我发现这个问题是因为我想要独立访问响应和内容，以便为用户添加一些错误处理。

Curl允许您自定义输出。您可以打印HTTP状态代码以std输出并将内容写入另一个文件。

curl -s -o response.txt -w "%{http_code}" http://example.com

这允许您检查返回代码，然后决定是否值得打印、处理、记录响应等。

http_response=$(curl -s -o response.txt -w "%{http_code}" http://example.com)
if [ $http_response != "200" ]; then
    # handle error
else
    echo "Server returned:"
    cat response.txt    
fi

%{http_code}是一个由curl代替的变量。你可以做更多的事情，或者发送代码到stderr，等等。参见curl手册和——write-out选项。

-w, --write-out Make curl display information on stdout after a completed transfer. The format is a string that may contain plain text mixed with any number of variables. The format can be specified as a literal "string", or you can have curl read the format from a file with "@filename" and to tell curl to read the format from stdin you write "@-". The variables present in the output format will be substituted by the value or text that curl thinks fit, as described below. All variables are specified as %{variable_name} and to output a normal % you just write them as %%. You can output a newline by using \n, a carriage return with \r and a tab space with \t. The output will be written to standard output, but this can be switched to standard error by using %{stderr}.

https://man7.org/linux/man-pages/man1/curl.1.html

-i选项是你想要的:

curl -i http://localhost

在输出中包含协议头(H/F)

或者你可以使用verbose选项:

curl -v http://localhost

-v，——verbose使操作更健谈

要获得响应代码和响应:

$ curl -kv https://www.example.org

要得到响应代码:

$ curl -kv https://www.example.org 2>&1 | grep -i 'HTTP/1.1 ' | awk '{print $3}'| sed -e 's/^[ \t]*//'

2>&1:错误存储在输出中，用于解析 Grep:从输出中过滤响应代码行 Awk:从响应代码行中过滤响应代码 Sed:删除前导空白

一个清晰的使用管道读取

function cg(){
    curl -I --silent www.google.com | head -n 1 | awk -F' ' '{print $2}'
}
cg
# 200

欢迎在这里使用我的dotfile脚本

解释

——silent:使用管道时不显示进度条 head -n 1:只显示第一行 -F' ':使用分隔符分隔文本 '{print $2}':显示第二列

哇，这么多答案，cURL开发人员肯定把它留给了我们作为家庭练习:)好吧，这是我的想法-一个脚本，使cURL工作，因为它应该是，即:

像cURL那样显示输出。 如果HTTP响应代码不在2XX范围内，则使用非零代码退出

保存为curl-wrapper.sh:

#!/bin/bash

output=$(curl -w "\n%{http_code}" "$@")
res=$?

if [[ "$res" != "0" ]]; then
  echo -e "$output"
  exit $res
fi

if [[ $output =~ [^0-9]([0-9]+)$ ]]; then
    httpCode=${BASH_REMATCH[1]}
    body=${output:0:-${#httpCode}}

    echo -e "$body"

    if (($httpCode < 200 || $httpCode >= 300)); then
        # Remove this is you want to have pure output even in 
        # case of failure:
        echo
        echo "Failure HTTP response code: ${httpCode}"
        exit 1
    fi
else
    echo -e "$output"
    echo
    echo "Cannot get the HTTP return code"
    exit 1
fi

所以它就像往常一样，但不是curl do ./curl-wrapper.sh:

所以当结果在200-299范围内时:

./curl-wrapper.sh www.google.com 
# ...the same output as pure curl would return...
echo $?
# 0

当结果超出200-299范围时:

./curl-wrapper.sh www.google.com/no-such-page
# ...the same output as pure curl would return - plus the line
#    below with the failed HTTP code, this line can be removed if needed:
#
# Failure HTTP response code: 404
echo $?
# 1

只是不要传递“-w|——write-out”参数，因为这是脚本中添加的内容

我可以通过查看curl文档来得到一个解决方案，该文档指定使用-来将输出输出到stdout。

curl -o - -I http://localhost

要用http返回代码获得响应，我可以这样做

curl -o /dev/null -s -w "%{http_code}\n" http://localhost