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4

0

I have a directory filled with thousands of files in the format LastnameFirstnameYYYYMMDD.pdf. The last and first name will always been in title case.

I'd like to extract the last name so I can move these files to a directory structure of first letter of last name/lastname/full filename. Example: DoeJohn20190327 would be moved to D/Doe/DoeJohn20190327

regex

share|improve this question

edited Mar 30 at 5:21

Peter Mortensen

1,03421016

asked Mar 29 at 4:14

Joseph MoorJoseph Moor

211

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Check out 'sed' and 'awk'
  
  – Tintin
  Mar 29 at 4:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I'm not sure if the final destination of the file DoeJohn20190327.pdf is like this: D/Doe/DoeJohn20190327/DoeJohn20190327.pdf or D/Doe/DoeJohn20190327/DoeJohn20190327.pdf ?
  
  – Philippe Delteil
  Mar 29 at 15:37
  
  

  
  
  
  

add a comment | 

4

0

I have a directory filled with thousands of files in the format LastnameFirstnameYYYYMMDD.pdf. The last and first name will always been in title case.

I'd like to extract the last name so I can move these files to a directory structure of first letter of last name/lastname/full filename. Example: DoeJohn20190327 would be moved to D/Doe/DoeJohn20190327

regex

share|improve this question

edited Mar 30 at 5:21

Peter Mortensen

1,03421016

asked Mar 29 at 4:14

Joseph MoorJoseph Moor

211

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Check out 'sed' and 'awk'
  
  – Tintin
  Mar 29 at 4:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I'm not sure if the final destination of the file DoeJohn20190327.pdf is like this: D/Doe/DoeJohn20190327/DoeJohn20190327.pdf or D/Doe/DoeJohn20190327/DoeJohn20190327.pdf ?
  
  – Philippe Delteil
  Mar 29 at 15:37
  
  

  
  
  
  

add a comment | 

I have a directory filled with thousands of files in the format LastnameFirstnameYYYYMMDD.pdf. The last and first name will always been in title case.

I'd like to extract the last name so I can move these files to a directory structure of first letter of last name/lastname/full filename. Example: DoeJohn20190327 would be moved to D/Doe/DoeJohn20190327

regex

share|improve this question

edited Mar 30 at 5:21

Peter Mortensen

1,03421016

asked Mar 29 at 4:14

Joseph MoorJoseph Moor

211

share|improve this question

edited Mar 30 at 5:21

Peter Mortensen

1,03421016

asked Mar 29 at 4:14

Joseph MoorJoseph Moor

211

share|improve this question

edited Mar 30 at 5:21

Peter Mortensen

1,03421016

asked Mar 29 at 4:14

Joseph MoorJoseph Moor

211

edited Mar 30 at 5:21

Peter Mortensen

1,03421016

edited Mar 30 at 5:21

Peter Mortensen

1,03421016

asked Mar 29 at 4:14

Joseph MoorJoseph Moor

211

asked Mar 29 at 4:14

Joseph MoorJoseph Moor

211

2 Answers
2

active

oldest

votes

5

Here you have a solution. I tested it an it creates the folders as you explained.

for filename in *.pdf
do
 echo "Processing file $filename "
 first_letter="$filename:0:1"
 mkdir -p $first_letter #if already exists won't print error
 last_name=$(echo $filename | sed 's/([^[:blank:]])([[:upper:]])/1 2/g' |awk 'print $1')
 mkdir -p $first_letter/$last_name
 mv $filename $first_letter/$last_name
done

share|improve this answer

edited Mar 29 at 15:44

answered Mar 29 at 4:57

Philippe DelteilPhilippe Delteil

6201522

add a comment | 

4

If the lastname is always the shortest trailing string staring with an upper case letter (there are no compound lastnames for example) you could use a shell parameter expansion of the form $parameter%pattern in place of a regex solution.

Ex.

for f in [[:upper:]]*[[:upper:]]*; do 
 d="$f:0:1/$f%[[:upper:]]*/"
 echo mkdir -p "$d"
 echo mv "$f" "$d"
done
mkdir -p D/Doe/
mv DoeJohn20190327 D/Doe/

Remove the echos when you are satisfied that it is doing the right thing.

