警告:mysqli_query():无法获取 mysqli

Ric*_*378 4 php mysql mysqli

我是 PHP 的新手,不熟悉它的许多规则,所以这可能是一个愚蠢的问题。

我有一个将顶级类别和子类别合并到一个表中的数据库。我想首先打印出所有顶级类别,然后打印出与该类别相关联的子类别。

这是我的代码:

<?php
session_start();

include_once "/mysqli_connect.php";

$categories0 = mysqli_query($conn, "SELECT * FROM categories WHERE type = 'category'");


mysqli_close($conn);
?>

<html>
<head>
<meta charset="UTF-8" />
</head>
<body>

<div id="wrap" class="animate">

<?php 

while ($categories = mysqli_fetch_array($categories0, MYSQLI_ASSOC)) {

    $catecory_name = $categories['category'];
    echo '
    <div class="content">
    <div class="content_container no_padding">
    <div class="content_container header">
    <p>'.$categories['category'].'</p>
    </div>
    ';

    $subcategories0 = mysqli_query($conn, "SELECT * FROM categories WHERE type = 'subcategory'");

    while ($subcategories = mysqli_fetch_array($subcategories0, MYSQLI_ASSOC)) {
        echo $subcategories['category'];
    //mysqli_free_result($subcategories0);
    }

    echo '
    </div>
    </div>
    ';

}

 ?>

</div>
</div>


</body>
</html>
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这是连接脚本:

<?php

DEFINE ('DB_USER', '*');
DEFINE ('DB_PASSWD', '*');
DEFINE ('DB_HOST', '*');
DEFINE ('DB_NAME', '*');

$conn = mysqli_connect(DB_HOST, DB_USER, DB_PASSWD, DB_NAME);

if(!$conn){
    die('Database connection error');
}

echo '<!-- Connected to database -->'

?>
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它返回以下错误:

Warning: mysqli_query(): Couldn't fetch mysqli
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当两个查询都在 doctype 之上时,一切都很好,但是当第二个查询低于 doctype 时,就会发生错误。

第一个查询总是运行没有问题,它是返回错误的第二个。

我似乎无法弄清楚发生了什么,如果有人可以帮助我,将不胜感激。

Sat*_*aty 7

您忘记关闭 while 循环。在需要关闭它的地方检查注释。

<?php
$categories0 = mysqli_query($conn, "SELECT * FROM categories WHERE type = 'category'");
?>
<!DOCTYPE html>

<?php
while ($categories = mysqli_fetch_array($categories0, MYSQLI_ASSOC)) {// need to close your loop

$catecory_name = $categories['category'];
echo '
<div class="content">
<div class="content_container header">
<p>'.$categories['category'].'</p>
</div>
';
}// close here
$subcategories0 = mysqli_query($conn, "SELECT * FROM categories WHERE type = 'subcategory'");
// The above line is where the error occurs 

while ($subcategories = mysqli_fetch_array($subcategories0, MYSQLI_ASSOC)) {
    echo $subcategories['category'];

}


?>
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更新

从顶部删除关闭连接,因为在它之后您的查询将不会执行。连接关闭后,您的连接变量将消失。

<?php
session_start();

include_once "/mysqli_connect.php";

$categories0 = mysqli_query($conn, "SELECT * FROM categories WHERE type = 'category'");


?>
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