Question

在这里有一些人的帮助下，我一直在为我的工作网站建立一个在线目录，今天我把它从开发服务器移到生产服务器上，这样其他几个人就可以开始为我测试一些东西了。我似乎遇到了菜单查询的问题。我收到以下错误

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result,
boolean given in /websites/store/includes/menu.php 
on line 15

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result,
boolean given in /websites/store/includes/menu.php on line 34

Warning: Invalid argument supplied for foreach() 
in /websites/store/includes/menu.php on line 34

我花了一整天时间试图弄清楚这里发生了什么，代码在开发服务器上完美运行但只是在生产服务器上抛出错误

任何人都可以向我解释发生了什么或如何解决它

以下是页面的代码

<?php
// Category Listing From Database
// Open MySql Database Connection
    include ('sqlopen.php');

// SQL Query for Category Listing
$sql = mysqli_query($conn, "    SELECT CategoryName, SubcategoryName, SubcategoryID 
                                FROM Products GROUP BY SubcategoryID ORDER BY CategoryName");



// Create Array from Data

$menu = array();
while ($row = mysqli_fetch_assoc($sql)) {
    // Creates First Level Array Items For Parent IDs 
    if (!in_array($row['CategoryName'], $menu['CategoryName'])) {
            $menu['CategoryName'][] = $row['CategoryName'];
            }
    if (!empty($row['SubcategoryName']))
                // Creates Second Level Array for Child IDs
                $menu['SubcategoryName'][$row['CategoryName']][] = $row['SubcategoryName'];
                // Creates Third Level Array for Category IDs
                $menu['SubcategoryID'][$row['SubcategoryName']][$row['CategoryName']][] = $row['SubcategoryID'];    
}

 //__________________________________________________________________________________________________________
// Category Menu
?>

<ul id="storenav">
<?php

foreach ($menu['CategoryName'] as $cat) { ?>
    <li><a class="sub" tabindex="1"><?php echo $cat ?></a><ul>

    <?php
    foreach ($menu['SubcategoryName'][$cat] as $subcat) {
        foreach($menu['SubcategoryID'][$subcat][$cat] as $id) ?>
            <li><a href='/store/prodlist.php?subcatid=<?php echo $id ?>'><?php echo $subcat ?></a></li>
    <?php } ?>  
    </ul></li>
<?php } ?>  

</ul>
<?php




//__________________________________________________________________________________________________________
// Open MySql Database Connection
    include ('sqlclose.php');


?>

Answer 1

你假设这条线 $sql = mysqli_query()

返回结果集。但是，如果查询失败，则返回FALSE，一个布尔值。

这就是这种情况。

编写代码如下： `

 if($sql = mysqli_query($conn, "your query here")){ 

      $menu = array();
      while($row = ...) {

      }

 }
else {
   echo 'something went wrong'; // and more error handling.
}

＆GT？; `

-

以及查询失败的原因？ part group by在此查询中没什么关系，因为没有aggragation函数（min（），max（），count（）等）即便如此：所有列都应在“group by”部分中提及

请参阅mysqli_error（）让Mysql告诉您查询失败的原因。

Answer 2

首先，您没有正确处理错误。更好的说，你甚至没有检查它们。请尝试下面的代码。它将更好地描述出现了什么问题。

<?php
/**
 * Created by: PhpStorm.
 * Project: Stackoverflow
 * File name: .php.
 * User: Deathstorm
 * Date: 23-6-2017.
 * Time: 10:14.
 * File Description: ...
 */
?>
<?php
// Category Listing From Database
// Open MySql Database Connection
include ('sqlopen.php');

// SQL Query for Category Listing
$sql = mysqli_query($conn, "SELECT CategoryName, SubcategoryName, SubcategoryID
FROM Products GROUP BY SubcategoryID ORDER BY CategoryName");


if ($conn->query($sql) === TRUE){
// Create Array from Data

$menu = array();
while ($row = mysqli_fetch_assoc($sql)) {
// Creates First Level Array Items For Parent IDs
if (!in_array($row['CategoryName'], $menu['CategoryName'])) {
$menu['CategoryName'][] = $row['CategoryName'];
}
if (!empty($row['SubcategoryName']))
// Creates Second Level Array for Child IDs
$menu['SubcategoryName'][$row['CategoryName']][] = $row['SubcategoryName'];
// Creates Third Level Array for Category IDs
$menu['SubcategoryID'][$row['SubcategoryName']][$row['CategoryName']][] = $row['SubcategoryID'];
}
}

else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}


//__________________________________________________________________________________________________________
// Category Menu
?>

<ul id="storenav">
    <?php

    foreach ($menu['CategoryName'] as $cat) { ?>
        <li><a class="sub" tabindex="1"><?php echo $cat ?></a><ul>

                <?php
                foreach ($menu['SubcategoryName'][$cat] as $subcat) {
                    foreach($menu['SubcategoryID'][$subcat][$cat] as $id) ?>
                        <li><a href='/store/prodlist.php?subcatid=<?php echo $id ?>'><?php echo $subcat ?></a></li>
                <?php } ?>
            </ul></li>
    <?php } ?>

</ul>
<?php




//__________________________________________________________________________________________________________
// Open MySql Database Connection
include ('sqlclose.php');


?>

PHP7 - MySQLI在将站点移动到新服务器后出错处理问题

2 个答案: