在utx awk中將utc時間戳轉換爲est

Question

我在unix中搜索一系列文件，但問題在於我的輸入數據是UTC格式我必須將冷杉轉換爲EST，然後應用過濾器是否有辦法在unix awk中？

awk '$2FS >="2017-02-21T11:09:41.000Z" && $2FS <="2017-02-23T11:09:41.000Z"' 

有什麼一些事情，如下面很簡單：

awk 'convert_to_date($2FS,'EST') >="2017-02-21T11:09:41.000Z" 
&& convert_to_date($2FS,'EST') <="2017-02-23T11:09:41.000Z"' 

Answer 1

你可以試試下面的bash解決方案 -

貓TIME.TXT

腳本

1 2017-02-22T11:09:41.000Z 
2 2017-02-23T11:09:41.000Z 
3 2017-02-24T11:09:41.000Z 
4 2017-02-25T11:09:41.000Z 
5 2017-02-26T11:09:41.000Z 



    cat ./time.sh 
#!/bin/bash 
d="2017-02-21T11:09:41.000Z" 
b="2017-02-23T11:09:41.000Z" 
USERTIME1=`awk -F'[T]|[-]|[:]|[.]' '{datetime=mktime($1 " "$2 " " $3 " " $4 " " $5 " " $6); print datetime}' <<< $d` 
USERTIME2=`awk -F'[T]|[-]|[:]|[.]' '{datetime=mktime($1 " "$2 " " $3 " " $4 " " $5 " " $6); print datetime}' <<< $b` 
for i in `cat time.txt|awk '{print $2}'` 
do 
var=`awk 'BEGIN{FS="[T]|[-]|[:]|[.]"} {print mktime($1 " "$2 " " $3 " " $4 " " $5 " " $6)}' <<< $i` 
var1=`expr $var - 21600` 
if [ $var1 -ge $USERTIME1 ] && [ $var1 -le $USERTIME2 ] 
then 
echo "$i" 
fi 
done 

輸出：

./time.sh 
2017-02-22T11:09:41.000Z 
2017-02-23T11:09:41.000Z 

我試圖用awk解決這個問題，但沒有成功，如果任何人都可以幫忙。這裏是代碼 -

### 1:11 AM Friday, Eastern Time (ET) is 
### 6:11 AM Friday, Coordinated Universal Time (UTC) 
### UTC to EST is 6 hour of difference means (-21600) 
BEGIN{FS="[T]|[-]|[:]|[.]";d="2017-02-25T11:09:41.000Z";b="2017-02-23T11:09:41.000Z"} 
{ 
USERTIME1=(substr(d,0,4) " " substr(d,6,2) " "substr(d,9,2) " " substr(d,12,2) " " substr(d,15,2) " "substr(d,18,2) " " substr(d,21,2)) 
USERTIME2=(substr(b,0,4) " " substr(b,6,2) " "substr(b,9,2) " " substr(b,12,2) " " substr(b,15,2) " "substr(b,18,2) " " substr(b,21,2)) 
UTCTIME=($1 " "$2 " " $3 " " $4 " " $5 " " $6) 
ESTTIME=(UTCTIME-21600) 
if((mktime(ESTTIME) >= mktime(USERTIME1)) && mktime((ESTTIME) <= mktime(USERTIME2)) 
print 
}