<?php
foreach ($recorded as $rows )
{
echo "<pre>";
print_r ($rows);
echo "</pre>";
echo "<img src='".base_url()."upload/thumbs/".$rows->file_name."'/>";
}
?>
当我打印$rows
时,结果是:
Array
(
[0] => stdClass Object
(
[file_name] => steve-big.jpg
)
[1] => stdClass Object
(
[file_name] => steve1-big.jpg
)
[2] => stdClass Object
(
[file_name] => steve2-big.jpg
)
[3] => stdClass Object
(
[file_name] => steve3-big.jpg
)
)
Array
(
[0] => stdClass Object
(
[file_name] => size-5-5-width-m.jpg
)
[1] => stdClass Object
(
[file_name] => 3316783.jpg
)
[2] => stdClass Object
(
[file_name] => 3316786.jpg
)
[3] => stdClass Object
(
[file_name] => 3316792.jpg
)
)
但是为什么我不能回应它呢?
我正在使用代码点火器,这是我收到的错误消息:
遇到 PHP 错误 |严重性:通知 |消息:尝试获取非对象的属性 |文件名: 视图/母店.php |行号:17
但是只有给我一个特定的数字,我才能回声它:$rows[0]->file_name
、$rows[1]
等等。
拜托,帮帮忙,这真的让我感到困惑。
这应该适合您:
foreach ($recorded as $rows)
array_walk($rows, function($v){
echo "<img src='" . base_url() . "upload/thumbs/" . $v->file_name . "'/>";
});
正如我们所看到的$rows
再次包含数组。所以你不能不echo
它。相反,您必须使用 print_r
,echo
您必须迭代$row
数组。尝试下面的代码。
<?php
foreach ($recorded as $rows )
{
// echo "<pre>";
// print_r ($rows);
// echo "</pre>";
foreach ($rows as $r)
{
echo "<img src='".base_url()."upload/thumbs/".$r->file_name."'/>";
}
}
?>
这可能会对您有所帮助。
试试
<?php
foreach ($recorded as $rows=>$val )
{
echo $val['file_name'].'<br/>';
}
?>
检查错误:
A PHP Error was encountered | Severity: Notice | Message: Trying to get property of non-object | Filename: views/mothershop.php | Line Number: 17
因为$rows
是一个数组,没有名为 file_name
的属性。
试试这个代码
<?php
foreach ($recorded as $rows )
{
echo "<img src='".base_url()."upload/thumbs".$rows['file_name']."'/>";
}
?>
尝试在行上再次重申由于结果的数组结构是二维数组
foreach ($recorded as $rows ) {
foreach ($rows as $key => $value) {
echo "<img src='".base_url()."upload/thumbs/".$value->file_name."'/>";
}
}
试试这个:
$baseUrl = base_url();
foreach ($recorded as &$rows) {
foreach ($rows as &$row) {
echo '<img src="'.$baseUrl.'upload/thumbs/'.$row->file_name.'"/>';
}
}