bash / sed / awk从文本文件中删除或gsub时间戳模式

Question

I have a text file like this:

1/7/2017 12:53  DROP TABLE table1                                                   
1/7/2017 12:53  SELECT  

1/7/2017 12:55  --UPDATE #dat_recency SET
Select * from table 2
into table 3;

I'd like to remove all of the timestamp patterns ( M/D/YYYY HH:MM , M/DD/YYYY HH:MM , MM/D/YYYY HH:MM , MM/DD/YYYY HH:MM ). I can find the patterns using grep but can't figure out how to use gsub . Any suggestions?

DESIRED OUTPUT:

DROP TABLE table1                                                   
SELECT  

--UPDATE #dat_recency SET
Select * from table 2
into table 3;

Answer 1

You can use this sed command to remove data/time stamps from line start:

sed -i.bak -E 's~([0-9]{1,2}/){2}[0-9]{4} [0-9]{2}:[0-9]{2} *~~' file

cat file

DROP TABLE table1
SELECT

--UPDATE #dat_recency SET
Select * from table 2
into table 3;

Answer 2

Use the default space separator, make first and second columns to empty string and then print the whole line.

awk '/^[0-9]/{$1=$2="";gsub(/^[ \t]+|[ \t]+$/, "")} !/^[0-9]/{print}' sample.csv

the command checks each line whether starts with numeric or not, if it is replace the first 2 columns with empty strings and remove leading spaces; otherwise print the original line.

output:

DROP TABLE table1
SELECT

--UPDATE #dat_recency SET
Select * from table 2
into table 3;