[英]How to extract string field value from a URL using regular expression in shell script?
我正在一个项目中,需要从bash shell脚本对我的服务器之一进行url调用。
http://hostname.domain.com:8080/beat
点击上面的URL后,我将得到下面的响应,我需要解析它并从中提取state
值
num_retries_allowed: 3 count: 30 count_behind: 100 state: POST_INIT num_rounds: 60 hour_col: 2 day_col: 0
现在,我想使用正则表达式提取state
变量值。 我能够从中提取count
和count_behind
值,但不确定如何从中提取state
值。
#send the request, put response in variable
DATA=$(wget -O - -q -t 1 http://hostname.domain.com:8080/beat)
#grep $DATA for count and count_behind
COUNT=$(echo $DATA | grep -oE 'count: [0-9]+' | awk '{print $2}')
COUNT_BEHIND=$(echo $DATA | grep -oE 'count_behind: [0-9]+' | awk '{print $2}')
# how to extract state variable value here?
STATE= what do I add here?
另外如果在$DATA
如果state
变量不存在,那么我想将0
赋给STATE
变量。 之后,我想验证条件并根据该条件退出脚本。
如果STATE
等于POST_INIT
则退出shell脚本成功,或者STATE
等于0,然后也成功退出。
#verify conditionals
if [[ $STATE -eq "POST_INIT" || $STATE -eq "0" ]]; then exit 0; fi
您可以使用此grep -P
:
state=$(grep -oP 'state: \K\S+' <<< "$DATA")
[[ -z "$state" ]] && state=0
echo "$state"
POST_INIT
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