[英]Regular Expressions in Perl
I am trying to write a regex to match a particular line and perform action on the line below it . 我试图编写一个正则表达式来匹配特定的行,并在它下面的行上执行操作。 Reading the file
a.txt
The contents of a.txt
读文件
a.txt
的内容a.txt
I am from Melbourne .
Aussie rocks #The text can be anything below the first line
I am writing a regular expression to read the file a.txt
and trying to replace the text below line 1
. 我正在写一个正则表达式来读取文件
a.txt
并尝试替换第line 1
下面的文本。 Snippet :- 片段:-
open($fh,"a.txt") or die "cannot open:$!\n";
while(<$fh>){
if($_=~/^I am from\s+.*/){
#I have to replace the line below it .
}
Can anyone please help me. 谁能帮帮我吗。 I just have to
replace a line below the line that matches my regex with an empty line or anything
. 我只需
replace a line below the line that matches my regex with an empty line or anything
。 $line =~ s/<Line below line1>//;
. 。 How can I do that .?
我怎样才能做到这一点 。?
open(my $fh, "<", "a.txt") or die $!;
my $replace;
while(<$fh>){
$_ = "\n" if $replace;
$replace = /^I am from.*/;
print;
}
or by reading file at once, 或一次读取文件,
open(my $fh, "<", "a.txt") or die $!;
my $str = do { local $/; <$fh> };
$str =~ s/^I am from.*\n \K .*//xm;
print $str;
There are a variety of ways. 有多种方法。
while (<$fh>) {
print;
if (/^I am from/) {
<$fh> // die "Expected line"; # discard next line
print "Foo Blargh\n"; # output something else
}
}
which is my preferred solution. 这是我的首选解决方案。
my $replace = 0;
while (<$fh>) {
if ($replace) {
print "Foo Blargh\n";
$replace = 0;
}
else {
print;
$replace = 1 if /^I am from/;
}
}
my $contents = do { local $/; <$fh> };
$contents =~ s/^I am from.*\ņ\K.*/Foo Blargh/m;
print $contents;
That regex needs an explanation: ^
matches a line start under /m
. 该正则表达式需要说明:
^
匹配/m
下的行开头。 .*\\n
matches the rest of the line. .*\\n
与该行的其余部分匹配。 \\K
doesn't include the preceding pattern in the matched substring. \\K
在匹配的子字符串中不包含前面的模式。 The .*
matches the next line, which is then replaced by Foo Blargh
. .*
匹配下一行,然后由Foo Blargh
替换。
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