'];
$preview = $row['preview'];
$author = $row['name'];
$person_id = $row['author'];
$id = $row['id'];
$username = $row['username'];
include('secondary-article.php');
}
?>
</div>
基本上我的问题是部分抓取信息.php和页脚.php没有包含在主页上。请记住，当我在笔记本电脑（不是我当前使用的服务器）上使用它时，这完全有效。谢谢。
如果它真的是一个包含麻烦，你可以尝试以这种方式包含
<?php include(__DIR__.'/head.php'); ?>