PHP Arrays - 试图获取非对象的属性
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PHP Arrays - Trying to get property of non-object
提问by Aero Chocolate
I am still new to PHP so please bear with me.
我还是 PHP 新手，所以请耐心等待。
So I am getting this error: Notice: Trying to get property of non-object on this line:
所以我收到这个错误：注意：试图在这一行获取非对象的属性：
echo (
"<tr>".
"<td>".$row->last_name. "</td>".
"<td>".$row->first_name. "</td>".
"<td>".$row->phone_no. "</td>".
"<td>".$row->date_of_birth. "</td>".
"<td>".$row->membership. "</td>".
"</tr></table>");
I've used print_r on my function and I get:
我在我的函数上使用了 print_r，我得到：
Array
(
[0] => Array
(
[0] => Lee
[last_name] => Lee
[1] => Lian
[first_name] => Lian
[2] => 39025823
[phone_no] => 39025823
[3] => 1967-09-19
[date_of_birth] => 1967-09-19
[4] => T
[membership] => T
[5] =>
[status] =>
[6] => 0
[room_no] => 0
)
)
So there are elements within the array.
所以数组中有元素。
The actual code falls under:
实际代码属于：
foreach($array as $row)
{
echo (
"<tr>".
"<td>".$row->last_name. "</td>".
"<td>".$row->first_name. "</td>".
"<td>".$row->phone_no. "</td>".
"<td>".$row->date_of_birth. "</td>".
"<td>".$row->membership. "</td>".
"</tr></table>");
}
I was thinking - how would I convert an array into an object? Maybe that would be my solution.
我在想 - 如何将数组转换为对象？也许这就是我的解决方案。
回答by Phil
I was thinking - how would I convert an array into an object? Maybe that would be my solution.
我在想 - 如何将数组转换为对象？也许这就是我的解决方案。
That indeed would be one solution.
这确实是一种解决方案。
$row = (object) $row;
Another would be to use the right syntax for the data-type in question, in this case an array.
另一种方法是对所讨论的数据类型使用正确的语法，在这种情况下是数组。
Instead of
代替
$row->last_name
You should use
你应该使用
$row['last_name']
回答by Pascal MARTIN
As you are working with an array, you should use []to access the array's items :
当您使用数组时，您应该使用[]访问数组的项：
echo $row['last_name'];
Use the right syntax, and the error will go away ;-)
使用正确的语法，错误就会消失;-)
Still, if you really want to convert an array to an object (not really sure why you'd do that, though, in this specific case), you can use this :
尽管如此，如果您真的想将数组转换为对象（但不确定为什么要这样做，但在这种特定情况下），您可以使用以下命令：
$row = (object)$row;
echo $row->last_name;
Here's the relevant section of the manual : Type Casting
这是手册的相关部分：Type Casting
回答by Anthony Hyman
Try this...
尝试这个...
foreach($array as $row)
{
echo (
"<tr>".
"<td>".$row['last_name']. "</td>".
"<td>".$row['first_name']. "</td>".
"<td>".$row['phone_no']. "</td>".
"<td>".$row['date_of_birth']. "</td>".
"<td>".$row['membership']. "</td>".
"</tr></table>");
}
