如何匹配awk中的变量中给定的模式?

I want to extract a substring where certain pattern exist from pipe separated file, thus I used below command,

我想从管道分离的文件中提取一个存在特定模式的子字符串，因此我使用了下面的命令，

awk -F ":" '/REWARD REQ. SERVER HEADERS/{print $1, $2, $3, $4}' sample_profile.txt

Here, 'REWARD REQ. SERVER HEADERS' is a pattern which is to be searched in the file, and print its first 4 parts on a colon separated line.

在这里,“奖励申请。服务器头是一个在文件中搜索的模式，并在一个冒号分隔的线上打印前4个部分。

Now, I want to send bash variable to act as a pattern. thus I used below command, but it's not working.

现在，我想发送bash变量作为一个模式。因此，我使用了下方命令，但它不起作用。

awk -v pat="$pattern" -F ":" '/pat/{print $1, $2 , $3, $4 } sample_profile.txt

How can I use -v and -F in a single awk command?

如何在一个awk命令中使用-v和-F ?

3 个解决方案

#1

The problem here does not have to do with -F.

这里的问题与-F无关。

The problem is the usage of /pat/ when you want pat to be a variable. If you say /pat/, awk understands it as a literal "pat", so it will try to match those lines containing the string "pat".

问题是当你想让pat成为变量时/pat/的用法。如果您输入/pat/， awk将其理解为一个字面的“pat”，因此它将尝试匹配包含字符串“pat”的行。

If you want to provide the pattern through a variable, you need to use ~ as this:

如果您想通过变量提供模式，您需要使用~

 awk -v pat="$pattern" '$0 ~ pat'

All together, your code should be:

总之，您的代码应该是:

awk -v pat="$pattern" -F ":" '$0~pat{print $1, $2, $3, $4 }' file
#                             ^^^^^^

See an example:

看一个例子:

Given this file:

鉴于这种文件:

$ cat file
hello
this is a var
hello bye

Let's look for lines containing "hello":

让我们寻找包含“hello”的行:

$ awk '/hello/' file
hello
hello bye

Let's now try looking for "pat", contained in a variable, the way you were doing it:

现在，让我们尝试寻找包含在变量中的“pat”，看看您是怎么做的:

$ awk -v pat="hello" '/pat/' file
$                                    # NO MATCHES!

Let's now use the $0 ~ pat expression:

现在我们使用$0 ~ pat表达式:

$ awk -v pat="hello" '$0~pat' file
hello                                 # WE MATCH!
hello bye

Of course, you can use such expressions to match just one field and say awk -v pat="$pattern" '$2 ~ pat' file and so on.

当然，您可以使用这样的表达式来匹配一个字段，比如awk -v pat=“$pattern”“$2 ~ pat”文件等等。

From GNU Awk User's Guide → 3.1 How to Use Regular Expressions:

从GNU Awk用户指南→3.1如何使用正则表达式:

When a regexp is enclosed in slashes, such as /foo/, we call it a regexp constant, much like 5.27 is a numeric constant and "foo" is a string constant.

当regexp包含在斜线中(如/foo/)时，我们将它称为regexp常量，就像5.27是一个数值常量，而“foo”是一个字符串常量。

And GNU Awk User's Guide → 3.6 Using Dynamic Regexps:

用户指南和GNU Awk→3.6使用动态regexp:

The righthand side of a ‘~’ or ‘!~’ operator need not be a regexp constant (i.e., a string of characters between slashes). It may be any expression. The expression is evaluated and converted to a string if necessary; the contents of the string are then used as the regexp. A regexp computed in this way is called a dynamic regexp or a computed regexp:

“~”或“!”的右手边。~ '运算符不必是regexp常数(即。，斜线之间的一串字符)。它可以是任何一种表达。如果需要，表达式将被计算并转换为字符串;然后将字符串的内容用作regexp。以这种方式计算的regexp称为动态regexp或计算regexp:
BEGIN { digits_regexp = "[[:digit:]]+" }
$0 ~ digits_regexp    { print }
This sets digits_regexp to a regexp that describes one or more digits, and tests whether the input record matches this regexp.

这将digits_regexp设置为regexp，该regexp描述一个或多个数字，并测试输入记录是否与该regexp匹配。

#2

awk -v pat="$pattern" -F":" '$0 ~ pat { print $1, $2, $3, $4 }' sample_profile.txt

You can't use the variable inside the regex // notation (there's no way to distinguish it from searching for pat); you have to specify that the variable is a regex with the ~ (matching) operator.

您不能使用regex //符号内的变量(没有办法区分它与搜索pat);您必须指定该变量是具有~(匹配)操作符的regex。

#3

This is kind of a hack but it makes things a little simpler for me.

这是一种技巧，但对我来说更简单。

cmd="awk '/$pattern/'"
eval $cmd

making it a string first lets you manipulate it past the boundaries of awk

使它成为一个字符串首先允许您操作它通过awk的边界

#1