Question

我在使用curl命令时遇到了一些问题。例如

$ curl --connect-timeout 2000  -v -X POST ....
* About to connect() to ************************
*   Trying ::1... Connection refused
*   Trying 127.0.0.1... connected
* Connected to ********************
> POST ********************** HTTP/1.1
> User-Agent: curl/7.19.7 (x86_64-redhat-linux-gnu) libcurl/7.19.7 NSS/3.12.9.0 zlib/1.2.3 libidn/1.18 libssh2/1.2.2
> Host: ******
> Accept: */*
> Content-Type: application/json
> Content-Length: 92
>
< HTTP/1.1 200 OK
< ****************:
< X-b: **************
< X-a: ***************
< Content-Type: application/json
< Content-Length: 8228
< Server: Jetty(6.1.26)


....
\"f1\":\"v1\"},\"f2\":{\"f3\":false},\"f4\":\"v4\",\"f5\":fa* Connection #0 to host localhost left intact
* Closing connection #0
lse,\"f6\":25}}"}}

注意，问题是：

json引用了\
内容已被Connection #0 to host localhost left intact Closing connection #0

我的问题是：如何摆脱这些问题？

Answer 1

出现错误消息，因为您可能加入了标准输出和标准错误。尝试重定向标准错误：

curl 2>/dev/null

要取消查询字符串，请参阅此处：Linux/Bash: How to unquote?

关于linux的curl命令打破json文本并获得反斜杠

1 个答案: