如何在CURL Shell脚本中捕获超时/错误? [英] How to catch timeout/errors in a CURL shell script?
问题描述
我想向URL发出cURL GET请求,并根据返回的HTTP代码确定是否对返回的HTTP数据进行处理。
I want to issue a cURL GET request to a URL and based on the return HTTP code decide whether to do something with the returned HTTP data.
例如,如果通过cURL进行的某个url请求的HTTP代码有效(没有超时或错误),则将请求返回的数据保存在我的系统中。
For example, if the HTTP code for a certain url request through cURL is valid (not timed out or error), then keep the returned data from the request somewhere in my system.
如何我可以实际捕获返回的HTTP代码(或超时)并基于此做出决定吗?
How can I actually 'catch' the returned HTTP code (or timeout) and do the decision based on that?
推荐答案
执行以下命令 script.sh http://www.google.com/
。
-D
-将标头转储到文件
-o
-将响应写入文件
-s
-保持沉默
-w
-显示指定变量的值
Execute following as script.sh http://www.google.com/
.
-D
- dump headers to file
-o
- write response to file
-s
- be silent
-w
- display value of specified variables
#!/bin/bash
RESPONSE=response.txt
HEADERS=headers.txt
status=$(curl -s -w %{http_code} $1 -o $RESPONSE)
# or
#curl -s -D $HEADERS $1 -o $RESPONSE
#status=$(cat $HEADERS | head -n 1 | awk '{print $2}')
echo $status
使用 $ status
和 $ RESPONSE
进行进一步处理。
Use $status
and $RESPONSE
for further processing.
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