bash with curl not return nothing

API Spot/Margin API
1

hi i have this script

#!/bin/bash
urlencode() {
local string=“$1”
local length=“${#string}”
local result=“”

for (( i = 0; i < length; i++ )); do
    local c="${string:$i:1}"
    case $c in
        [-_.~a-zA-Z0-9]) result+="$c" ;;
        *) result+="%$(printf '%02X' "'$c")" ;;
    esac
done

echo "$result"

}

_Bin_SymbolPriceTicker() {
local apiUrl=“$ServerBinance/ticker/price”

# Verifica la presenza di symbol o symbols
if [[ "$1" == *","* ]]; then
    IFS=',' read -ra symbols <<< "$1"
    local encodedSymbols=$(IFS=,; printf "\"%s\"," "${symbols[@]//\"/%22}")
    apiUrl+="?symbols=[$(echo "${encodedSymbols%,}" | sed 's/,$//')]"
    shift
elif [ -n "$1" ]; then
    apiUrl+="?symbol=%22$1%22"
    shift
fi

# Visualizza l'URL costruito prima della chiamata di curl
echo "URL: $apiUrl"

response=$(curl -s -X GET "$apiUrl")

}
echo curl return curl -s -X GET ‘https://api.binance.com/api/v3/ticker/price?symbols=[“BTCUSDT”,“BNBBTC”]’
but if i call also in shell not work why ? but if i insert url in browser work o_O anyone can help me ? thanks

2

Hey,
what happens if you replace curl -s -X GET with curl -s --globoff?

3

thanks for rply i resolve in this mode i encode " and , with %22 … thanks