php - 给出警告 : mysqli_fetch_array() expects parameter 1 to be mysqli_result, bool 值

Question

什么鬼？ - 这正在成为一场噩梦，似乎一个又一个错误。我愿意支付某人的时间只是为了帮助我实现似乎无法实现的目标，拜托。我想要的只是一个下拉菜单，可以从中选择数据根据其类别创建一个表格并显示它..帮助! :(

<form action="portfolio.php" method="post">
 <select onload="displayProject(this.value);" onchange="displayProject(this.value);">
  <option value='none'>All</option>
  <option value='1'>Fencing</option>
  <option value='2'>Driveway</option>
 </select>
</form>

<?php 

$db = new mysqli('localhost', 'wlarter_user', 'pw', 'wlarter_portfolio');

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$option= $_POST['option'];

$queries = "SELECT * FROM image"; 
if ($option != 'none'){
 $queries = "SELECT * FROM image where category=".$option; 
}

$result=@mysqli_query($db,"$queries");

while($row = mysqli_fetch_array($result))
{
?>

<div class="box-portfolio"> <?php echo $row['Img']; ?> </div>


<?php
}                      
mysqli_close($db);
?>

Answer 1

简单地扭转它，可能就是你想要的:

$queries=$query;

进入

$query = $queries;

完整代码:

<FORM action="portfolio.php" method="post">
    <SELECT onload="displayProject(this.value);" onchange="displayProject(this.value);">
        <OPTION VALUE='none'>All</OPTION>
        <OPTION VALUE='1'>Fencing</OPTION>
        <OPTION VALUE='2'>Driveway</OPTION>
    </SELECT>
</FORM>

<?php

$db = new mysqli('localhost', 'wlarter_user', 'pw', 'wlarter_portfolio');

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$option= $_POST['option'];

$queries = "SELECT * FROM image";
if ($option != 'none'){
    $queries = "SELECT * FROM image where category=".$option;
}

$query=$queries;
$result=mysqli_query($db,"$query");

while($row = mysqli_fetch_array($result))
{
    ?>

    <div class="box-portfolio"> <?php echo $row['Img']; ?> </div>


<?php
}
mysqli_close($db);
?>