如何在 bash 中的某些字符之间获取 2 个字符串

Question

String:

echo "40125512|abcd32External_SOC=ALPHA3;PCRFabcran"

I want to grab everything before the first instance of |and everything between External_SOC and;PCRF. And store them as 2 different variables if possible.

x=40125512
y=ALPHA3

This gives me the following:

sed -e 's/|.*External_SOC=\(.*\);PCRF.*/\1/'

40125512ALPHA3

Answer 1

EDIT: As per OP it needs to be done in a single line creation 2 of variables if this is the case then try following.

read -r x y <<<$(echo "40125512|abcd32External_SOC=ALPHA3;PCRFabcran" | sed 's/\([^|]*\).*=\([^;]*\).*/\1 \2/')
echo "$x"
40125512
echo "$y"
ALPHA3

OR use following as per anubhav sir's comment:

read x y < <(sed -E 's~^([^|]+)\|.*External_SOC=(.+);PCRF.*~\1 \2~' <<< "40125512|abcd32External_SOC=ALPHA3;PCRFabcran")

---

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Could you please try following. One could use it from 2 separate commands to create 2 separate variables.

x=$(echo "40125512|abcd32External_SOC=ALPHA3;PCRFabcran" | sed 's/\([^|]*\).*/\1/')
echo "$x"
40125512

y=$(echo "40125512|abcd32External_SOC=ALPHA3;PCRFabcran" | sed 's/.*=\([^;]*\).*/\1/')
echo "$y"
ALPHA3