bash腳本 - 檢索每一個充滿*

Question

我有一個bash腳本，就像

for i in /path/to/file/*.in; do 
    ./RunCode < "$i" 
done 

我希望能夠捕捉到類似* .OUT輸出值，帶*號是一樣的*。在。我如何檢索*被擴展到的文本，以便我可以重用它？

Answer 1

更改文件後綴使用bash：

for i in /path/to/file/*.in; do 
    ./RunCode < "$i" > "${i%.in}.out" 
done 

從man bash：

${parameter%word} 
${parameter%%word} 

Remove matching suffix pattern. The word is expanded to produce a 
pattern just as in pathname expansion. If the pattern matches a 
trailing portion of the expanded value of parameter, then the result 
of the expansion is the expanded value of parameter with the shortest 
matching pattern (the ``%'' case) or the longest matching pattern 
(the ``%%'' case) deleted. If parameter is @ or *, the pattern removal 
operation is applied to each positional parameter in turn, and the 
expansion is the resultant list. If parameter is an array variable 
subscripted with @ or *, the pattern removal operation is applied 
to each member of the array in turn, and the expansion is the resultant 
list. 

Answer 2

通過在你的問題的措辭（可能是更清晰），我想你想刪除前導路徑。

您可以使用parameter expansion來完成你想要的東西：

out_dir="/path/out" 
for i in /path/to/file/*.in; do 
    name="${i##*/}" 
    ./RunCode < "$i" > "$out_dir/${name%.in}.out" 
done 

這將刪除前面的路徑和.in擴展，名稱的所有輸出文件與.out擴展，並將其放置在該目錄/path/out。

• ${i##*/} - 刪除通過/中最後出現的前導字符，以獲得與.in擴展名的文件的名稱。

• ${name%.in}.out - 從name刪除尾隨.in擴展名，並替換爲.out。