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

一行代码,只是为了获得状态代码:

curl -s -i https://www.google.com | head -1

将其更改为head -2也会给出时间。


如果你想要一个while-true循环,它会是:

URL="https://www.google.com"

while true; do
    echo "------"
    curl -s -i $URL | head -2
    sleep 2;
done

直到您执行cmd+C(或Windows中的ctrl+C)。

------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:38 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:41 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:43 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:45 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:47 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:49 GMT

其他回答

一行代码,只是为了获得状态代码:

curl -s -i https://www.google.com | head -1

将其更改为head -2也会给出时间。


如果你想要一个while-true循环,它会是:

URL="https://www.google.com"

while true; do
    echo "------"
    curl -s -i $URL | head -2
    sleep 2;
done

直到您执行cmd+C(或Windows中的ctrl+C)。

------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:38 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:41 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:43 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:45 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:47 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:49 GMT

根据我的经验,我们通常这样使用旋度

curl -f http://localhost:1234/foo || exit 1

curl: (22) The requested URL returned error: 400 Bad Request

这样我们就可以在curl失败时使用管道,并且它还显示状态代码。

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

在末尾追加一行“http_code:200”,然后grep关键字“http_code:”并提取响应代码。

result=$(curl -w "\nhttp_code:%{http_code}" http://localhost)

echo "result: ${result}"   #the curl result with "http_code:" at the end

http_code=$(echo "${result}" | grep 'http_code:' | sed 's/http_code://g') 

echo "HTTP_CODE: ${http_code}"  #the http response code

在这种情况下,您仍然可以使用非静默模式/ verbose模式来获取有关请求的更多信息,例如curl响应体。

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