如何在shell中转换变量中的字符串

Question

set lamdrills = ("drill_1-2" drill_2-3")
set drill_1_2_fill_layers = ("cvf1") #input from other files
set span = `echo "$lam_drl" | cut -f2 -d'_'`
set startspan = `echo "$span" | cut -f1 -d'-'`
set endspan = `echo "$span" | cut -f2 -d'-'`
set fill_layers = `echo drill_"$startspan"_"$endspan"_fill_layers`

我想将drill_1_2_fill_layers分配给fill_layers

Answer 1

看起来有点像你在寻找

set lamdrills = ("drill_1-2" "drill_2-3")  # typo fixed
foreach drill ($lamdrills)                 # loop over array
set fill_layers = drill_`echo $drill | sed 's/.*_//;s/-/_/'`_fill_layers
# ... use this variable
end

C外壳很狡猾而且不受欢迎(有充分的理由(；也许您想使用Bourne兼容的shell？

for drill in "drill_1-2" "drill_2-3"; do
fill_layers="drill_$(echo "$drill" | sed 's/.*_//;s/-/_/')_fill_layers"
# ... use "$fill_layers"
done

当然，如果你使用的标签能始终如一地使用相同的约定和标点符号模式，那就更好了。