Bash：如何获取变量内容中出现的第一个数字

Question

如何获得第一个变量数

我有一个变量：

STR="My horse weighs 3000 kg but the car weighs more"
STR="Maruska found 000011 mushrooms but only 001 was not with meat"
STR="Yesterday I almost won the lottery 0000020 CZK but in the end it was only 05 CZK"

我需要得到数字：

3000
11
20

Answer 1

这是一种方法：

echo $STR | grep -o -E '[0-9]+' | head -1 | sed -e 's/^0\+//'

测试：

$ STR="My horse weighs 3000 kg but the car weighs more"
$ echo $STR | grep -o -E '[0-9]+' | head -1 | sed -e 's/^0\+//'
3000

$ STR="Maruska found 000011 mushrooms but only 001 was not with meat"
$ echo $STR | grep -o -E '[0-9]+' | head -1 | sed -e 's/^0\+//'
11

$ STR="Yesterday I almost won the lottery 0000020 CZK but in the end it was only 05 CZK"
$ echo $STR | grep -o -E '[0-9]+' | head -1 | sed -e 's/^0\+//'
20

Answer 2

使用 gawk，将记录分隔符设置RS为数字序列。RS可以通过 检索与模式匹配的文本RT。添加0到RT强制它为一个数字（从而删除前导零）。打印第一个实例后立即退出

awk -v RS=[0-9]+ '{print RT+0;exit}' <<< "$STR"

或者这里是一个 bash 解决方案

shopt -s extglob
read -r Z _ <<< "${STR//[^[:digit:] ]/}"
echo ${Z##+(0)}

Answer 3

查找正则表达式和man grep.

echo $STR | grep -o [0-9]*

要删除前导零，请将其视为数字：

LIT=$(echo $STR | grep -o [0-9]*)
VAL=$(expr $LIT + 0)
echo $VAL