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How to grep the following lines from a file?
grep list of names and information from bigger fileHow to find strings using regular expressions in grepHow to pass variables to awk command with conditions?Search all xml files recursively in directory for a specific tag and grep the tag's valuegrep multiple patterns and print results with the match patternGrep multiple files and output to multiple files in a single commandHow do you match a line which starts with a special character using grep?Copying list of file from grep command to target directoryGrep latest file for string and alert / email if foundShell script to create one file and append the results
I have a file which has the names of many files in a directory in the following format:
A20150824.0950-0955_jambala_CcnActiveSessionCounterJob
A20150824.0945-0950_jambala_CcnActiveSessionCounterJob
A20150824.0940-0945_jambala_CcnActiveSessionCounterJob
A20150824.0935-0940_jambala_CcnActiveSessionCounterJob
A20150824.0955-1000_jambala_CcnActiveSessionCounterJob
A20150824.0000-0005_jambala_CcnActiveSessionCounterJob
A20150824.0100-0105_jambala_CcnActiveSessionCounterJob
A20150824.0105-0110_jambala_CcnActiveSessionCounterJob
A20150824.0110-0115_jambala_CcnActiveSessionCounterJob
A20150824.0115-0120_jambala_CcnActiveSessionCounterJob
A20150824.0120-0125_jambala_CcnActiveSessionCounterJob
A20150824.1400-1405_jambala_CcnActiveSessionCounterJob
The naming convention of the above files is A<YYYYMMDD>.HHMM-HHMM_<city>_CcnActiveSessionCounterJob.
These files are created for every 5 minute for all the hours each day such that, for every hour i get 12 files and for every day 12X24 files. I have generated a file which has got the names of all the 12x24 files and have a while loop in bash script where I am trying to do some processing per hour. I want to find a way by which I could create another file which contains the 12 files for each hour. For this, I am having a while outer loop which gives the hour value and inner minute loop which gives the minutes value. These files has the time information in their names. Ex:
A20150824.0950-0955_jambala_CcnActiveSessionCounterJob
gives the information of 09-50 AM to 09-55 AM.
How do I use grep to extract the filenames from the file which contains all the 12X24 file names and put them in a separate file such that a new file contains the following file names:
A20150824.0900-0905_jambala_CcnActiveSessionCounterJob
A20150824.0905-0910_jambala_CcnActiveSessionCounterJob
A20150824.0910-0915_jambala_CcnActiveSessionCounterJob
A20150824.0915-0920_jambala_CcnActiveSessionCounterJob
.
.
.
A20150824.0955-1000_jambala_CcnActiveSessionCounterJob
I already have a variable hour which has the currently being processed hour information, I was trying to use the following but it doesn't work:
grep -E '.$hour' FILE_with_ALL_FILENAMES
where .$hour is targeted for .09 in the above file name. How do i fix this?
shell-script command-line grep regular-expression
edited 4 hours ago
Rui F Ribeiro
41.2k1481140
asked Aug 25 '15 at 6:37
Ankit VashisthaAnkit Vashistha
85682130
add a comment |
I have a file which has the names of many files in a directory in the following format:
A20150824.0950-0955_jambala_CcnActiveSessionCounterJob
A20150824.0945-0950_jambala_CcnActiveSessionCounterJob
A20150824.0940-0945_jambala_CcnActiveSessionCounterJob
A20150824.0935-0940_jambala_CcnActiveSessionCounterJob
A20150824.0955-1000_jambala_CcnActiveSessionCounterJob
A20150824.0000-0005_jambala_CcnActiveSessionCounterJob
A20150824.0100-0105_jambala_CcnActiveSessionCounterJob
A20150824.0105-0110_jambala_CcnActiveSessionCounterJob
A20150824.0110-0115_jambala_CcnActiveSessionCounterJob
A20150824.0115-0120_jambala_CcnActiveSessionCounterJob
A20150824.0120-0125_jambala_CcnActiveSessionCounterJob
A20150824.1400-1405_jambala_CcnActiveSessionCounterJob
The naming convention of the above files is A<YYYYMMDD>.HHMM-HHMM_<city>_CcnActiveSessionCounterJob.
These files are created for every 5 minute for all the hours each day such that, for every hour i get 12 files and for every day 12X24 files. I have generated a file which has got the names of all the 12x24 files and have a while loop in bash script where I am trying to do some processing per hour. I want to find a way by which I could create another file which contains the 12 files for each hour. For this, I am having a while outer loop which gives the hour value and inner minute loop which gives the minutes value. These files has the time information in their names. Ex:
A20150824.0950-0955_jambala_CcnActiveSessionCounterJob
gives the information of 09-50 AM to 09-55 AM.
How do I use grep to extract the filenames from the file which contains all the 12X24 file names and put them in a separate file such that a new file contains the following file names:
A20150824.0900-0905_jambala_CcnActiveSessionCounterJob
A20150824.0905-0910_jambala_CcnActiveSessionCounterJob
A20150824.0910-0915_jambala_CcnActiveSessionCounterJob
A20150824.0915-0920_jambala_CcnActiveSessionCounterJob
.
