I have a file with the following entries:
MySql-DataBase-2020-09-22_183748.zip
MySql-DataBase-2020-09-22_184023.zip
MySql-DataBase-2020-09-23_205331.zip
MySql-DataBase-2020-09-23_205606.zip
MySql-DataBase-2020-09-24_200123.zip
MySql-DataBase-2020-09-24_200358.zip
MySql-DataBase-2020-09-25_115839.zip
MySql-DataBase-2020-09-25_120115.zip
MySql-DataBase-2020-09-26_094608.zip
MySql-DataBase-2020-09-26_094843.zip
MySql-DataBase-2020-09-27_122523.zip
MySql-DataBase-2020-09-27_122758.zip
MySql-DataBase-2020-10-01_230024.zip
MySql-DataBase-2020-10-01_230300.zip
MySql-DataBase-2020-10-02_120944.zip
MySql-DataBase-2020-10-02_121219.zip
MySql-DataBase-2020-10-03_151414.zip
MySql-DataBase-2020-10-03_151649.zip
MySql-DataBase-2020-10-04_211059.zip
MySql-DataBase-2020-10-04_211334.zip
MySql-DataBase-2020-10-05_064049.zip
MySql-DataBase-2020-10-05_064324.zip
I want to extract Files which are between today's date & 3 days back.
For today's date 2020/10/05, 3 days back is 2020/10/02.
The output must be:
MySql-DataBase-2020-10-02_120944.zip
MySql-DataBase-2020-10-02_121219.zip
MySql-DataBase-2020-10-03_151414.zip
MySql-DataBase-2020-10-03_151649.zip
MySql-DataBase-2020-10-04_211059.zip
MySql-DataBase-2020-10-04_211334.zip
MySql-DataBase-2020-10-05_064049.zip
I tried using this command to gets 3days back date
date --date='-3 day' '+%Y/%m/%d'
And then used these command to get output between date range
sed -n '/3day=date --date='-3 day' '+%Y/%m/%d'/,/date/p' s.txt
I am getting this error:
sed: -e expression #1, char 20: unterminated address regex
Please Help me to fix this issue. I'll be using this in a bash script.
Using a process substitution to generate the dates and feed that to grep as the patterns to search for:
grep -F -f <(for d in {0..3}; do date -d "$d days ago" "+%F"; done) file
sed -n "/$(date --date='-3 day' '+%Y-%m-%d')/,/$(date +'%y-%m-%d')/p"
MySql-DataBase-2020-10-02_120944.zip
MySql-DataBase-2020-10-02_121219.zip
MySql-DataBase-2020-10-03_151414.zip
MySql-DataBase-2020-10-03_151649.zip
MySql-DataBase-2020-10-04_211059.zip
MySql-DataBase-2020-10-04_211334.zip
MySql-DataBase-2020-10-05_064049.zip
Notice, the use of double quotes at the outermost level. Also, notice the same format of date in both boundaries.
Related
I have following file:(A sample file with filename: 2015_09_22_processedPartnumList.txt, Location: /a/b/c/itemreport)
DataLoader_trace_2015_09_22_02_01_32.0956.log:INFO: 2015-09-22
Data Processing Starts : 12345678
I just want to get all the ids from the above file i.e. 12345678 .... (each id in a separate line, not comma separated) in a file /a/b/c/d/ids_date +%d_%m_%Y_%H_%M_%S.log
I have written the following script, but the file I am getting is empty. Without showing any exception or anything. So, it is very difficult for me to identify the errors. Please tell me what is wrong in the script.
LOGDIR=/a/b/logdir
tr=`date +%p`
echo $tr
if [ $tr = "PM" ];
then
date=`date +%Y-%m-%d`
echo "considering today's date for grepping logs"
else
date=`date -d '1 day ago' +%Y-%m-%d`
echo "considering yesterday's date for grepping logs as job run is delayed"
fi
ITEM_FILE=/a/b/c/d/ids_`date +%d_%m_%Y_%H_%M_%S`.log
After implementing grep in PCRE, I am getting this and not any ids are being copied into the new file.
If your grep supports PCRE, you can do:
grep -Po '.*:\s\K\d+$' /a/b/c/itemreport/2015_09_22_processedPartnumList.txt \
>/apps/feeds/out/catalog/ItemPartnumbers_"$(date '+%d_%m_%Y_%H_%M_%S')".log
.*:\s will match upto the space after :, \K will discard the match
\d+$ will match our desired portion i.e. the digits till the end of the line
Example:
% grep -Po '.*:\s\K\d+$' 2015_09_22_processedPartnumList.txt \
>ItemPartnumbers_"$(date '+%d_%m_%Y_%H_%M_%S')".log
% cat ItemPartnumbers_09_11_2015_11_30_49.log
13982787
14011550
13984790
13984791
14176509
14902623
14924193
14924194
13982787
46795670
46795671
That's not very good solution, but it's working.
cat your\ file | cut -d ':' -f2-2 | tr -d INFO
I 'm using bash
I have a date string now, for example:
2015111301
(yyyymmddHH)
how to get the date string of next hour?
that is:
2015111302
Try this:
in="2015111301"
out="$(date -d "${in:0:8} ${in:8:2}:00:00 +1hour" '+%Y%m%d%H')"
echo "$out"
Output:
2015111302
See: 3.5.3 Shell Parameter Expansion
This below script worked for me
#!/bin/bash
tempval=$(echo 2015111301 | sed 's/\(.\{8\}\)/\1 /g')
tempval=$(echo "$tempval 1 hour")
date -d "$tempval" +%Y%m%d%H
I have a file named Try1.txt that contains the string
"End Date : "
I have a script named Try1.sh which has the string assigned to a variable
STR1="End Date : "
I have a replacement string also assigned to the following variable
STR2="End Date : Tuesday 05/06/2014"
I want to edit in place the file and replace STR1 with STR2.
