php 注意：尝试在 C:\xampp\htdocs\ 中获取非对象的属性
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Notice: Trying to get property of non-object in C:\xampp\htdocs\
提问by user2279037
While executing this code it gives an error message which i don't have any idea about. I did goooooogle, but I didn't find any solution.
在执行此代码时，它给出了一条我不知道的错误消息。我做了goooooogle，但我没有找到任何解决方案。
**Notice: Trying to get property of non-object in C:\xampp\htdocs**
**注意：尝试在 C:\xampp\htdocs 中获取非对象的属性**
<?php
$_REQUEST['PropertyID'];
$_REQUEST['PropertyImageID'];
$propertyImageID=$_GET['PropertyImageID'];
$PropertyID=$_GET['PropertyID'];
require_once('db.php');
$propertyquery = "SELECT PropertyImageID, PropertyName, PropertyStatus, PropertyLocation, PropertyRegion FROM properties WHERE PropertyImageID =$propertyImageID";
$propertyquery_run = $connection->query($propertyquery);
if ($propertyquery_run-> num_rows > 0)
{
while ($propertyrow = $propertyquery_run-> fetch_assoc())
{
$imagequery = "SELECT PropertyImageID, ImagePath FROM propertyimages WHERE PropertyImageID = $PropertyID";
$imagequery_run = $connection->query($imagequery);
?>
<table style="margin-left:348px;">
<tr><td>PropertyName:</td><td><span style="color:#0F0; font-weight:bold;"><?php echo htmlentities($propertyrow['PropertyName']) ?></span></td></tr>
<tr><td>PropertyStatus:</td><td><span style="color:#0F0; font-weight:bold;"><?php echo htmlentities($propertyrow['PropertyStatus']) ?></span></td></tr>
<tr><td>PropertyLocation:</td><td><span style="color:#0F0; font-weight:bold;"><?php echo htmlentities($propertyrow['PropertyLocation']) ?></span></td></tr>
<tr><td>PropertyRegion:</td><td><span style="color:#0F0; font-weight:bold;"><?php echo htmlentities($propertyrow['PropertyRegion']) ?></span></td></tr>
</table>
<center>
<?php
if($imagequery_run -> num_rows > 0)
{
while ($imagerow = $imagequery_run-> fetch_assoc())
{
?>
<div style="border:solid 2px #9F0; border-radius:5px; height:222px; width:544px;">
<img src="<?php echo htmlentities($imagerow['ImagePath']) ?>" width="544" height="222" >
</div><br />
<?php
}
}
}
}
?>
回答by ibi0tux
This notice occurs when you're using a non-object variable as an object.
You're probably using : $a -> b?like syntax on a variable $a?which is not an object.
当您将非对象变量用作对象时，会出现此通知。您可能$a -> b在变量上使用 : ?like 语法$a？它不是一个对象。
Make sure $propertyquery_run, $imagequery_run... are objects which have properties you're using.
确保$propertyquery_run, $imagequery_run... 是具有您正在使用的属性的对象。
Edit :
编辑 ：
If the error is here $imagequery_run -> num_rows > 0. This means $imagequery_runis not an object that have num_rowsproperties. Maybe it's NULL. Check its value before the test.
如果错误在这里$imagequery_run -> num_rows > 0。这意味着$imagequery_run不是具有num_rows属性的对象。也许它是NULL。在测试前检查其值。
$imagequery = "SELECT PropertyImageID, ImagePath FROM propertyimages WHERE PropertyImageID = $PropertyID";
$imagequery_run = $connection->query($imagequery);
if(!$imagequery_run)
{
// do what you want
}