.
.
A20150824.0955-1000_jambala_CcnActiveSessionCounterJob
I already have a variable hour which has the currently being processed hour information, I was trying to use the following but it doesn't work:
grep -E '.$hour' FILE_with_ALL_FILENAMES
where .$hour is targeted for .09 in the above file name. How do i fix this?
shell-script command-line grep regular-expression
edited 4 hours ago
Rui F Ribeiro
41.2k1481140
asked Aug 25 '15 at 6:37
Ankit VashisthaAnkit Vashistha
85682130
add a comment |
I have a file which has the names of many files in a directory in the following format:
A20150824.0950-0955_jambala_CcnActiveSessionCounterJob
A20150824.0945-0950_jambala_CcnActiveSessionCounterJob
A20150824.0940-0945_jambala_CcnActiveSessionCounterJob
A20150824.0935-0940_jambala_CcnActiveSessionCounterJob
A20150824.0955-1000_jambala_CcnActiveSessionCounterJob
A20150824.0000-0005_jambala_CcnActiveSessionCounterJob
A20150824.0100-0105_jambala_CcnActiveSessionCounterJob
A20150824.0105-0110_jambala_CcnActiveSessionCounterJob
A20150824.0110-0115_jambala_CcnActiveSessionCounterJob
A20150824.0115-0120_jambala_CcnActiveSessionCounterJob
A20150824.0120-0125_jambala_CcnActiveSessionCounterJob
A20150824.1400-1405_jambala_CcnActiveSessionCounterJob
The naming convention of the above files is A<YYYYMMDD>.HHMM-HHMM_<city>_CcnActiveSessionCounterJob.
These files are created for every 5 minute for all the hours each day such that, for every hour i get 12 files and for every day 12X24 files. I have generated a file which has got the names of all the 12x24 files and have a while loop in bash script where I am trying to do some processing per hour. I want to find a way by which I could create another file which contains the 12 files for each hour. For this, I am having a while outer loop which gives the hour value and inner minute loop which gives the minutes value. These files has the time information in their names. Ex:
A20150824.0950-0955_jambala_CcnActiveSessionCounterJob
gives the information of 09-50 AM to 09-55 AM.
How do I use grep to extract the filenames from the file which contains all the 12X24 file names and put them in a separate file such that a new file contains the following file names:
A20150824.0900-0905_jambala_CcnActiveSessionCounterJob
A20150824.0905-0910_jambala_CcnActiveSessionCounterJob
A20150824.0910-0915_jambala_CcnActiveSessionCounterJob
A20150824.0915-0920_jambala_CcnActiveSessionCounterJob
.
.
.
A20150824.0955-1000_jambala_CcnActiveSessionCounterJob
I already have a variable hour which has the currently being processed hour information, I was trying to use the following but it doesn't work:
grep -E '.$hour' FILE_with_ALL_FILENAMES
where .$hour is targeted for .09 in the above file name. How do i fix this?
shell-script command-line grep regular-expression
edited 4 hours ago
Rui F Ribeiro
41.2k1481140
asked Aug 25 '15 at 6:37
Ankit VashisthaAnkit Vashistha
85682130
I have a file which has the names of many files in a directory in the following format:
A20150824.0950-0955_jambala_CcnActiveSessionCounterJob
A20150824.0945-0950_jambala_CcnActiveSessionCounterJob
A20150824.0940-0945_jambala_CcnActiveSessionCounterJob
A20150824.0935-0940_jambala_CcnActiveSessionCounterJob
A20150824.0955-1000_jambala_CcnActiveSessionCounterJob
A20150824.0000-0005_jambala_CcnActiveSessionCounterJob
A20150824.0100-0105_jambala_CcnActiveSessionCounterJob
A20150824.0105-0110_jambala_CcnActiveSessionCounterJob
A20150824.0110-0115_jambala_CcnActiveSessionCounterJob
A20150824.0115-0120_jambala_CcnActiveSessionCounterJob
A20150824.0120-0125_jambala_CcnActiveSessionCounterJob
A20150824.1400-1405_jambala_CcnActiveSessionCounterJob
The naming convention of the above files is A<YYYYMMDD>.HHMM-HHMM_<city>_CcnActiveSessionCounterJob.
These files are created for every 5 minute for all the hours each day such that, for every hour i get 12 files and for every day 12X24 files. I have generated a file which has got the names of all the 12x24 files and have a while loop in bash script where I am trying to do some processing per hour. I want to find a way by which I could create another file which contains the 12 files for each hour. For this, I am having a while outer loop which gives the hour value and inner minute loop which gives the minutes value. These files has the time information in their names. Ex:
A20150824.0950-0955_jambala_CcnActiveSessionCounterJob
gives the information of 09-50 AM to 09-55 AM.
