Shell script return value with which output

I want to return to one false that will do , The result is a node Path out , The script is as follows ：

#!/bin/bash
# Tips “ Please enter a name ” And wait for 30 second , Save user input into variables name in 
read -t 30 -p " Please enter your mobile number :" phone

# Tips “ Please input a password ” And wait for 30 second , Save user input into variables pwd in , Input content is hidden 
read -t 30 -s -p " Please enter the user password :" password
echo -e "\n"

# Login interface 
loginRes=`curl -X POST  "'Content-type':'application/json'" -d '{"api":"login.pw","params":{"loginId":'$phone',"password":'$password',"appId":"yptb"},"nonce":"uuid","sign":"sign","version":"1.0","timestamp":"230993499430010"}' https://apigw.xxxx.com/request`
echo " Login return information ：$loginRes"

function getJsonValueByNode() {
    if which node; then
        local json="$1"
        local key="$2"
        node -pe "JSON.stringify(JSON.parse(process.argv[1]).$key)" "$json" | awk '{print ($0 == "undefined" ? "null" : $0)}'
        return 0
    else
        return 1
    fi
}

value=$(getJsonValueByNode $loginRes "result.success")

echo " Variable ：$value"

Solution ：

take which Of output to /dev/null in , Reference resources ：How can I check if a program exists from a Bash script? - Stack Overflow