我将要解决以下问题:
我有一个文件,其内容如下:
1521471079313,219,HTTP Request 14,200,OK,PROD 50 rpm 1-10,text,true,,17665,204,1,1,177,0,35
1521471080337,263,HTTP Request 11,200,OK,PROD 50 rpm 1-10,text,true,,30268,202,1,1,169,0,0
1521471081404,245,HTTP Request 12,200,OK,PROD 50 rpm 1-10,text,true,,5134,201,1,1,210,0,37
1521471082453,125,HTTP Request 13,200,OK,PROD 50 rpm 1-10,text,true,,8910,201,1,1,106,0,0
1521471083381,217,HTTP Request 14,200,OK,PROD 50 rpm 1-10,text,true,,17665,204,1,1,188,0,0
1521471084402,303,HTTP Request 11,200,OK,PROD 50 rpm 1-10,text,true,,30268,202,1,1,226,0,41
- 列表中的第一项是epoc时间戳,我想将其转换为可读格式。
- 我尝试了以下命令:
cat file.csv|sed -E "s/^([0-9]*)(,.*)/$(date -r \1 '+%m-%d-%Y:%H:%M:%S')\2/p"
看起来它可以工作,但是我发现它会将其转换为:
01-01-1970:01:00:01,245,HTTP Request 13,200,OK,PROD 50 rpm 1-10,text,true,,8910,201,1,1,219,0,43
01-01-1970:01:00:01,276,HTTP Request 14,200,OK,PROD 50 rpm 1-10,text,true,,17665,204,1,1,217,0,0
01-01-1970:01:00:01,276,HTTP Request 14,200,OK,PROD 50 rpm 1-10,text,true,,17665,204,1,1,217,0,0
01-01-1970:01:00:01,242,HTTP Request 11,200,OK,PROD 50 rpm 1-10,text,true,,30268,202,1,1,216,0,34
01-01-1970:01:00:01,242,HTTP Request 11,200,OK,PROD 50 rpm 1-10,text,true,,30268,202,1,1,216,0,34
01-01-1970:01:00:01,147,HTTP Request 12,200,OK,PROD 50 rpm 1-10,text,true,,5134,201,1,1,119,0,0
01-01-1970:01:00:01,147,HTTP Request 12,200,OK,PROD 50 rpm 1-10,text,true,,5134,201,1,1,119,0,0
所有的时间戳都看起来像是“时光之初 :-)”,而不是我想要的格式。
我知道在sed命令中有一个命令替换,并且还有两个基于前面正则表达式的组引用,但为什么它不起作用让我感到困惑。
awk:调用未定义的函数strftime
。 - Ivonet