[英]perl regular expression substitution with groups
I have the following json input 我有以下json输入
... "somefield":"somevalue", "time":"timevalue", "anotherfield":"value" ...
inside my ksh script I wish to replace timevalue with my value. 在我的ksh脚本中,我希望将timevalue替换为我的值。 So I created this regular expression using groups with works just fine
所以我用工作组创建了这个正则表达式
data=`cat somefile.json`
echo $data | perl -pe "s|(.*time\"\s*\:\s*\").*?(\".*)|\1%TIME%\2|g" | another-script.sh
... "somefield":"somevalue", "time":"%TIME%", "anotherfield":"value" ...
However ... I cannot use number as substitution because perl uses numbers to define groups .. so this one obviously doen't work 但是...我不能用数字代替,因为perl使用数字来定义组..所以这个显然不起作用
perl -pe "s|(.*time\"\s*\:\s*\").*?(\".*)|\120:00:00\2|g"
I can overcome this by doing two step substitution 我可以通过两步替换来克服这个问题
perl -pe "s|(.*time\"\s*\:\s*\").*?(\".*)|\1%TIME%\2|g" | perl -pe "s|%TIME%|20:00:00|"
... "somefield":"somevalue", "time":"20:00:00", "anotherfield":"value" ...
but I am sure there is a better and more elegant way doing it 但我敢肯定,有更好,更优雅的方法
Whilst you could do this with regexes, it would be so much easier with the right tool 虽然您可以使用正则表达式执行此操作,但是使用正确的工具会容易得多
jq '.time="20:00:00"' somefile.json
If you particularly wish to use Perl, the core Perl distribution has included a JSON parser since 2011, so you could do something like: 如果您特别希望使用Perl,Perl的核心发行版自2011年起就包含JSON解析器,因此您可以执行以下操作:
perl -MJSON::PP=decode_json,encode_json -0 -E '$j = decode_json(<>); $j->{time} = "20:00:00"; say encode_json($j)' somefile.json
Perl doesn't use \\1
for substitution. Perl不使用
\\1
进行替换。 If you had enabled warnings (eg with perl -w
), perl would have told you it's $1
. 如果您启用了警告(例如,使用
perl -w
),则perl会告诉您它是$1
。 Which can be disambiguated from surrounding digits by adding {
}
: 可以通过添加
{
}
与周围的数字区分开来:
perl -pe 's|(.*time"\s*:\s*").*?(".*)|${1}20:00:00$2|g'
(I also removed all the redundant backslashes from the regex.) (我还从正则表达式中删除了所有多余的反斜杠。)
On another note, what's the point of matching .*
if you're just going to replace it by itself? 另一方面,如果只想自己替换它,那么匹配
.*
有什么意义呢? Couldn't it just be 不能只是
perl -pe 's|(time"\s*:\s*").*?(")|${1}20:00:00$2|g'
? ?
I'm not a big fan of .*
or .*?
我不是
.*
或.*?
忠实粉丝.*?
. 。 If you're trying to match the inside of a quoted string, it would be better to be specific:
如果您尝试匹配带引号的字符串的内部,则最好进行具体说明:
perl -pe 's|(time"\s*:\s*")[^"]*(")|${1}20:00:00$2|g'
We're not trying to validate the input string, so now there's really no reason to match that final "
(and replace it by itself) either: 我们并没有尝试验证输入字符串,因此,现在实际上也没有理由匹配最后一个
"
(并用它自己替换)”:
perl -pe 's|(time"\s*:\s*")[^"]*|${1}20:00:00|g'
If your perl is not ancient (5.10+), you can use \\K
to "keep" leading parts of the string, ie not include it in the match: 如果您的perl不是古老的(5.10+),则可以使用
\\K
来“保留”字符串的前导部分,即不要将其包括在匹配项中:
perl -pe 's|time"\s*:\s*"\K[^"]*|20:00:00|g'
Now only the [^"]*
part will be substituted, saving us from having to do any capturing. 现在只有
[^"]*
部分将被替换,从而使我们不必进行任何捕获。
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