Question

正如标题所述？可能吗？我使用下面的代码，但它无法确定变量是否有空格。希望有人能帮助我

for file in `ls *.[Pp][Dd][Ff]`
do
  var1=`echo "$file" | sed -e "s/.*-\(.*\)-.*/\1/"`
  var2="document.num=.*$var1"
  var3=`grep -l ${var2} *xml`

case "$var3" in
*\ *)
    echo $var3 >> haha
    ;;
*)
    var4=`echo "$var3" | sed -e "s/-.*//"`
        varName="$var4.$file"
        echo ${var2}
        echo $var3
        mv $file $varName
    ;;
esac

Answer 1

适合我：

var="ab"
#var="a b"

case "$var" in
*\ * ) echo "Has space" ;;
* ) echo "No space" ;;
esac

根据注释掉的变量，它会打印Has space或No space

Answer 2

使用Bash正则表达式测试

如果您想要如何灵活定义空格，可以使用Bash的内置正则表达式测试。例如：

# Variable has a space.
foo=' '
[[ $foo =~ [[:space:]] ]]; echo $?
0

# Variable has a tab.
foo=$'\t'
[[ $foo =~ [[:space:]] ]]; echo $?
0

# Variable has no whitespace because it's empty.
foo=''
[[ $foo =~ [[:space:]] ]]; echo $?
1

是否可以在变量中捕获空格？

2 个答案:

使用Bash正则表达式测试