如何让AWK使用在Bash脚本中创建的变量[重复]

时间:2022-05-23 23:52:41

This question already has an answer here:

这个问题在这里已有答案:

I have script that looks like this

我的脚本看起来像这样

#!/bin/bash
#exampel inputfile is "myfile.txt"
inputfile=$1
basen=`basename $inputfile .txt`  # create basename

cat $inputfile | 
awk '{print $basen "\t" $3}  # this doesn't print "myfile" but the whole content of it.

What I want to do above is to print out in AWK the variable called 'basen' created before. But somehow it failed to do what I hoped it will.

我上面要做的是在AWK中打印之前创建的名为“basen”的变量。但不知怎的,它没能按照我的希望做到。

So for example myfile.txt contain these lines

例如,myfile.txt包含这些行

foo bar bax
foo qux bar

With the above bash script I hope to get

使用上面的bash脚本我希望得到

myfile bax
myfile bar

What's the right way to do it?

什么是正确的方法呢?

6 个解决方案

#1


12  

You can use it like this.

你可以像这样使用它。

for i in `find $1 -name \*.jar`
do
jar tvf $i| awk -F '/' '/class/{print "'${i}'" " " $NF }' >> $classFile
done

You should use

你应该用

"'${i}'"

“ '$ {I}'”

in AWK to use the

在AWK中使用

$i

$ I

created in Bash Script.

在Bash脚本中创建。

#2


29  

The -v flag is for setting variables from the command line. Try something like this:

-v标志用于从命令行设置变量。尝试这样的事情:

awk -v "BASEN=$basen" '{print BASEN "\t" $3}'

#3


4  

you can just do everything in awk

你可以在awk中做所有事情

awk '{gsub(".txt","",ARGV[1]);print ARGV[1] "\t" $3}' inputfile.txt

#4


4  

Assuming you run awk as the sub process of the shell you declared the vars
Within the shell

假设你运行awk作为shell的子进程,你在shell中声明了vars

export MY_VARS="whatever"; #// IT NEEDS to be exported, to allow the sub process awk read access.
echo ""| awk '{
    print "Reading values from within awk : "ENVIRON["MY_VARS"];
}'

Result:

结果:

Reading values from within awk : whatever

从awk中读取值:无论如何

notice the importance of export. With out it, the vars from the shell is considered local and does not get passed to the co-processes.

注意出口的重要性。在外壳中,来自shell的变量被认为是本地的,并且不会传递给协同进程。

#5


2  

The reason is that bash variables (environment variables) are not expanded within single-quoted strings. Try replacing

原因是bash变量(环境变量)不会在单引号字符串中展开。尝试更换

'{print $basen "\t" $3}'

with

"{print \"$basen\" \"\t\" \$3}"

#6


1  

The easiest way is to make an awk variable. awk -v awkvar=$bashvar 'awkscript'.

最简单的方法是创建一个awk变量。 awk -v awkvar = $ bashvar'awkscript'。

#1


12  

You can use it like this.

你可以像这样使用它。

for i in `find $1 -name \*.jar`
do
jar tvf $i| awk -F '/' '/class/{print "'${i}'" " " $NF }' >> $classFile
done

You should use

你应该用

"'${i}'"

“ '$ {I}'”

in AWK to use the

在AWK中使用

$i

$ I

created in Bash Script.

在Bash脚本中创建。

#2


29  

The -v flag is for setting variables from the command line. Try something like this:

-v标志用于从命令行设置变量。尝试这样的事情:

awk -v "BASEN=$basen" '{print BASEN "\t" $3}'

#3


4  

you can just do everything in awk

你可以在awk中做所有事情

awk '{gsub(".txt","",ARGV[1]);print ARGV[1] "\t" $3}' inputfile.txt

#4


4  

Assuming you run awk as the sub process of the shell you declared the vars
Within the shell

假设你运行awk作为shell的子进程,你在shell中声明了vars

export MY_VARS="whatever"; #// IT NEEDS to be exported, to allow the sub process awk read access.
echo ""| awk '{
    print "Reading values from within awk : "ENVIRON["MY_VARS"];
}'

Result:

结果:

Reading values from within awk : whatever

从awk中读取值:无论如何

notice the importance of export. With out it, the vars from the shell is considered local and does not get passed to the co-processes.

注意出口的重要性。在外壳中,来自shell的变量被认为是本地的,并且不会传递给协同进程。

#5


2  

The reason is that bash variables (environment variables) are not expanded within single-quoted strings. Try replacing

原因是bash变量(环境变量)不会在单引号字符串中展开。尝试更换

'{print $basen "\t" $3}'

with

"{print \"$basen\" \"\t\" \$3}"

#6


1  

The easiest way is to make an awk variable. awk -v awkvar=$bashvar 'awkscript'.

最简单的方法是创建一个awk变量。 awk -v awkvar = $ bashvar'awkscript'。