如何从 bash grep 输出中删除最后一个字符
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How to remove the last character from a bash grep output
提问by liv2hak
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2`
outputs something like this
输出这样的东西
"Abc Inc";
What I want to do is I want to remove the trailing ";" as well. How can i do that? I am a beginner to bash. Any thoughts or suggestions would be helpful.
我想要做的是删除尾随的“;” 以及。我怎样才能做到这一点？我是 bash 的初学者。任何想法或建议都会有所帮助。
采纳答案by Laurence Gonsalves
I'd use sed 's/;$//'. eg:
我会用sed 's/;$//'. 例如：
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | sed 's/;$//'`
回答by ztank1013
This will remove the last character contained in your COMPANY_NAME var regardless if it is or not a semicolon:
这将删除 COMPANY_NAME 变量中包含的最后一个字符，无论它是否是分号：
echo "$COMPANY_NAME" | rev | cut -c 2- | rev
回答by grokster
foo="hello world"
echo ${foo%?}
hello worl
回答by Aaron J Lang
I'd use head --bytes -1, or head -c-1for short.
我会使用head --bytes -1，或head -c-1简称。
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | head --bytes -1`
headoutputs only the beginning of a stream or file. Typically it counts lines, but it can be made to count characters/bytes instead. head --bytes 10will output the first ten characters, but head --bytes -10will output everything except the last ten.
head仅输出流或文件的开头。通常它计算行数，但可以改为计算字符/字节数。head --bytes 10将输出前十个字符，但head --bytes -10将输出除最后十个字符以外的所有字符。
NB: you may have issues if the final character is multi-byte, but a semi-colon isn't
注意：如果最后一个字符是多字节的，你可能会遇到问题，但分号不是
I'd recommend this solution over sedor cutbecause
我会推荐这个解决方案sed或者cut因为
- It's exactly what
headwas designed to do, thus less command-line options and an easier-to-read command - It saves you having to think about regular expressions, which are cool/powerful but often overkill
- It saves your machine having to think about regular expressions, so will be imperceptibly faster
- 这正是
head设计的目的，因此命令行选项更少，命令更易于阅读 - 它使您不必考虑正则表达式，这些表达式很酷/功能强大但通常矫枉过正
- 它使您的机器不必考虑正则表达式，因此会不知不觉地更快
回答by mightypile
I believe the cleanest way to strip a single character from a string with bash is:
我相信用 bash 从字符串中去除单个字符的最干净的方法是：
echo ${COMPANY_NAME:: -1}
but I haven't been able to embed the grep piece within the curly braces, so your particular task becomes a two-liner:
但我无法将 grep 嵌入花括号中，因此您的特定任务变成了两行：
COMPANY_NAME=$(grep "company_name" file.txt); COMPANY_NAME=${COMPANY_NAME:: -1}
This will strip any character, semicolon or not, but can get rid of the semicolon specifically, too. To remove ALL semicolons, wherever they may fall:
这将删除任何字符，分号与否，但也可以专门删除分号。要删除所有分号，无论它们在哪里：
echo ${COMPANY_NAME/;/}
To remove only a semicolon at the end:
只删除末尾的分号：
echo ${COMPANY_NAME%;}
Or, to remove multiple semicolons from the end:
或者，从末尾删除多个分号：
echo ${COMPANY_NAME%%;}
For great detail and more on this approach, The Linux Documentation Project covers a lot of ground at http://tldp.org/LDP/abs/html/string-manipulation.html
有关此方法的详细信息和更多信息，Linux 文档项目在http://tldp.org/LDP/abs/html/string-manipulation.html 上涵盖了很多内容
回答by Det
Using sed, if you don't know what the last character actually is:
使用sed, 如果您不知道最后一个字符实际上是什么：
$ grep company_name file.txt | cut -d '=' -f2 | sed 's/.$//'
"Abc Inc"
回答by Has QUIT--Anony-Mousse
Don't abuse cats. Did you know that grepcan read files, too?
不要滥用cats。你知道它grep也可以读取文件吗？
The canonical approach would be this:
规范的方法是这样的：
grep "company_name" file.txt | cut -d '=' -f 2 | sed -e 's/;$//'
the smarter approach would use a single perlor awkstatement, which can do filter and different transformations at once. For example something like this:
更聪明的方法是使用单个perlorawk语句，它可以一次进行过滤和不同的转换。例如这样的事情：
COMPANY_NAME=$( perl -ne '/company_name=(.*);/ && print ' file.txt )
回答by kurumi
don't have to chain so many tools. Just one awk command does the job
不必链接这么多工具。只需一个 awk 命令即可完成这项工作
COMPANY_NAME=$(awk -F"=" '/company_name/{gsub(/;$/,"",) ;print }' file.txt)
回答by Jens Bodal
Assuming the quotation marks are actually part of the output, couldn't you just use the -o switch to return everything between the quote marks?
假设引号实际上是输出的一部分，难道您不能只使用 -o 开关来返回引号之间的所有内容吗？
COMPANY_NAME="\"ABC Inc\";" | echo $COMPANY_NAME | grep -o "\"*.*\""
回答by Paused until further notice.
In Bash using only one external utility:
在仅使用一个外部实用程序的 Bash 中：
IFS='= ' read -r discard COMPANY_NAME <<< $(grep "company_name" file.txt)
COMPANY_NAME=${COMPANY_NAME/%?}
