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我使用它来获取所有图像具有特定通配符名称:如何使用PHP glob()正确命令?
<?php
$images = glob("/var/www/user/html/images/".$row['id']."@*.jpg");
foreach($images as $image) {
echo "<img src=\"".str_replace("/var/www/user/html/images/", "http://www.example.com/images/", $image)."\">\n";
?>
如果有例如四象匹配的glob模式()的输出是:
<img src="http://www.example.com/images/[email protected]">
<img src="http://www.example.com/images/[email protected]">
<img src="http://www.example.com/images/[email protected]">
<img src="http://www.example.com/images/[email protected]">
他们正确排序:[email protected], [email protected], [email protected], [email protected]
。
但是,如果有例如12图像输出是这样的:
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
正如你看到它是不是正确排序:[email protected], [email protected], [email protected], [email protected], [email protected], [email protected], […]
我能做些什么来解决这个问题?有任何想法吗?
'natsort($ images)'' – apokryfos