cut not outputting anything
I created a script by using the command line initially:
grep -c "Author" reviews_folder/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
It worked, however I then needed to make it a shell script which allowed the user to enter a directory for it to search, so I changed it to:
grep -c "Author" $1/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
After this was complete, rather than the script working, it just printed lots of blank lines.
Any help would be greatly appreciated. Thanks!
command-line bash scripts cut-command
edited Mar 14 at 17:04
Ravexina
33.3k1488115
asked Mar 14 at 16:34
Matthew EvansMatthew Evans
61
add a comment |
I created a script by using the command line initially:
grep -c "Author" reviews_folder/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
It worked, however I then needed to make it a shell script which allowed the user to enter a directory for it to search, so I changed it to:
grep -c "Author" $1/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
After this was complete, rather than the script working, it just printed lots of blank lines.
Any help would be greatly appreciated. Thanks!
command-line bash scripts cut-command
edited Mar 14 at 17:04
Ravexina
33.3k1488115
asked Mar 14 at 16:34
Matthew EvansMatthew Evans
61
Add anecho $1to see what's going on, and any spaces in the directory names?
– Xen2050
Mar 14 at 22:45
add a comment |
I created a script by using the command line initially:
grep -c "Author" reviews_folder/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
It worked, however I then needed to make it a shell script which allowed the user to enter a directory for it to search, so I changed it to:
grep -c "Author" $1/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
After this was complete, rather than the script working, it just printed lots of blank lines.
Any help would be greatly appreciated. Thanks!
command-line bash scripts cut-command
edited Mar 14 at 17:04
Ravexina
33.3k1488115
asked Mar 14 at 16:34
Matthew EvansMatthew Evans
61
I created a script by using the command line initially:
grep -c "Author" reviews_folder/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
It worked, however I then needed to make it a shell script which allowed the user to enter a directory for it to search, so I changed it to:
grep -c "Author" $1/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
After this was complete, rather than the script working, it just printed lots of blank lines.
Any help would be greatly appreciated. Thanks!
command-line bash scripts cut-command
command-line bash scripts cut-command
edited Mar 14 at 17:04
Ravexina
33.3k1488115
asked Mar 14 at 16:34
Matthew EvansMatthew Evans
61
edited Mar 14 at 17:04
Ravexina
33.3k1488115
asked Mar 14 at 16:34
Matthew EvansMatthew Evans
61
edited Mar 14 at 17:04
Ravexina
33.3k1488115
edited Mar 14 at 17:04
Ravexina
33.3k1488115
edited Mar 14 at 17:04
Ravexina
33.3k1488115
33.3k1488115
asked Mar 14 at 16:34
Matthew EvansMatthew Evans
61
asked Mar 14 at 16:34
Matthew EvansMatthew Evans
61
asked Mar 14 at 16:34
Matthew EvansMatthew Evans
61
61
Add anecho $1to see what's going on, and any spaces in the directory names?
– Xen2050
Mar 14 at 22:45
add a comment |
Add anecho $1to see what's going on, and any spaces in the directory names?
– Xen2050
Mar 14 at 22:45
Add an
echo $1 to see what's going on, and any spaces in the directory names?– Xen2050
Mar 14 at 22:45
Add an
echo $1 to see what's going on, and any spaces in the directory names?– Xen2050
Mar 14 at 22:45
add a comment |
1 Answer
1
active
oldest
votes
I guess you are running your script like: script.sh dir/ or bash script.sh dir/.
The way you pass the directory is causing the problem.
In your script you got a grep command:
grep -c "Author" $1/*
If you pass the directory with tailing slash, the grep command would be run as:
grep -c "Author" dir//*
Example:
grep -c "Author" dir//somefile.dat dir//someother-file.dat
And its result would be similar to:
dir//somefile.dat:5
And your cut command would only return an empty string and pass it to sort:
dir//somefile.dat:5 # < You are asking for second part which is empty
So either run your script like: script.sh dir or change it to something like this:
#!/bin/bash
DIR=${1%/}
grep -c "Author" ${DIR}/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
answered Mar 14 at 17:01
RavexinaRavexina
33.3k1488115
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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votes
I guess you are running your script like: script.sh dir/ or bash script.sh dir/.
The way you pass the directory is causing the problem.
In your script you got a grep command:
grep -c "Author" $1/*
If you pass the directory with tailing slash, the grep command would be run as:
grep -c "Author" dir//*
Example:
grep -c "Author" dir//somefile.dat dir//someother-file.dat
And its result would be similar to:
dir//somefile.dat:5
And your cut command would only return an empty string and pass it to sort:
dir//somefile.dat:5 # < You are asking for second part which is empty
So either run your script like: script.sh dir or change it to something like this:
#!/bin/bash
DIR=${1%/}
grep -c "Author" ${DIR}/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
answered Mar 14 at 17:01
RavexinaRavexina
33.3k1488115
add a comment |
I guess you are running your script like: script.sh dir/ or bash script.sh dir/.
The way you pass the directory is causing the problem.
In your script you got a grep command:
grep -c "Author" $1/*
If you pass the directory with tailing slash, the grep command would be run as:
grep -c "Author" dir//*
Example:
grep -c "Author" dir//somefile.dat dir//someother-file.dat
And its result would be similar to:
dir//somefile.dat:5
And your cut command would only return an empty string and pass it to sort:
dir//somefile.dat:5 # < You are asking for second part which is empty
So either run your script like: script.sh dir or change it to something like this:
#!/bin/bash
DIR=${1%/}
grep -c "Author" ${DIR}/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
answered Mar 14 at 17:01
RavexinaRavexina
33.3k1488115
add a comment |
I guess you are running your script like: script.sh dir/ or bash script.sh dir/.
The way you pass the directory is causing the problem.
In your script you got a grep command:
grep -c "Author" $1/*
If you pass the directory with tailing slash, the grep command would be run as:
grep -c "Author" dir//*
Example:
grep -c "Author" dir//somefile.dat dir//someother-file.dat
And its result would be similar to:
dir//somefile.dat:5
And your cut command would only return an empty string and pass it to sort:
dir//somefile.dat:5 # < You are asking for second part which is empty
So either run your script like: script.sh dir or change it to something like this:
#!/bin/bash
DIR=${1%/}
grep -c "Author" ${DIR}/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
answered Mar 14 at 17:01
RavexinaRavexina
33.3k1488115
I guess you are running your script like: script.sh dir/ or bash script.sh dir/.
The way you pass the directory is causing the problem.
In your script you got a grep command:
grep -c "Author" $1/*
If you pass the directory with tailing slash, the grep command would be run as:
grep -c "Author" dir//*
Example:
grep -c "Author" dir//somefile.dat dir//someother-file.dat
And its result would be similar to:
dir//somefile.dat:5
And your cut command would only return an empty string and pass it to sort:
dir//somefile.dat:5 # < You are asking for second part which is empty
So either run your script like: script.sh dir or change it to something like this:
#!/bin/bash
DIR=${1%/}
grep -c "Author" ${DIR}/* | cut -d / -f 2 | sort -nt':' -k2 | sed 's/.dat:/ /g'
answered Mar 14 at 17:01
RavexinaRavexina
33.3k1488115
edited Mar 18 at 20:48
answered Mar 14 at 17:01
RavexinaRavexina
33.3k1488115
answered Mar 14 at 17:01
RavexinaRavexina
33.3k1488115
answered Mar 14 at 17:01
RavexinaRavexina
33.3k1488115
33.3k1488115
add a comment |
add a comment |
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Add an
echo $1to see what's going on, and any spaces in the directory names?– Xen2050
Mar 14 at 22:45