如何让 curl 仅输出 http 响应正文（json）而没有其他标头等
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how to get curl to output only http response body (json) and no other headers etc
提问by Rishi
I am using curl in a bash script to fetch the response of a service as below,
我在 bash 脚本中使用 curl 来获取服务的响应，如下所示，
response=$(curl -isb -H "Accept: application/json" "http://host:8080/some/resource")
Service response is of json type and on browser I could perfectly fine response.
However curl response has other unwanted things (such as set-cookie, content-length header in this case) and sometimes the actual response is eaten up.
服务响应是 json 类型，在浏览器上我可以完美响应。
但是 curl 响应还有其他不需要的东西（例如在这种情况下的 set-cookie、content-length 标头），有时实际的响应会被吃掉。
Here is the output of echo $response>
这是echo $response>的输出
Set-Cookie: rack.session=BAh7CEkiD3Nlc3Npb25faWQGOgZFVEkiRWJlY2JiOTE2M2Q1ZWI4NThjMDdi%0AYjRiOWRjMGMxMGEwYTBkMjE3NmJhZDVjYzY4YjY4ZTlmMTE2ZGVkYWE3MTMG%0AOwBGS
SIJY3NyZgY7AEZJIiVhZmQ2MmUyZGMxMzFmOGEwMjg3NDlhNWM3YmVm%0AN2FjNwY7AEZJIg10cmFja2luZwY7AEZ7B0kiFEhUVFBfVVNFUl9BR0VOVAY7%0AAFRJIi00MTc0OGM2MWNkMzljZTYxNzY3ZjU0
Y2I5OTdiYWRkN2MyNTBkYmU4%0ABjsARkkiGUhUVFBfQUNDRVBUX0xBTkdVQUdFBjsAVEkiLWRhMzlhM2VlNWU2%0AYjRiMGQzMjU1YmZlZjk1NjAxODkwYWZkODA3MDkGOwBG%0A--ee97a62095e7d42129
tontent-Length: 354c8; path=/; HttpOnly
This is breaking my response parsing logic.
I have seen this happening intermittently which is weird.
这打破了我的响应解析逻辑。
我已经看到这种情况间歇性地发生，这很奇怪。
Is there a way to get "only"json response from curl output?
I went through the curl documentation but could not see any thing/ or I could have missed it.
Appreciate any help! Thx
有没有办法从 curl 输出中获得“唯一”的json 响应？
我浏览了 curl 文档，但看不到任何东西/或者我可能会错过它。
感谢任何帮助！谢谢
回答by Remy Lebeau
You are specifying the -ioption:
您正在指定-i选项：
-i, --include
(HTTP) Include the HTTP-header in the output. The HTTP-header includes things like server-name, date of the document, HTTP-version and more...
-i, --include
(HTTP)在输出中包含 HTTP 标头。HTTP 标头包括服务器名称、文档日期、HTTP 版本等内容...
Simply remove that option from your command line:
只需从命令行中删除该选项：
response=$(curl -sb -H "Accept: application/json" "http://host:8080/some/resource")
回答by Victor VIet
#!/bin/bash
req=$(curl -s -X GET http://host:8080/some/resource -H "Accept: application/json") 2>&1
echo "${req}"
