BASH shell 使用正则表达式从文件中获取值到参数中
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BASH shell use regex to get value from file into a parameter
提问by Jay Gilford
I've got a file that I need to get a piece of text from using regex. We'll call the file x.txt. What I would like to do is open x.txt, extract the regex match from the file and set that into a parameter. Can anyone give me some pointers on this?
我有一个文件，我需要使用正则表达式获取一段文本。我们将调用文件x.txt. 我想要做的是 open x.txt，从文件中提取正则表达式匹配并将其设置为参数。任何人都可以给我一些指示吗？
EDIT
编辑
So in x.txtI have the following line
所以在x.txt我有以下行
$variable = '1.2.3';
I need to extract the 1.2.3from the file into my bash script to then use for a zip file
我需要1.2.3将文件从文件中提取到我的 bash 脚本中，然后用于 zip 文件
回答by vladr
Use sedto do it efficiently?in a single pass:
用sed它高效地做吗？一次通过：
var=$(sed -ne "s/\$variable *= *['\"]\([^'\"]*\)['\"] *;.*//p" file)
The above works whether your value is enclosed in single ordouble quotes.
无论您的值是用单引号还是双引号括起来，以上都适用。
Also see Can GNU Grep output a selected group?.
另请参阅GNU Grep 可以输出选定的组吗？.
$ cat dummy.txt
$bla = '1234';
$variable = '1.2.3';
blabla
$variable="hello!"; #comment
$ sed -ne "s/\$variable *= *['\"]\([^'\"]*\)['\"] *;.*//p" dummy.txt
1.2.3
hello!
$ var=$(sed -ne "s/^\$variable *= *'\([^']*\)' *;.*//p" dummy.txt)
$ echo $var
1.2.3 hello!
? or at least as efficiently as sedcan churn through data when compared to grepon your platform of choice. :)
? 或者至少sed与grep在您选择的平台上相比可以有效地处理数据。:)
回答by doubleDown
You can use the grep-chop-chop technique
您可以使用 grep-chop-chop 技术
var="$(grep -F -m 1 '$variable =' file)"; var="${var#*\'}"; var="${var%\'*}"
回答by higuaro
If all the file lines have that format ($<something> = '<value>'), the you can use cutlike this:
如果所有文件行都具有该格式 ( $<something> = '<value>')，则可以cut像这样使用：
value=$(cut -d"'" -f2 file)
