我使用curl来获取http报头以查找http状态代码并返回响应。我使用命令获取http头信息

curl -I http://localhost

为了得到响应,我使用命令

curl http://localhost

一旦使用了-I标志,我就只得到了头信息,响应就不再存在了。是否有一种方法可以同时获得http响应和头/http状态码在一个命令?


当前回答

我的方法是:

为了获得两者(标题和正文),我通常执行curl - d - <url>,如:

$ curl -D- http://localhost:1234/foo
HTTP/1.1 200 OK
Connection: Keep-Alive
Transfer-Encoding: chunked
Content-Type: application/json
Date: Wed, 29 Jul 2020 20:59:21 GMT

{"data":["out.csv"]}

这将转储标题(- d)到stdout(-)(在man curl中查找——dump-header)。

恕我直言,在这种情况下也很方便:

我经常使用jq来获得json数据(例如从一些其他api)格式化。但是由于jq不需要HTTP标头,诀窍是使用-D/dev/stderr将标头打印到stderr。注意,这次我们还使用-sS(——silent,——show-errors)来抑制进度表(因为我们写入管道)。

$ curl -sSD/dev/stderr http://localhost:1231/foo | jq .
HTTP/1.1 200 OK
Connection: Keep-Alive
Transfer-Encoding: chunked
Content-Type: application/json
Date: Wed, 29 Jul 2020 21:08:22 GMT

{
  "data": [
    "out.csv"
  ]
}

我想这也可以很方便,如果你想打印头部(快速检查)控制台,但重定向主体到一个文件(例如,当它的某种二进制不混乱你的终端):

$ curl -sSD/dev/stderr http://localhost:1231 > /dev/null
HTTP/1.1 200 OK
Connection: Keep-Alive
Transfer-Encoding: chunked
Content-Type: application/json
Date: Wed, 29 Jul 2020 21:20:02 GMT

注意:这与curl -I <url>!As -I将执行HEAD请求而不是GET请求(在man curl中查找——HEAD。是:对于大多数HTTP服务器,这将产生相同的结果。但是我知道很多商业应用程序根本没有实现HEAD请求;-P

其他回答

详细模式会告诉你一切

curl -v http://localhost
while : ; do curl -sL -w "%{http_code} %{url_effective}\\n" http://host -o /dev/null; done

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

这对我来说很管用:

curl -Uri 'google.com' | select-object StatusCode

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