Count the number of files in a directory Post: 302132228

10 More Discussions You Might Find Interesting

1. UNIX for Dummies Questions & Answers

Count number of files in subdirectories

Hello, I am new to unix and would like to have a count of all the files in the current directory as well as all the files in a subdirectory. The command I used was ls -R | wc -l but the number returned wasn't correct. Can someone please help? Thanks

2. Shell Programming and Scripting

count number of files in a directory

what's the script to do that? i want to only count the number of files in that directory, not including any sub directories at all

3. UNIX for Dummies Questions & Answers

Read directory files and count number of lines

Hello, I'm trying to create a BASH file that can read all the files in my working directory and tell me how many words and lines are in that file. I wrote the following code: FILES="*" for f in "$FILES" do echo -e `wc -l -w $f` done My issue is that my file is outputting in one...

4. Shell Programming and Scripting

perl script on how to count the total number of lines of all the files under a directory

how to count the total number of lines of all the files under a directory using perl script.. I mean if I have 10 files under a directory then I want to count the total number of lines of all the 10 files contain. Please help me in writing a perl script on this.

5. UNIX for Dummies Questions & Answers

Count number of files in directory excluding existing files

Hi, Please let me know how to find out number of files in a directory excluding existing files..The existing file format will be unknown..each time.. Thanks

6. Shell Programming and Scripting

How to count the number of files starting with a pattern in a Directory

Hi! In our current directory there are around 35000 files. Out of these a few thousands(around 20000) start with, "testfiles9842323879838". I want to count the number of files that have filenames starting with the above pattern. Please help me with the command i could use. Thank...

7. Shell Programming and Scripting

How to count number of files in directory and write to new file with number of files and their name?

Hi! I just want to count number of files in a directory, and write to new text file, with number of files and their name output should look like this,, assume that below one is a new file created by script Number of files in directory = 25 1. a.txt 2. abc.txt 3. asd.dat...

8. Shell Programming and Scripting

Count the number of subset of files in a directory

hi I am trying to write a script to count the number of files, with slightly different subset name, in a directory for example, in directory /data, there are a subset of files that are name as follow /data/data_1_(1to however many).txt /data/data_2_(1 to however many).txt...

9. Shell Programming and Scripting

Count number of files

Hi All! I need to have a script that counts the number of files arriving in a landing directory, them some app pick these files to be processed and load to a DB. But this process is so fast that I am not able to count all the files arriving on a landing directory. Please can you help? My...

10. Shell Programming and Scripting

Count number of pattern matches per line for all files in directory

I have a directory of files, each with a variable (though small) number of lines. I would like to go through each line in each file, and print the: -file name -line number -number of matches to the pattern /comp/ for each line. Two example files: cat...

LEARN ABOUT REDHAT

zipgrep

ZIPGREP(1L)															       ZIPGREP(1L)

NAME

       zipgrep - search files in a ZIP archive for lines matching a pattern

SYNOPSIS

       zipgrep [egrep_options] pattern file[.zip] [file(s) ...]  [-x xfile(s) ...]

DESCRIPTION

       zipgrep	will  search  files  within  a ZIP archive for lines matching the given string or pattern.  zipgrep is a shell script and requires
       egrep(1) and unzip(1L) to function.  Its output is identical to that of egrep(1).

ARGUMENTS

       pattern
	      The pattern to be located within a ZIP archive.  Any string or regular expression accepted by egrep(1) may be used.  file[.zip] Path
	      of  the  ZIP archive.  (Wildcard expressions for the ZIP archive name are not supported.)  If the literal filename is not found, the
	      suffix .zip is appended.	Note that self-extracting ZIP files are supported, as with any other ZIP archive; just	specify  the  .exe
	      suffix (if any) explicitly.

       [file(s)]
	      An  optional list of archive members to be processed, separated by spaces.  If no member files are specified, all members of the ZIP
	      archive are searched.  Regular expressions (wildcards) may be used to match multiple members:

	      *      matches a sequence of 0 or more characters

	      ?      matches exactly 1 character

	      [...]  matches any single character found inside the brackets; ranges are specified by a beginning character, a hyphen, and an  end-
		     ing character.  If an exclamation point or a caret (`!' or `^') follows the left bracket, then the range of characters within
		     the brackets is complemented (that is, anything except the characters inside the brackets is considered a match).

	      (Be sure to quote any character that might otherwise be interpreted or modified by the operating system.)

       [-x xfile(s)]
	      An optional list of archive members to be excluded from processing.  Since wildcard characters  match  directory	separators  (`/'),
	      this  option  may  be used to exclude any files that are in subdirectories.  For example, ``zipgrep grumpy foo *.[ch] -x */*'' would
	      search for the string ``grumpy'' in all C source files in the main directory of the ``foo'' archive, but none in any subdirectories.
	      Without the -x option, all C source files in all directories within the zipfile would be searched.

OPTIONS

       All options prior to the ZIP archive filename are passed to egrep(1).

SEE ALSO

       egrep(1), unzip(1L), zip(1L), funzip(1L), zipcloak(1L), zipinfo(1L), zipnote(1L), zipsplit(1L)

URL

       The Info-ZIP home page is currently at
       http://www.info-zip.org/pub/infozip/
       or
       ftp://ftp.info-zip.org/pub/infozip/ .

AUTHORS

       zipgrep was written by Jean-loup Gailly.

Info-ZIP							 17 February 2002						       ZIPGREP(1L)

10 More Discussions You Might Find Interesting

1. UNIX for Dummies Questions & Answers

Count number of files in subdirectories

Discussion started by: cbeverly

2. Shell Programming and Scripting

count number of files in a directory

Discussion started by: finalight

3. UNIX for Dummies Questions & Answers

Read directory files and count number of lines

Discussion started by: jl487

4. Shell Programming and Scripting

perl script on how to count the total number of lines of all the files under a directory

Discussion started by: adityam