bash脚本 - 检索每一个充满*

Question

我有一个bash脚本，就像

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

我希望能够捕捉到类似* .OUT输出值，带*号是一样的*。在。我如何检索*被扩展到的文本，以便我可以重用它？

Answer 1

更改文件后缀使用bash：

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

从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

通过在你的问题的措辞（可能是更清晰），我想你想删除前导路径。

您可以使用parameter expansion来完成你想要的东西：

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

这将删除前面的路径和.in扩展，名称的所有输出文件与.out扩展，并将其放置在该目录/path/out。

• ${i##*/} - 删除通过/中最后出现的前导字符，以获得与.in扩展名的文件的名称。

• ${name%.in}.out - 从name删除尾随.in扩展名，并替换为.out。