如何将命令的输出分配给shell中的另一个变量[重复]

Question

如何将命令的输出分配给 shell 中的另一个变量， 这是我尝试过的。

summary = `grep -A 4 '#HM based Segregation FEP (WNS)(TNS)' some_file | awk '{print $2, $3}' | sed 's/,=,//g' | tr '\n' ',' | sed 's/based Segregation,//g' | sed 's/(\|\ (/\//g' | sed 's/)//g'`

summary1 = $(grep -A 4 '#HM based Segregation FEP (WNS)(TNS)' some_file | awk '{print $2, $3}' | sed 's/,=,//g' | tr '\n' ',' | sed 's/based Segregation,//g' | sed 's/(\|\ (/\//g' | sed 's/)//g')
echo "$summary"
echo "$summary1"

基本上，我正在从一些模式中寻找并删除不需要的数据。 输出是

summary: Command not found.
Illegal variable name.

我做错了什么？

Answer 1

Shell 变量赋值语法为var=value。删除等号周围的空格。例如：

summary="$(grep -A 4 '#HM based Segregation FEP (WNS)(TNS)' some_file | awk '{print $2, $3}' | sed 's/,=,//g' | tr '\n' ',' | sed 's/based Segregation,//g' | sed 's/(\|\ (/\//g' | sed 's/)//g')"