变量中的通配符，Shell bash，怎么办？变量=$变量2*
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Wildcard in variable, Shell bash, how to do ? Variable=$variable2*
提问by felipe Salomao
For me it worked:
对我来说它有效：
diretorio=$(echo 'test 123'*)
but not worked when i used variable in quotes
但当我在引号中使用变量时不起作用
Var2="test 123"
diretorio=$(echo '$Var2'*)
How to solve it?
如何解决？
回答by Gilles Quenot
The mistake in your globis that
你的glob 中的错误是
diretorio=$(echo '$Var2'*)
is a shot in /dev/null, because the shelldon't expand variables in single quotes.
是一个镜头/dev/null，因为外壳不会用单引号扩展变量。
So :
所以 ：
diretorio=$(echo "$Var2"*)
Learn the difference between ' and " and `. See http://mywiki.wooledge.org/Quotesand http://wiki.bash-hackers.org/syntax/words
了解 ' 和 " 和 ` 之间的区别。请参阅http://mywiki.wooledge.org/Quotes和http://wiki.bash-hackers.org/syntax/words
回答by Gordon Davisson
May I suggest an alternate approach? Instead of making a space-separated list of filenames (which will cause horrible confusion if any of the filenames contain spaces, e.g. "test 123"), use an array:
我可以建议另一种方法吗？而不是制作一个以空格分隔的文件名列表（如果任何文件名包含空格，这将导致可怕的混乱，例如“test 123”），使用数组：
diretorio=("${Var2}"*)
doSomethingWithAllFiles "${diretorio[@]}"
for umDiretorio in "${diretorio[@]}"; do
doSomethingWithASingleFile "$umDiretorio"
done
回答by jlliagre
Use double quotes:
使用双引号：
diretorio=$(echo "$Var2"*)
Single ones prevent variable substitution
单个防止变量替换
