如何preg_match_all以"http"开头并以(")或(')或空格(制表符

我建议您使用parse_url来获取部分url !看看php.net

EDIT:

$file = file_get_contents( YOUR FILE NAME );
$lines = explode("'r'n", $file);
foreach( $lines as $line ){
$urlParts = parse_url( $line );
if( $urlParts['scheme'] == 'http' ){
 // Do anything ...
}
}

CHANGE:

看，我不知道你的代码是什么!如果你想抓取HTML来找到链接，我建议你这样做，它会返回标签的href值给你:

preg_match_all ( "/<[ ]{0,}a[ 'n'r][^<>]{0,}(?<= |'n|'r)(?:href)[ 'n'r]{0,}=[ 'n'r]{0,}['"|']{0,1}([^'"'>< ]{0,})[^<>]{0,}>((?:(?!<[ 'n'r]*'/a[ 'n'r]*>).)*)<[ 'n'r]*'/a[ 'n'r]*>/ is", $source, $regs );
for ( $x = 0; $x < count ( $regs [ 1 ] ); $x ++ ) {
$tmp_array [ "link_raw" ] = trim ( $regs [ 1 ] [ $x ] );
}

然后使用parse_url检查

您的意思是要删除"Wupload"， "Fileserve"answers"上载" ?标题和捕获url在数组中?如果是，请尝试以下操作:

preg_match_all('!^http://.*'n!m', $string, $matches);
echo "<pre>" . print_r($matches, 1) . "</pre>";

这应该可以满足您的需求:

<?php
$matches = array();
preg_match_all('@https?://([-'w'.]+)+(:'d+)?(/([-'w/_'.]*('?'S+)?)?)?@', $string, $matches);
foreach ($matches[0] as $match) {
    // Do your processing here.
}
?>