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This is a painful curl --fail limitation. From man curl :

-f, --fail (HTTP) Fail silently (no output at all) on server errors

But there is no way to get both the non-zero return code AND the response body in stdout.

Based on pvandenberk's answer and this other very useful trick learned on SO, here is a workaround :

curl_with_error_code () {
    _curl_with_error_code "$@" | sed '$d'
}
_curl_with_error_code () {
    local curl_error_code http_code
    exec 17>&1
    http_code=$(curl --write-out '\n%{http_code}\n' "$@" | tee /dev/fd/17 | tail -n 1)
    curl_error_code=$?
    exec 17>&-
    if [ $curl_error_code -ne 0 ]; then
        return $curl_error_code
    fi
    if [ $http_code -ge 400 ] && [ $http_code -lt 600 ]; then
        echo "HTTP $http_code" >&2
        return 127
    fi
}

This function behaves exactly as curl, but will return 127 (a return code non-used by curl) in case of a HTTP code in the range [400, 600[.

This is a painful curl --fail limitation. From man curl :

-f, --fail (HTTP) Fail silently (no output at all) on server errors

But there is no way to get both the non-zero return code AND the response body in stdout.

Based on pvandenberk's answer and this other very useful trick learned on SO, here is a workaround :

curl_with_error_code () {
    _curl_with_error_code "$@" | sed '$d'
}
_curl_with_error_code () {
    local curl_error_code http_code
    exec 17>&1
    http_code=$(curl --write-out '\n%{http_code}\n' "$@" | tee /dev/fd/17 | tail -n 1)
    curl_error_code=$?
    exec 17>&-
    if [ $curl_error_code -ne 0 ]; then
        return $curl_error_code
    fi
    if [ $http_code -ge 400 ] && [ $http_code -lt 600 ]; then
        echo "HTTP $http_code" >&2
        return 127
    fi
}

This function behaves exactly as curl, but will return 127 (a return code non-used by curl) in case of a HTTP code in the range [400, 600[.

This is a painful curl --fail limitation. From man curl :

-f, --fail (HTTP) Fail silently (no output at all) on server errors

But there is no way to get both the non-zero return code AND the response body in stdout.

Based on pvandenberk's answer and this other very useful trick learned on SO, here is a workaround :

curl_with_error_code () {
    _curl_with_error_code "$@" | sed '$d'
}
_curl_with_error_code () {
    local curl_error_code http_code
    exec 17>&1
    http_code=$(curl --write-out '\n%{http_code}\n' "$@" | tee /dev/fd/17 | tail -n 1)
    curl_error_code=$?
    exec 17>&-
    if [ $curl_error_code -ne 0 ]; then
        return $curl_error_code
    fi
    if [ $http_code -ge 400 ] && [ $http_code -lt 600 ]; then
        echo "HTTP $http_code" >&2
        return 127
    fi
}

This function behaves exactly as curl, but will return 127 (a return code non-used by curl) in case of a HTTP code in the range [400, 600[.

This is a painful curl --fail limitation. From man curl :

-f, --fail (HTTP) Fail silently (no output at all) on server errors

But there is no way to get both the non-zero return code AND the response body in stdout.

Based on pvandenberk's answer and this other very useful trick learned on SO, here is a workaround :

curl_with_error_code () {
    _curl_with_error_code "$@" | sed '$d'
}
_curl_with_error_code () {
    local curl_error_code http_code
    exec 17>&1
    http_code=$(curl --write-out '\n%{http_code}\n' "$@" | tee /dev/fd/17 | tail -n 1)
    curl_error_code=$?
    exec 17>&-
    if [ $curl_error_code -ne 0 ]; then
        return $curl_error_code
    fi
    if [ $http_code -ge 400 ] && [ $http_code -lt 600 ]; then
        echo "HTTP $http_code" >&2
        return 127
    fi
}

This function behaves exactly as curl, but will return 127 (a return code non-used by curl) in case of a HTTP code in the range [400, 600[.

This is a painful curl --fail limitation. From man curl :

-f, --fail (HTTP) Fail silently (no output at all) on server errors

But there is no way to get both the non-zero return code AND the response body in stdout.

Based on pvandenberk's answer and this other very useful trick learned on SO, here is a workaround :

curl_with_error_code () {
    local curl_error_code http_code
    exec 17>&1
    http_code=$(curl --write-out '\n%{http_code}\n' "$@" | tee /dev/fd/17 | tail -n 1)
    curl_error_code=$?
    exec 17>&-
    if [ $curl_error_code -ne 0 ]; then
        return $curl_error_code
    fi
    if [ $http_code -ge 400 ] && [ $http_code -lt 600 ]; then
        echo "HTTP $http_code" >&2
        return 127
    fi
}

This function behaves exactly as curl, but will return 127 (a return code non-used by curl) in case of a HTTP code in the range [400, 600[.

This is a painful curl --fail limitation. From man curl :

-f, --fail (HTTP) Fail silently (no output at all) on server errors

But there is no way to get both the non-zero return code AND the response body in stdout.

Based on pvandenberk's answer and this other very useful trick learned on SO, here is a workaround :

curl_with_error_code () {
    _curl_with_error_code "$@" | sed '$d'
}
_curl_with_error_code () {
    local curl_error_code http_code
    exec 17>&1
    http_code=$(curl --write-out '\n%{http_code}\n' "$@" | tee /dev/fd/17 | tail -n 1)
    curl_error_code=$?
    exec 17>&-
    if [ $curl_error_code -ne 0 ]; then
        return $curl_error_code
    fi
    if [ $http_code -ge 400 ] && [ $http_code -lt 600 ]; then
        echo "HTTP $http_code" >&2
        return 127
    fi
}

This function behaves exactly as curl, but will return 127 (a return code non-used by curl) in case of a HTTP code in the range [400, 600[.