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我使用它來獲取所有圖像具有特定通配符名稱:如何使用PHP glob()正確命令?
<?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/", $image)."\">\n";
?>
如果有例如四象匹配的glob模式()的輸出是:
<img src="http://www.example.com/images/[email protected]">
<img src="http://www.example.com/images/[email protected]">
<img src="http://www.example.com/images/[email protected]">
<img src="http://www.example.com/images/[email protected]">
他們正確排序:[email protected], [email protected], [email protected], [email protected]
。
但是,如果有例如12圖像輸出是這樣的:
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
<img src="http://www.exapmple.com/images/[email protected]">
正如你看到它是不是正確排序:[email protected], [email protected], [email protected], [email protected], [email protected], [email protected], […]
我能做些什麼來解決這個問題?有任何想法嗎?
'natsort($ images)'' – apokryfos