See for example Parameter Expansion

share|improve this answer

answered Mar 29 at 7:03

steeldriversteeldriver

70.7k11115187

add a comment | 

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5

Here you have a solution. I tested it an it creates the folders as you explained.

for filename in *.pdf
do
 echo "Processing file $filename "
 first_letter="$filename:0:1"
 mkdir -p $first_letter #if already exists won't print error
 last_name=$(echo $filename | sed 's/([^[:blank:]])([[:upper:]])/1 2/g' |awk 'print $1')
 mkdir -p $first_letter/$last_name
 mv $filename $first_letter/$last_name
done

share|improve this answer

edited Mar 29 at 15:44

answered Mar 29 at 4:57

Philippe DelteilPhilippe Delteil

6201522

add a comment | 

5

Here you have a solution. I tested it an it creates the folders as you explained.

for filename in *.pdf
do
 echo "Processing file $filename "
 first_letter="$filename:0:1"
 mkdir -p $first_letter #if already exists won't print error
 last_name=$(echo $filename | sed 's/([^[:blank:]])([[:upper:]])/1 2/g' |awk 'print $1')
 mkdir -p $first_letter/$last_name
 mv $filename $first_letter/$last_name
done

share|improve this answer

edited Mar 29 at 15:44

answered Mar 29 at 4:57

Philippe DelteilPhilippe Delteil

6201522

add a comment | 

Here you have a solution. I tested it an it creates the folders as you explained.

for filename in *.pdf
do
 echo "Processing file $filename "
 first_letter="$filename:0:1"
 mkdir -p $first_letter #if already exists won't print error
 last_name=$(echo $filename | sed 's/([^[:blank:]])([[:upper:]])/1 2/g' |awk 'print $1')
 mkdir -p $first_letter/$last_name
 mv $filename $first_letter/$last_name
done

share|improve this answer

edited Mar 29 at 15:44

answered Mar 29 at 4:57

Philippe DelteilPhilippe Delteil

6201522

for filename in *.pdf
do
 echo "Processing file $filename "
 first_letter="$filename:0:1"
 mkdir -p $first_letter #if already exists won't print error
 last_name=$(echo $filename | sed 's/([^[:blank:]])([[:upper:]])/1 2/g' |awk 'print $1')
 mkdir -p $first_letter/$last_name
 mv $filename $first_letter/$last_name
done

share|improve this answer

edited Mar 29 at 15:44

answered Mar 29 at 4:57

Philippe DelteilPhilippe Delteil

6201522

answered Mar 29 at 4:57

Philippe DelteilPhilippe Delteil

6201522

answered Mar 29 at 4:57

Philippe DelteilPhilippe Delteil

6201522

4

If the lastname is always the shortest trailing string staring with an upper case letter (there are no compound lastnames for example) you could use a shell parameter expansion of the form $parameter%pattern in place of a regex solution.

Ex.

for f in [[:upper:]]*[[:upper:]]*; do 
 d="$f:0:1/$f%[[:upper:]]*/"
 echo mkdir -p "$d"
 echo mv "$f" "$d"
done
mkdir -p D/Doe/
mv DoeJohn20190327 D/Doe/

Remove the echos when you are satisfied that it is doing the right thing.

See for example Parameter Expansion

share|improve this answer

answered Mar 29 at 7:03

steeldriversteeldriver

70.7k11115187

add a comment | 

4

If the lastname is always the shortest trailing string staring with an upper case letter (there are no compound lastnames for example) you could use a shell parameter expansion of the form $parameter%pattern in place of a regex solution.

Ex.

for f in [[:upper:]]*[[:upper:]]*; do 
 d="$f:0:1/$f%[[:upper:]]*/"
 echo mkdir -p "$d"
 echo mv "$f" "$d"
done
mkdir -p D/Doe/
mv DoeJohn20190327 D/Doe/

Remove the echos when you are satisfied that it is doing the right thing.

See for example Parameter Expansion

share|improve this answer

answered Mar 29 at 7:03

steeldriversteeldriver

70.7k11115187

add a comment | 

If the lastname is always the shortest trailing string staring with an upper case letter (there are no compound lastnames for example) you could use a shell parameter expansion of the form $parameter%pattern in place of a regex solution.

Ex.

for f in [[:upper:]]*[[:upper:]]*; do 
 d="$f:0:1/$f%[[:upper:]]*/"
 echo mkdir -p "$d"
 echo mv "$f" "$d"
done
mkdir -p D/Doe/
mv DoeJohn20190327 D/Doe/

Remove the echos when you are satisfied that it is doing the right thing.

See for example Parameter Expansion

share|improve this answer

answered Mar 29 at 7:03

steeldriversteeldriver

70.7k11115187

for f in [[:upper:]]*[[:upper:]]*; do 
 d="$f:0:1/$f%[[:upper:]]*/"
 echo mkdir -p "$d"
 echo mv "$f" "$d"
done
mkdir -p D/Doe/
mv DoeJohn20190327 D/Doe/

share|improve this answer

answered Mar 29 at 7:03

steeldriversteeldriver

70.7k11115187

answered Mar 29 at 7:03

steeldriversteeldriver

70.7k11115187

answered Mar 29 at 7:03

steeldriversteeldriver

70.7k11115187