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- Using awk with variables 3 answers
- 使用awk和变量3的答案
I am trying to write a shell script to get certain data from below sample logs..
我正在尝试编写一个shell脚本，从下面的示例日志中获取某些数据。
Below is a sample log:
下面是一个样本日志:
2014-07-08 16:08:25,684: |ABC_130|1|10123ffffff2|P|489440201
2014-07-08 17:08:25,684: |ABC_130|1|aaaaaxxxxaab|P|489440201
2014-07-08 19:08:25,684: |ABC_130|1|aaaaababbaab|P|489440201
Below is a part of the script where I am facing issue, the issue I am facing is that the awk command doesn't give any output.
下面是我面临问题的脚本的一部分，我面临的问题是awk命令不提供任何输出。
#!/bin/sh
DATE_HOUR="`date -d '1 hour ago' "+%Y-%m-%d %H"`"
awk -F ":" '{if ($1='"$DATE_HOUR"') print $0}' log.txt
1 个解决方案
#1
3
Don't use shell variable like that in awk. Use -v name=val:
不要在awk中使用shell变量。使用- v名称= val:
awk -F ":" -v dt="$DATE_HOUR" '$1==dt' log.txt
btw I reduced your awk command to '$1==dt' since print $0 is default action and also if condition can be moved out of curly braces.
顺便说一下，我将awk命令减少到“$1== =dt”，因为打印$0是默认操作，如果条件可以从花括号中移出的话。
#1
3
Don't use shell variable like that in awk. Use -v name=val:
不要在awk中使用shell变量。使用- v名称= val:
awk -F ":" -v dt="$DATE_HOUR" '$1==dt' log.txt
btw I reduced your awk command to '$1==dt' since print $0 is default action and also if condition can be moved out of curly braces.
顺便说一下，我将awk命令减少到“$1== =dt”，因为打印$0是默认操作，如果条件可以从花括号中移出的话。