Json Shell脚本空格转义

Edit:本问题已根据背景情况进行了编辑.1.将原问题中的sh和2.sh脚本分别替换为它们的真实实现:monitor or.sh和api-check.sh.

Function:个

function checkAlive() {
    script="$1"
    args="${@:2}"
    startCheck=$(date +%s)
    echo "$@"
    tempResult="$( "$@" )"
    T_RC="$?"
    if [ "$T_RC" -ne 0 ]; then
        if [ "$result" = "" ]; then
            result="ERRORS: $tempResult"
        else
            result="$result ,, $tempResult"
        fi
    fi
    RC="$((RC > T_RC ? RC : T_RC))"
    timePassed
}

api-check.sh:个

method="$1"
url="https://$2"
jwt="$3"
body="$4"
echo "$body"

if ! response="$(curl -s --request "$method" "$url" -w "%{http_code}" --header \
    'Authorization: Bearer '"$jwt" --header 'Content-Type: application/json' --header 'Accept: application/json' --data-raw "$body")"; then
    echo "$url curl error getting code: $response"
    exit 2
fi

monitor.sh:个

api_url="foo"
oauth_token"bar"
body='{"address":"Test address"}'
checkAlive "api-check.sh" '"POST" "'$api_url'" "'$oauth_token'" '"'$body'"''

Execution Flow:个

• Sh将字符串json Body作为参数传递给check Alive函数(如上所定义).
• 在check Alive中，echo "$@"返回:

api-check.sh "POST" "foo" "bar" '{"address":"Test address"}'

• 由于再次出现转义问题，api-check.sh脚本永远不会正确执行

Question:个 我应该将什么json主体作为参数传递给check Alive才能正确转义OR我需要修改哪行代码以及在哪个脚本中才能正确地计算json主体？