Perl如何用正则来抽取一个长字符串里的一些相似字符?

milk_

$string = "(999999 :pref p1 8888 :type t1 7777777777777 66666666666 :comment ppp 5555555555555)";

我想把:XXXX 这样的内容抽取出来放入一个数组，该如何处理啊？

请指教

eyeqwer

什么意思?

ulmer

原帖由 milk_ 于 2006-3-6 17:26 发表
$string = "(999999 :pref p1 8888 :type t1 7777777777777 66666666666 :comment ppp 5555555555555)";

我想把:XXXX 这样的内容抽取出来放入一个数组，该如何处理啊？

Hi,

if you mean put all beginning with ":" into an array,
maybe use this pattern:
    my $pattern = ':.+[^:]';
    my @matches = $string =~ /$pattern/g;


Just 4 Fun, ulmer

milk_

原帖由 ulmer 于 2006-3-6 18:37 发表


Hi,

if you mean put all beginning with ":" into an array,
maybe use this pattern:
    my $pattern = ':.+[^:]';
    my @matches = $string =~ /$pattern/g;


Just 4 Fun, ulmer

    my @matches = $string =~ /$pattern/g;

great ,我就是想知道符合条件的string如何放到一个数组里头去.

apile

如果都是固定的..直接用split(/:/,$string);不就好了??

POLOGG

原帖由 ulmer 于 2006-3-6 18:37 发表


Hi,

if you mean put all beginning with ":" into an array,
maybe use this pattern:
    my $pattern = ':.+[^:]';
    my @matches = $string =~ /$pattern/g;


Just 4 Fun, ulmer

楼上的，能够满足要求么？

我写了一个

1. 
   
2. $string = "(999999 :pref p1 8888 :type t1 7777777777777 66666666666 :comment ppp 5555555555555)";
   
3. 
   
4. my $pattern = '((.)\2+)';
   
5. my @matches = $string =~ /$pattern/g;
   
6. my $count = 0;
   
7. 
   
8. while ( $count <= $#matches)
   
9. {
   
10.                 push @temp, $matches[$count];
    
11. 
    
12.                 $count = $count + 2;
    
13. }
    
14. 
    
15. print "@temp";
    

复制代码

[ 本帖最后由 POLOGG 于 2006-3-7 21:23 编辑 ]

dlms

汗俺都看不明白！