bash 在 AWK 中分配一个 Shell 变量
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Assign a Shell variable in AWK
提问by Ravi
I have strings like this
我有这样的字符串
/home/rm/home-scripts/originals/audicerttest/incoming/TEST040511.txt
/home/rm/home-scripts/originals/audicerttest2/incoming/TEST040512.txt
/home/rm/home-scripts/originals/audicerttest3/incoming/TEST040513.txt
etc..I want to extract strings 'audicerttest/incoming', audicerttest2/incoming' etc into a shell variable and use it later in the script. I tried something like this.
等等..我想将字符串 'audicerttest/incoming', audicerttest2/incoming' 等提取到一个 shell 变量中，然后在脚本中使用它。我试过这样的事情。
for file in `find ${ROOT}/* -type f | grep -v -f test.txt`
do
let count++
echo ${count}: ${file}
echo ${file} | eval $(awk '{split(/home/rm/home-scripts/originals/audicerttest/incoming/TEST040511.txt
,a,"/"); print "abc=a[6]/a[7]"}' < /dev/null)
echo abc====$abc
done
but its not giving any output for abc.
但它没有为 abc 提供任何输出。
回答by sampson-chen
There are a number of issues, but let's isolate problems and tackle them one at a time.
有很多问题，但让我们隔离问题并一次解决一个。
Given- Raw string:
给定- 原始字符串：
audicerttest/incoming
Desired Output- You want this part to be saved in a variable:
所需输出- 您希望将此部分保存在变量中：
string="/home/rm/home-scripts/originals/audicerttest/incoming/TEST040511.txt"
parsed=$(awk 'BEGIN{FS=OFS="/"} {print , }' <<< ${string})
How-To- This will do it:
操作方法- 这将做到：
/home/rm/home-scripts/originals/audicerttest/incoming/TEST040511.txt
/home/rm/home-scripts/originals/audicerttest2/incoming/TEST040512.txt
/home/rm/home-scripts/originals/audicerttest3/incoming/TEST040513.txt
In general, suppose you have a file called input_file.txt:
通常，假设您有一个名为 的文件input_file.txt：
awk 'BEGIN{FS=OFS="/"} {print , }' input_file.txt > parsed_vars.txt
You can do:
你可以做：
audicerttest/incoming
audicerttest2/incoming
audicerttest3/incoming
And parsed_vars.txtwill contain:
而parsed_vars.txt将包含：
echo ${file} | eval $(awk '{split(echo {$file} | awk ...
,a,"/"); print "abc=a[6]/a[7]"}' < /dev/null)
Some explanations:
一些解释：
parsed=$(...)- spawn a subshell and save the output tostdoutwithin that subshell to the variableparsedawk 'BEGIN{FS=OFS="/"}- invokeawkand set delimiter as/for both input and output.{print $6, $7}'- based on the format of your raw strings, you want to print the 6th (audicerttest) and the 7th (incoming) fields.<<< ${string}is the notation for using herestringinput_file.txt > parsed_vars.txt- read from specified input file, and redirect output to an output file.
parsed=$(...)- 生成一个子shell并将输出保存到stdout该子shell中的变量parsedawk 'BEGIN{FS=OFS="/"}-为输入和输出调用awk和设置分隔符/。{print $6, $7}'- 根据原始字符串的格式，您想要打印第 6 个（audicerttest）和第 7 个（传入）字段。<<< ${string}是使用herestring的符号input_file.txt > parsed_vars.txt- 从指定的输入文件读取，并将输出重定向到输出文件。
回答by Karoly Horvath
eval $(echo $file | awk '{split(abc=$(echo $file | awk '{split( cut -d / --output-delimiter / -f6,7
,a,"/"); print a[6]"/"a[7]}')
,a,"/"); print "abc="a[6]"/"a[7]""}')
Okay, a couple of problems here.
好的，这里有几个问题。
awkis reading from /dev/null, which doesn't make any sense, guess you want to process the filename:
awk正在读取/dev/null，这没有任何意义，猜测您要处理文件名：
string="/home/rm/home-scripts/originals/audicerttest/incoming/TEST040511.txt"
"a[6]"is the string a[6], no substitution is made.
"a[6]"是字符串a[6]，不进行替换。
Now eval the whole thing:
现在评估整个事情：
IFS=/ a=( $string ) printf -v parsed '%s/%s' "${a[@]:5:2}"
Finally, eval is evil, why don't you directly set the variable?
最后，eval是邪恶的，为什么不直接设置变量呢？
path=/home/rm/home-scripts/originals/audicerttest/incoming/TEST040511.txt
On a personal note, I thing this is a bit more clear:
就个人而言，我认为这更清楚一点：
IFS=/ split=(${path})
path_part="${split[5]}/${split[6]}"
回答by gniourf_gniourf
Here's a pure bash solution, not using any overkill awkand not using any evil evaland not using any subshells:
这是一个纯粹的 bash 解决方案，不使用任何矫枉过正awk，不使用任何邪恶eval，也不使用任何子shell：
You have this string:
你有这个字符串：
while read path; do
dirname=$(dirname $path)
abc=${dirname#/home/rm/home-scripts/originals/}
echo "abc===$abc"
done < filename
You want a string that contains the 5th and 6th fields (where separation is /) in a variable parsed. Let's go:
您需要一个字符串，其中包含/变量中的第 5 和第 6 个字段（其中分隔为）parsed。我们走吧：
回答by Mat
With only bash arrays, you could do the following:
仅使用 bash 数组，您可以执行以下操作：
Assuming:
假设：
##代码##Then:
然后：
##代码##$splitwill be an array, with element 0 empty, element one home, element 2 rm, etc. With that you just need to concatenate the elements you want to get the part you like.
$split将是一个数组，元素 0 为空，元素 1 home，元素 2rm等。这样你只需要连接你想要获得你喜欢的部分的元素。
回答by glenn Hymanman
Nobody has suggested simple parameter expansionto manipulate the strings. I'm assuming that the first 4 directories are constant
没有人建议使用简单的参数扩展来操作字符串。我假设前 4 个目录是不变的
##代码##