bash awk - 未终止的正则表达式
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Awk - unterminated regex
提问by BandyOrc
I am writing a shell script which needs to pull values out of a text file which looks like this:
我正在编写一个 shell 脚本，它需要从如下所示的文本文件中提取值：
app.full.name /warfilelocation/ warfilename
My shell script will be iterating over a list of application names and pulling out either the location or name using AWK. I have tested doing this on the command line using the following: awk "\$1 ~/app.full.name/ { print $2 }" applications.txt
我的 shell 脚本将遍历应用程序名称列表并使用 AWK 提取位置或名称。我已经使用以下命令在命令行上进行了测试： awk "\$1 ~/app.full.name/ { print $2 }" applications.txt
which returns what I would expect however when i put this in a shell script I start having issues.
这会返回我期望的结果，但是当我将其放入 shell 脚本时，我开始遇到问题。
I have a function that looks like this:
我有一个看起来像这样的函数：
function get_location() {
local application=
awk "$1 ~/^$application/ { print $2 }" applications.txt
}
But when i call this function i get the following error:
但是当我调用这个函数时，我收到以下错误：
awk: ~/^app.full.name
awk: ^ unterminated regexp
awk: cmd. line:1: app.full.name
awk: cmd. line:1: ^ syntax error
awk: cmd. line:2: app.full.name/ { print }
awk: cmd. line:2: ^ syntax error
Does anyone have any ideas what I am doing wrong here. I presume I am not escaping the variable correct but no matter what i try it doesnt seem to work.
有没有人知道我在这里做错了什么。我想我没有正确地转义变量，但无论我尝试什么，它似乎都不起作用。
Thanks in advance
提前致谢
回答by Fredrik Pihl
Use this approach to make awk recognize shell variables:
使用这种方法让 awk 识别 shell 变量：
awk -v "v1=$VAR1" -v "v2=$VAR2" '{print v1, v2}' input_file
Update
更新
$ cat input
tinky-winky
dipsy
laa-laa
noo-noo
po
$ teletubby='po'
$ awk -v "regexp=$teletubby" 'kent$ app=app.ful
kent$ echo "app.full.name /warfilelocation/ warfilename"|awk -v a=$app '{if(match(,a))print }'
/warfilelocation/
~ regexp' input
po
Note that anythingcould go into the shell-variable,
even a full-blown regexp, e.g ^d.*y. Just make sure to use single-quotes
to prevent the shell from doing any expansion.
请注意，任何东西都可以进入 shell 变量，即使是完整的正则表达式，例如^d.*y. 只要确保使用单引号来防止 shell 进行任何扩展。
回答by Kent
see this: using awk match() function
看到这个：使用 awk match() 函数
$ export application=foo/bar
$ awk "$1 ~/^$application/ { print $1 }"
gawk: cmd. line:1: ~/^foo/bar/ { print }
gawk: cmd. line:1: ^ parse error
回答by Matthew Farwell
It's hard to tell without knowing exactly the value of $application, but it seems like you have a strange character in $application, such as a " or a / or something like that.
在不确切知道 $application 的值的情况下很难判断，但似乎 $application 中有一个奇怪的字符，例如 " 或 / 或类似的东西。
$ export application=`echo foo/bar | sed -e 's;/;\\/;g'`
$ awk "$1 ~/^$application/ { print $1 }"
I would look at the exact value that you have in $application, and if it contains a /, escape it.
我会查看您在 $application 中的确切值，如果它包含 /，则将其转义。
One way to do this would be to use:
一种方法是使用：
##代码##回答by thiton
The error messages seem to indicate that there is a stray newline at the end of $application, which gives the "line 2" error messages.
错误消息似乎表明 $application 末尾有一个杂散的换行符，它给出了“第 2 行”错误消息。
