How can I strip out a numeric prefix from multiple filenames?

up vote
1
down vote

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I am new to Unix. I have >2000 files which are named like

123-FILENAME_TEXT_M101K_20150929.CSV

where 123 can be any three or four-digit number. The files are all within the same directory. I would like a script that removes the prefixing number and the dash. (i.e. the leading 123- should be removed from the example name)

I have tried mv **-FILENAME* FILENAME*.
rename is not an available function

Any advice would be greatly appreciated.

edited Jan 31 at 14:35

ilkkachu

49.8k674137

asked Jan 31 at 14:22

Kim

add a commentÂ |Â

up vote
1
down vote

favorite

I am new to Unix. I have >2000 files which are named like

123-FILENAME_TEXT_M101K_20150929.CSV

I have tried mv **-FILENAME* FILENAME*.
rename is not an available function

Any advice would be greatly appreciated.

edited Jan 31 at 14:35

ilkkachu

49.8k674137

asked Jan 31 at 14:22

Kim

add a commentÂ |Â

up vote
1
down vote

favorite

I am new to Unix. I have >2000 files which are named like

123-FILENAME_TEXT_M101K_20150929.CSV

I have tried mv **-FILENAME* FILENAME*.
rename is not an available function

Any advice would be greatly appreciated.

edited Jan 31 at 14:35

ilkkachu

49.8k674137

asked Jan 31 at 14:22

Kim

I am new to Unix. I have >2000 files which are named like

123-FILENAME_TEXT_M101K_20150929.CSV

I have tried mv **-FILENAME* FILENAME*.
rename is not an available function

Any advice would be greatly appreciated.

edited Jan 31 at 14:35

ilkkachu

49.8k674137

asked Jan 31 at 14:22

Kim

edited Jan 31 at 14:35

ilkkachu

49.8k674137

edited Jan 31 at 14:35

ilkkachu

49.8k674137

edited Jan 31 at 14:35

ilkkachu

49.8k674137

asked Jan 31 at 14:22

Kim

asked Jan 31 at 14:22

Kim

asked Jan 31 at 14:22

Kim

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

You could try something like this, it will remove everything before the first -

for file in *-*.CSV
do
 newName="$file#*-"
 mv -- "$file" "$newName"
done

edited Jan 31 at 14:51

Jeff Schaller

31.4k846105

answered Jan 31 at 14:32

mmoisse

584

1

The assignment is actually one of the few places that don't need the quotes (unless you use an unquoted $* there in Bash; It's had a bunch of bugs with that)
â€“Â ilkkachu
Jan 31 at 14:42

Quotes are indeed a source of bug, but I didn't know that it was not needed on assignment. Tx
â€“Â mmoisse
Jan 31 at 14:45

1

Quotes won't harm anyway. To clarify @ilkkachu's comment, in the case of bash's var=$*, it's the absence of quotes that cause problems.
â€“Â StÃ©phane Chazelas
Jan 31 at 15:12

Sorry I am a bit stuck, I have tried running this as a shell script but getting bad substitution - any ideas where I am going wrong?
â€“Â Kim
Jan 31 at 15:41

Solved, I was running as a .sh rather than .bash, even thou in my script I had #!/bin/bash. Any all work now tysm
â€“Â Kim
Jan 31 at 15:56

add a commentÂ |Â

up vote
0
down vote

I have tried by using combination of sed,awk and find . Tested and working fine

find . -type f -iname "*.CSV"| sed "s/^.///g"| sed -n '/^[0-9]4/p' | awk -F "-" 'print "mv" " " $0 " " $2'| sh

answered Jan 31 at 19:18

Praveen Kumar BS

1,010128

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

You could try something like this, it will remove everything before the first -

for file in *-*.CSV
do
 newName="$file#*-"
 mv -- "$file" "$newName"
done

edited Jan 31 at 14:51

Jeff Schaller

31.4k846105

answered Jan 31 at 14:32

mmoisse

584

1

The assignment is actually one of the few places that don't need the quotes (unless you use an unquoted $* there in Bash; It's had a bunch of bugs with that)
â€“Â ilkkachu
Jan 31 at 14:42

Quotes are indeed a source of bug, but I didn't know that it was not needed on assignment. Tx
â€“Â mmoisse
Jan 31 at 14:45

1

Quotes won't harm anyway. To clarify @ilkkachu's comment, in the case of bash's var=$*, it's the absence of quotes that cause problems.
â€“Â StÃ©phane Chazelas
Jan 31 at 15:12

Sorry I am a bit stuck, I have tried running this as a shell script but getting bad substitution - any ideas where I am going wrong?
â€“Â Kim
Jan 31 at 15:41

Solved, I was running as a .sh rather than .bash, even thou in my script I had #!/bin/bash. Any all work now tysm
â€“Â Kim
Jan 31 at 15:56

add a commentÂ |Â

up vote
4
down vote

accepted

You could try something like this, it will remove everything before the first -

for file in *-*.CSV
do
 newName="$file#*-"
 mv -- "$file" "$newName"
done

edited Jan 31 at 14:51

Jeff Schaller

31.4k846105

answered Jan 31 at 14:32

mmoisse

584

1

The assignment is actually one of the few places that don't need the quotes (unless you use an unquoted $* there in Bash; It's had a bunch of bugs with that)
â€“Â ilkkachu
Jan 31 at 14:42

Quotes are indeed a source of bug, but I didn't know that it was not needed on assignment. Tx
â€“Â mmoisse
Jan 31 at 14:45

1

Quotes won't harm anyway. To clarify @ilkkachu's comment, in the case of bash's var=$*, it's the absence of quotes that cause problems.
â€“Â StÃ©phane Chazelas
Jan 31 at 15:12

