如何从Perl字符串中提取文件路径? [英] How can I extract a file path from a Perl string?

问题描述

我想从该字符串中找到具有全路径的文件名

I want to find the file name with fullpath from this string

"[exec] /snschome/sns/ccyp/mb_ccsns/bb/cbil5/v85_8/amdocs/iamcust/bil/cl/business/handlers/ClReportsHandler.java:233: cannot resolve symbol"

我要提取

/snschome/sns/ccyp/mb_ccsns/bb/cbil5/v85_8/amdocs/iamcust/bil/cl/business/handlers/ClReportsHandler.java

我正在Perl中尝试此操作

and I am trying this in Perl

    $_=$string_from_to_match
    my @extract_file=split(/.*(\/.*)\:.*/);
    print $extract_file[1],"\n";`

但是我得到以下输出:

/ClReportsHandler.java:233:

它与最后一个/和最后一个:相匹配.如何将其更改为第一个/和第一个:?

It is matching the last / and the last :. How can I change it to first / and first :?

推荐答案

在这种情况下，粘性和伸展性"很有用.您知道，左边是[exec]，后跟空格，右边是冒号，后跟行号.您想要介于两者之间的内容:

This is a case where "tacking and stretching" is useful. You know that [exec] followed by whitespace is on the left and colon followed by a line number is on the right. You want what's in between:

#! /usr/bin/perl

use warnings;
use strict;

$_ = "[exec] /snschome/sns/ccyp/mb_ccsns/bb/cbil5/v85_8/amdocs/iamcust/bil/cl/business/handlers/ClReportsHandler.java:233: cannot resolve symbol";

if (/\[exec\]\s*(.+?):\d+/) {
  print $1, "\n";
}

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