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apile |
发表于： 2002/12/27 01:09pm
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#!/usr/bin/perl
$abc = 'Paul and his kids want to go to @peiking@ and live there.';$abc=~/\@peiking\@/; print $&."\n";
$abc=$`."\@beijing\@".$';
print $abc."\n"; you can't use tr to replace a string....
because in your case the whole char. int string will be replaced by the rule.
p---->b
k---->j
t---->d
s---->x so you need to use var. of regular expression. $` && $' |
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apile |
发表于： 2002/12/27 01:11pm
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or another way...
$abc = 'Paul and his kids want to go to @peiking@ and live there.';
$abc=~s/\@peiking\@/\@beijing\@/g;
is easier... |
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halihan |
发表于： 2002/12/27 02:23pm
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谢谢您的回复, 对不起我可能说得不清楚:
我的主要目标是使用此规则:
p->bk->jt->ds->x…
替换文件中所有在 @...@ 之内的字符
而规则共有数十条之多, 所以才想到使用 tr/pkts/bjdx/
但又不是替换整个文件, 而是只有在 @...@ 之内的字符
所以想不出解决方案…
当然使用 s///g 也可以, 但是数十条规则非常不好用,
请指点迷津,谢谢! |
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apile |
发表于： 2002/12/27 10:02pm
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you can use..
#!/usr/bin/perl
$abc = 'Paul and his kids want to go to @peiking@ and live there.';
$abc=~/\@peiking\@/;
$cde = $&;
$cde = tr/pkts/bjdx/;$abc=$`.$cde.$';
print $abc."\n"; this is what you want...
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apile |
发表于： 2002/12/27 10:05pm
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#!/usr/bin/perl
$abc = 'Paul and his kids want to go to @peiking@ and live there.';
$abc=~/\@(\w+)\@/;
$cde = $1;
$cde = tr/pkts/bjdx/;
$abc=$`.$cde.$';
print $abc."\n";this is an example for general case.......
best regards....
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halihan |
发表于： 2003/01/02 10:08pm
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啊! 真是太强了!! 非常感谢大侠专业的回复!!
小弟另有一问:
如果 $abc = 'Paul and his kids want to go to @peiking@ and live there……
不是只有一行, 而是整个文件, 而且一行中可能出现数个 @...@ 部分,
那么是否必须使用循环, 依序处理每一行才能办到?
还是有更高明的办法??
恳请指点迷津, 感激不尽!! |
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apile |
发表于： 2003/01/03 08:18am
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| #!/usr/bin/perl $abc = 'Paul and his kids want to go to @peiking@ and live there @peiking@';
$abc =~s/(\@\w+\@)/&fun1($1)/ge;
print $abc."\n";
sub fun1($){
local($aa) = @_;
$aa =~ tr/pkts/bjdx/;
return $aa;
} this is what you want.......
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halihan |
发表于： 2003/01/04 10:59pm
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哇!!! 这就是我想要的!!! :em02:
大侠武功高强, 神乎其技, 行侠仗义, 在下万分佩服!!
请受在下一拜!!! :em15: |
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