Sorry I am a bit stuck, I have tried running this as a shell script but getting bad substitution - any ideas where I am going wrong?
â€“Â Kim
Jan 31 at 15:41

Solved, I was running as a .sh rather than .bash, even thou in my script I had #!/bin/bash. Any all work now tysm
â€“Â Kim
Jan 31 at 15:56

add a commentÂ |Â

up vote
4
down vote

accepted

You could try something like this, it will remove everything before the first -

for file in *-*.CSV
do
 newName="$file#*-"
 mv -- "$file" "$newName"
done

edited Jan 31 at 14:51

Jeff Schaller

31.4k846105

answered Jan 31 at 14:32

mmoisse

584

You could try something like this, it will remove everything before the first -

for file in *-*.CSV
do
 newName="$file#*-"
 mv -- "$file" "$newName"
done

edited Jan 31 at 14:51

Jeff Schaller

31.4k846105

answered Jan 31 at 14:32

mmoisse

584

edited Jan 31 at 14:51

Jeff Schaller

31.4k846105

edited Jan 31 at 14:51

Jeff Schaller

31.4k846105

edited Jan 31 at 14:51

Jeff Schaller

31.4k846105

answered Jan 31 at 14:32

mmoisse

584

answered Jan 31 at 14:32

mmoisse

584

answered Jan 31 at 14:32

mmoisse

584

1

The assignment is actually one of the few places that don't need the quotes (unless you use an unquoted $* there in Bash; It's had a bunch of bugs with that)
â€“Â ilkkachu
Jan 31 at 14:42

Quotes are indeed a source of bug, but I didn't know that it was not needed on assignment. Tx
â€“Â mmoisse
Jan 31 at 14:45

1

Quotes won't harm anyway. To clarify @ilkkachu's comment, in the case of bash's var=$*, it's the absence of quotes that cause problems.
â€“Â StÃ©phane Chazelas
Jan 31 at 15:12

Sorry I am a bit stuck, I have tried running this as a shell script but getting bad substitution - any ideas where I am going wrong?
â€“Â Kim
Jan 31 at 15:41

Solved, I was running as a .sh rather than .bash, even thou in my script I had #!/bin/bash. Any all work now tysm
â€“Â Kim
Jan 31 at 15:56

add a commentÂ |Â

1

The assignment is actually one of the few places that don't need the quotes (unless you use an unquoted $* there in Bash; It's had a bunch of bugs with that)
â€“Â ilkkachu
Jan 31 at 14:42

Quotes are indeed a source of bug, but I didn't know that it was not needed on assignment. Tx
â€“Â mmoisse
Jan 31 at 14:45

1

Quotes won't harm anyway. To clarify @ilkkachu's comment, in the case of bash's var=$*, it's the absence of quotes that cause problems.
â€“Â StÃ©phane Chazelas
Jan 31 at 15:12

Sorry I am a bit stuck, I have tried running this as a shell script but getting bad substitution - any ideas where I am going wrong?
â€“Â Kim
Jan 31 at 15:41

Solved, I was running as a .sh rather than .bash, even thou in my script I had #!/bin/bash. Any all work now tysm
â€“Â Kim
Jan 31 at 15:56

The assignment is actually one of the few places that don't need the quotes (unless you use an unquoted $* there in Bash; It's had a bunch of bugs with that)
â€“Â ilkkachu
Jan 31 at 14:42

Quotes are indeed a source of bug, but I didn't know that it was not needed on assignment. Tx
â€“Â mmoisse
Jan 31 at 14:45

Quotes won't harm anyway. To clarify @ilkkachu's comment, in the case of bash's var=$*, it's the absence of quotes that cause problems.
â€“Â StÃ©phane Chazelas
Jan 31 at 15:12

Sorry I am a bit stuck, I have tried running this as a shell script but getting bad substitution - any ideas where I am going wrong?
â€“Â Kim
Jan 31 at 15:41

Solved, I was running as a .sh rather than .bash, even thou in my script I had #!/bin/bash. Any all work now tysm
â€“Â Kim
Jan 31 at 15:56

add a commentÂ |Â

up vote
0
down vote

I have tried by using combination of sed,awk and find . Tested and working fine

find . -type f -iname "*.CSV"| sed "s/^.///g"| sed -n '/^[0-9]4/p' | awk -F "-" 'print "mv" " " $0 " " $2'| sh

answered Jan 31 at 19:18

Praveen Kumar BS

1,010128

add a commentÂ |Â

up vote
0
down vote

I have tried by using combination of sed,awk and find . Tested and working fine

find . -type f -iname "*.CSV"| sed "s/^.///g"| sed -n '/^[0-9]4/p' | awk -F "-" 'print "mv" " " $0 " " $2'| sh

answered Jan 31 at 19:18

Praveen Kumar BS

1,010128

add a commentÂ |Â

up vote
0
down vote

I have tried by using combination of sed,awk and find . Tested and working fine

find . -type f -iname "*.CSV"| sed "s/^.///g"| sed -n '/^[0-9]4/p' | awk -F "-" 'print "mv" " " $0 " " $2'| sh

answered Jan 31 at 19:18

Praveen Kumar BS

1,010128

I have tried by using combination of sed,awk and find . Tested and working fine

find . -type f -iname "*.CSV"| sed "s/^.///g"| sed -n '/^[0-9]4/p' | awk -F "-" 'print "mv" " " $0 " " $2'| sh

answered Jan 31 at 19:18

Praveen Kumar BS

1,010128

answered Jan 31 at 19:18

Praveen Kumar BS

1,010128

answered Jan 31 at 19:18

Praveen Kumar BS

1,010128

answered Jan 31 at 19:18

Praveen Kumar BS

1,010128

add a commentÂ |Â

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