0
我想发送$ c2 = @'http://marvelconcepts.net/fb3/02.php?c ='。$ c。' & &'.'c1 ='。$ c1;在02.php当我得到$ C ON 02.php它没有显示完整的URL我想通过网址在另一页上发送网址
<?php
$c='Flower';
$c1='http://aux.iconpedia.net/uploads/1337412470.png';
[email protected]'http://marvelconcepts.net/fb3/02.php?c='.$c.'&&'.'c1='.$c1;
?>
<a href='02.php?c=<?php echo $c; ?>&c1=<?php echo $c1; ?>&c2="<?php echo $c2; ?>"'>share</a>
对不起,你到底想干什么? – Bojangles 2012-08-10 10:05:37
您需要对字符串进行urlencode。 – 2012-08-10 10:05:44
你为什么在那里有'@'? – PeeHaa 2012-08-10 10:10:01