I tried several different sed commands but could not figure it out.
I tried sed -i -e "s/$STR1/$STR2/g" <Try1.txt >Try1.out.txt
but it gives me the following error:
sed.exe: -e expression #1, char 49: unknown option to `s'
Your problem is that $STR2 contains a slash /.
Assuming that neither $STR1 nor $STR2 contain an underscore, _, the following will work:
$ sed "s_${STR1}_${STR2}_g" <Try1.txt >Try1.out.txt
Test:
$ STR1="End Date : "
$ STR2="End Date : Tuesday 05/06/2014"
$ echo "$STR1" > Try1.txt
$ sed "s_${STR1}_${STR2}_g" <Try1.txt
End Date : Tuesday 05/06/2014
If you have no guarantee about not having underscores in your strings, the following also seems to work:
sed "s^A$STR1^A$STR2^Ag" <Try1.txt
where the three ^A's are entered as Cntrl-V Cntrl-A. (Using \x01 doesn't work for me.)
See also, for example, sed search and replace strings containing / and Sed replacement not working when using variables.
I have strftime format of time, let's say (%Y-%m-%d %H:%M:%S) and a file which should contain this kind of data e.g. (2012-02-11 17:15:00). I need to check if given pattern actually matches the data.
How to approach this? awk, date?
EDIT:
More info: The user enters the strftime format, let's say on input. Then he enters a file which should contain those dates. I need to make sure, that those data are valid (he didn't make a mistake). So I need to check the rows in the input file and see, if there are data that matches the given pattern. Example:
user enters strftime format: (%Y-%m-%d %H:%M:%S)
input file: (2012-02-11 17:15:00) long sentence
VALID
user enters strftime format: Date[%Y.%m.%d %H:%M:%S]
input file: Date-2012.02.11 17:15:00- long sentence
INVALID
If you allow an external helper binary, I've written dateutils to batch process date and time data.
dconv -q -i '(%Y-%m-%d %H:%M:%S)' <<EOF
not a match: 2012-04-10 12:00:00
a match: (2012-04-10 13:00:00)
EOF
will give
2012-04-10T13:00:00
-i is the input format, -q suppresses warnings. And dconv tries to convert input lines to output lines (in this case it converts matching lines to ISO standard format.
So using this, a file matches completely if the number of input lines equals the number of output lines.
If you want to check current datetime:
echo "(2012-02-11 17:15:00)" | grep "$(date "+%Y-%m-%d %H:%M:%S")"
If some other you need GNU date (-d option). This works for me:
echo "(2012-02-11 17:15:00)" |
grep "$(date -d "2012-02-11 17:15:00" "+%Y-%m-%d %H:%M:%S")"
I would take a brute force approach to this: replace any %X specifier with a corresponding regular expression, then you can filter out lines that don't match the resulting generated regex:
user_format="%Y-%m-%d"
awk -v fmt_string="$user_format" '
BEGIN {
gsub(/[][(){}?|*+.]/ "\\&", fmt_string) # protect any regex-special chars
gsub(/%Y/, "([0-9]{4})", fmt_string)
gsub(/%m/, "(0[1-9]|1[012])", fmt_string)
gsub(/%d/, "(0[1-9]|[12][0-9]|3[01])", fmt_string)
# and so on
}
$0 !~ "^" fmt_string {print "line " NR " does not match: " $0}
' filename
I am trying to create a script to convert two columns in a .csv file which are date and time into unix timestamps. So i need to get the date and time column from each row, convert it and insert it into an additional column at the end containing the timestamp.
Could anyone help me? So far i have discovered the unix command to convert any give time and date to unixstamp:
date -d "2011/11/25 10:00:00" "+%s"
1322215200
I have no experience with bash scripting could anyone get me started?
Examples of my columns and rows:
Columns: Date, Time,
Row 1: 25/10/2011, 10:54:36,
Row 2: 25/10/2011, 11:15:17,
Row 3: 26/10/2011, 01:04:39,
Thanks so much in advance!
You don't provide an exerpt from your csv-file, so I'm using this one:
[foo.csv]
2011/11/25;12:00:00
2010/11/25;13:00:00
2009/11/25;19:00:00
Here's one way to solve your problem:
$ cat foo.csv | while read line ; do echo $line\;$(date -d "${line//;/ }" "+%s") ; done
2011/11/25;12:00:00;1322218800
2010/11/25;13:00:00;1290686400
2009/11/25;19:00:00;1259172000
(EDIT: Removed an uneccessary variable.)
(EDIT2: Altered the date command so the script actually works.)
this should do the job:
awk 'BEGIN{FS=OFS=", "}{t=$1" "$2; "date -d \""t"\" +%s"|getline d; print $1,$2,d}' yourCSV.csv
note
you didn't give any example. and you mentioned csv, so I assume that the column separator in your file should be "comma".
test
kent$ echo "2011/11/25, 10:00:00"|awk 'BEGIN{FS=OFS=", "}{t=$1" "$2; "date -d \""t"\" +%s"|getline d; print $1,$2,d}'
2011/11/25, 10:00:00, 1322211600
Now two imporvements:
First: No need for cat foo.csv, just stream that via < foo.csv into the while loop.
Second: No need for echo & tr to create the date stringformat. Just use bash internal pattern and substitute and do it inplace
while read line ; do echo ${line}\;$(date -d "${line//;/ }" +'%s'); done < foo.csv