Bash：非法变量名错误
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Bash: illegal variable name error
提问by Jeff
I am trying to extract a field from a text file using grep. I want to store the line number into a bash variable for later, but I am getting an illegal variable name error. This is part of my script:
我正在尝试使用 grep 从文本文件中提取一个字段。我想将行号存储到 bash 变量中以备后用，但我收到了非法变量名错误。这是我的脚本的一部分：
#!/bin/csh
set echo
grep -n -m 1 "HR" Tossed/length/TLD2R-TT.txt | cut -d : -f 1
To_Start=$((grep -n -m 1 "HR" Tossed/length/TLD2R-TT.txt | cut -d : -f 1))
This is the output:
这是输出：
[maurerj1@rucc-headnode Tenengolts_Generate]$ ./flow_LBBH.sh 7 0 0 0
grep --color=auto -n -m 1 HR0 Tossed/length7/TL7D2R0-0TT.txt
cut -d : -f 1
1 #This is the right number
Illegal variable name. #why is this not working?
From what I've read, uppercase, lowercase and underscores are allowed in bash variable names, so what am I doing wrong?
从我读过的内容来看，bash 变量名中允许使用大写、小写和下划线，那么我做错了什么？
回答by Jean-Fran?ois Fabre
the $(( expr ))syntax is used for calculations hence the confusing message.
该$(( expr ))语法用于计算因此混淆消息。
ex: echo $((4+4))yields 8
例如：echo $((4+4))产量 8
you want to evaluate the result of a command, simple parenthesis will do:
你想评估一个命令的结果，简单的括号就可以了：
To_Start=$(grep -n -m 1 "HR" Tossed/length/TLD2R-TT.txt | cut -d : -f 1)
Simple reproducer to prove my point:
简单的复制器来证明我的观点：
To_Start=$(echo a:b | cut -d : -f 1)
echo $To_Start
yields:
产量：
a
回答by Jeff
I fixed the issue by changing from csh to bash, removing set echo (caused some weird issue by making all input variables into "echo")and changing from $(()) to $().
我通过从 csh 更改为 bash、删除 set echo（通过将所有输入变量设置为“echo”而导致一些奇怪的问题）以及从 $(()) 更改为 $() 来解决该问题。
