我尝试从我的数据库中获取一些信息到我的网页。一切似乎都很好,但有一件事不想做对。我把我的数据库中的所有信息都放到$ data中。当我这样做时
print_r($data);
我的网页给了我这个:
(
[0] => stdClass Object
(
[reparatie_id] => 19
[customer_id] => 4
[medewerker] => 4
[name] => Joost
)
)
一切似乎都很好但是当我尝试这样做时:
echo $data->voornaam;
我一直收到此错误
A PHP Error was encountered
Severity: Notice
Message: Trying to get property of non-object
Filename: reparaties/cases.php
Line Number: 7
Backtrace:
File: C:\Ampps\www\beco\application\views\reparaties\cases.php
Line: 7
Function: _error_handler
File: C:\Ampps\www\beco\application\controllers\Reparaties.php
Line: 57
Function: view
File: C:\Ampps\www\beco\public\index.php
Line: 315
Function: require_once
答案 0 :(得分:3)
由于您的$data
是一维数组,因此需要 -
$data[0]->reparatie_id;
$data[0]->customer_id;
$data[0]->medewerker;
$data[0]->name;//so on for other indexes
答案 1 :(得分:2)
Actually the $data array has an object at 0 position. So you need to any property of object. do like this:
<?php
$data[0]->reparatie_id;
$data[0]->customer_id;
$data[0]->medewerker;
$data[0]->name; ?>
output will be:
19, 4, 4, joost
答案 2 :(得分:0)
首先,您的$data
数组中没有voornaam
元素。因此,假设您想要回显数组中的元素,您将使用foreach
数组,如下所示:
foreach($data as $value) {
echo $value->name;
echo $value->voorname; //if it exists
}
但是,如果您只想从数组中访问该单个元素,那么您可以这样做:
echo $data[0]->name;