Curl返回http状态代码以及响应

I use curl to get http headers to find http status code and also return response. I get the http headers with the command

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

curl -I http://localhost

To get the response, I use the command

要获得响应，我使用命令

curl http://localhost

As soon as use the -I flag, I get only the headers and the response is no longer there. Is there a way to get both the http response and the headers/http status code in in one command?

一旦使用-I标志，我只获得标题，响应不再存在。有没有办法在一个命令中同时获取http响应和headers / http状态代码？

4 个解决方案

#1

I was able to get a solution by looking at the curl doc which specifies to use - for the output to get the output to stdout.

我能够通过查看指定使用的curl doc获得解决方案 - 输出将输出输出到stdout。

curl -o - http://localhost

To get the response with just the http return code, I could just do

要获得只有http返回码的响应，我可以这么做

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

#2

I use this command to print the status code without any other output. Additionally, it will only perform a HEAD request and follow the redirection (respectively -I and -L).

我使用此命令打印状态代码而没有任何其他输出。此外，它只执行HEAD请求并遵循重定向（分别为-I和-L）。

curl -o -I -L -s -w "%{http_code}" http://localhost

This makes it very easy to check the status code in a health script:

这使得检查运行状况脚本中的状态代码变得非常容易：

sh -c '[ $(curl -o -I -L -s -w "%{http_code}" http://localhost) -eq 200 ]'

#3

I have used this :

我用过这个：

    request_cmd="$(curl -i -o - --silent -X GET --header 'Accept: application/json' --header 'Authorization: _your_auth_code==' 'https://example.com')"

To get the HTTP status

获取HTTP状态

    http_status=$(echo "$request_cmd" | grep HTTP |  awk '{print $2}')
    echo $http_status

To get the response body I've used this

为了获得响应主体我已经使用过了

    output_response=$(echo "$request_cmd" | grep body)
    echo $output_response

#4

the verbose mode will tell you everything

详细模式会告诉你一切

curl -v http://localhost

#1