How do I use grep to extract the filenames from the file which contains all the 12X24 file names and put them in a separate file such that a new file contains the following file names:
A20150824.0900-0905_jambala_CcnActiveSessionCounterJob
A20150824.0905-0910_jambala_CcnActiveSessionCounterJob
A20150824.0910-0915_jambala_CcnActiveSessionCounterJob
A20150824.0915-0920_jambala_CcnActiveSessionCounterJob
.
.
.
A20150824.0955-1000_jambala_CcnActiveSessionCounterJob
I already have a variable hour which has the currently being processed hour information, I was trying to use the following but it doesn't work:
grep -E '.$hour' FILE_with_ALL_FILENAMES
where .$hour is targeted for .09 in the above file name. How do i fix this?
shell-script command-line grep regular-expression
shell-script command-line grep regular-expression
edited 4 hours ago
Rui F Ribeiro
41.2k1481140
asked Aug 25 '15 at 6:37
Ankit VashisthaAnkit Vashistha
85682130
edited 4 hours ago
Rui F Ribeiro
41.2k1481140
asked Aug 25 '15 at 6:37
Ankit VashisthaAnkit Vashistha
85682130
edited 4 hours ago
Rui F Ribeiro
41.2k1481140
edited 4 hours ago
Rui F Ribeiro
41.2k1481140
edited 4 hours ago
Rui F Ribeiro
41.2k1481140
41.2k1481140
asked Aug 25 '15 at 6:37
Ankit VashisthaAnkit Vashistha
85682130
asked Aug 25 '15 at 6:37
Ankit VashisthaAnkit Vashistha
85682130
asked Aug 25 '15 at 6:37
Ankit VashisthaAnkit Vashistha
85682130
85682130
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Assuming:
hour=09
Just use that:
grep ".$hour" file
With the single quotes in your example, the variable is not interpreted as variable. Therefore the pattern searches for $hour. Also the dot has to be escaped, else it would match any character.
answered Aug 25 '15 at 6:46
chaoschaos
35.8k974118
Maybe also use$(printf %02d $hour)as it might be integer values without leading zeros. This would also allow using e.g. a simpleforloop from 1 to 12.
– Fiximan
Aug 25 '15 at 8:02
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
Assuming:
hour=09
Just use that:
grep ".$hour" file
With the single quotes in your example, the variable is not interpreted as variable. Therefore the pattern searches for $hour. Also the dot has to be escaped, else it would match any character.
answered Aug 25 '15 at 6:46
chaoschaos
35.8k974118
Maybe also use$(printf %02d $hour)as it might be integer values without leading zeros. This would also allow using e.g. a simpleforloop from 1 to 12.
– Fiximan
Aug 25 '15 at 8:02
add a comment |
Assuming:
hour=09
Just use that:
grep ".$hour" file
With the single quotes in your example, the variable is not interpreted as variable. Therefore the pattern searches for $hour. Also the dot has to be escaped, else it would match any character.
answered Aug 25 '15 at 6:46
chaoschaos
35.8k974118
Maybe also use$(printf %02d $hour)as it might be integer values without leading zeros. This would also allow using e.g. a simpleforloop from 1 to 12.
– Fiximan
Aug 25 '15 at 8:02
add a comment |
Assuming:
hour=09
Just use that:
grep ".$hour" file
With the single quotes in your example, the variable is not interpreted as variable. Therefore the pattern searches for $hour. Also the dot has to be escaped, else it would match any character.
answered Aug 25 '15 at 6:46
chaoschaos
35.8k974118
Assuming:
hour=09
Just use that:
grep ".$hour" file
With the single quotes in your example, the variable is not interpreted as variable. Therefore the pattern searches for $hour. Also the dot has to be escaped, else it would match any character.
answered Aug 25 '15 at 6:46
chaoschaos
35.8k974118
answered Aug 25 '15 at 6:46
chaoschaos
35.8k974118
answered Aug 25 '15 at 6:46
chaoschaos
35.8k974118
answered Aug 25 '15 at 6:46
chaoschaos
35.8k974118
35.8k974118
Maybe also use$(printf %02d $hour)as it might be integer values without leading zeros. This would also allow using e.g. a simpleforloop from 1 to 12.
– Fiximan
Aug 25 '15 at 8:02
add a comment |
Maybe also use$(printf %02d $hour)as it might be integer values without leading zeros. This would also allow using e.g. a simpleforloop from 1 to 12.
– Fiximan
Aug 25 '15 at 8:02
Maybe also use
$(printf %02d $hour) as it might be integer values without leading zeros. This would also allow using e.g. a simple for loop from 1 to 12.– Fiximan
Aug 25 '15 at 8:02
Maybe also use
$(printf %02d $hour) as it might be integer values without leading zeros. This would also allow using e.g. a simple for loop from 1 to 12.– Fiximan
Aug 25 '15 at 8:02
add a comment |
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