python - File Manipulation: Scripting Question

Question

I have a script which connects to database and gets all records which statisfy the query. These record results are files present on a server, so now I have a text file which has all file names in it.

I want a script which would know:

1. What is the size of each file in the output.txt file?
2. What is the total size of all the files present in that text file?

Update: I would like to know how can I achieve my task using Perl programming language, any inputs would be highly appreciated.

Note: I do not have any specific language constraint, it could be either Perl or Python scripting language which I can run from the Unix prompt. Currently I am using the bash shell and have sh and py script. How can this be done?

My scripts:

#!/usr/bin/ksh
export ORACLE_HOME=database specific details
export PATH=$ORACLE_HOME/bin:path information
sqlplus database server information<<EOF
SET HEADING OFF
SET ECHO OFF
SET PAGESIZE 0
SET LINESIZE 1000
SPOOL output.txt
select * from my table_name;
SPOOL OFF
EOF

I know du -h would be the command which I should be using but I am not sure how should my script be, I have tried something in python. I am totally new to Python and it's my first time effort.

Here it is:

import os

folderpath='folder_path'
file=open('output file which has all listing of query result','r')

for line in file:
 filename=line.strip()
 filename=filename.replace(' ', ' ')
 fullpath=folderpath+filename
# print (fullpath)
 os.system('du -h '+fullpath)

File names in the output text file for example are like: 007_009_Bond Is Here_009_Yippie.doc

Any guidance would be highly appreciated.

Update:

1. How can I move all the files which are present in output.txt file to some other folder location using Perl ?
2. After doing step1, how can I delete all the files which are present in output.txt file ?

Any suggestions would be highly appreciated.

Answer 1

Eyeballing, you can make YOUR script work this way:

1) Delete the line filename=filename.replace(' ', ' ') Escaping is more complicated than that, and you should just quote the full path or use a Python library to escape it based on the specific OS;

2) You are probably missing a delimiter between the path and the file name;

3) You need single quotes around the full path in the call to os.system.

This works for me:

#!/usr/bin/python
import os

folderpath='/Users/andrew/bin'
file=open('ft.txt','r')

for line in file:
    filename=line.strip()
    fullpath=folderpath+"/"+filename
    os.system('du -h '+"'"+fullpath+"'")

The file "ft.txt" has file names with no path and the path part is '/Users/andrew/bin'. Some of the files have names that would need to be escaped, but that is taken care of with the single quotes around the file name.

That will run du -h on each file in the .txt file, but does not give you the total. This is fairly easy in Perl or Python.

Here is a Python script (based on yours) to do that:

#!/usr/bin/python
import os

folderpath='/Users/andrew/bin/testdir'
file=open('/Users/andrew/bin/testdir/ft.txt','r')

blocks=0
i=0
template='%d total files in %d blocks using %d KB
'

for line in file:
    i+=1
    filename=line.strip()
    fullpath=folderpath+"/"+filename
    if(os.path.exists(fullpath)):
        info=os.stat(fullpath)
        blocks+=info.st_blocks
        print `info.st_blocks`+""+fullpath
    else:
        print '"'+fullpath+"'"+" not found"

print `blocks`+"Total"
print " "+template % (i,blocks,blocks*512/1024)

Notice that you do not have to quote or escape the file name this time; Python does it for you. This calculates file sizes using allocation blocks; the same way that du does it. If I run du -ahc against the same files that I have listed in ft.txt I get the same number (well kinda; du reports it as 25M and I get the report as 24324 KB) but it reports the same number of blocks. (Side note: "blocks" are always assumed to be 512 bytes under Unix even though the actual block size on larger disc is always larger.)

Finally, you may want to consider making your script so that it can read a command line group of files rather than hard coding the file and the path in the script. Consider:

#!/usr/bin/python
import os, sys

total_blocks=0
total_files=0
template='%d total files in %d blocks using %d KB
'

print
for arg in sys.argv[1:]: 
    print "processing: "+arg
    blocks=0
    i=0
    file=open(arg,'r')
    for line in file:
        abspath=os.path.abspath(arg)
        folderpath=os.path.dirname(abspath)
        i+=1
        filename=line.strip()
        fullpath=folderpath+"/"+filename
        if(os.path.exists(fullpath)):
           info=os.stat(fullpath)
           blocks+=info.st_blocks
           print `info.st_blocks`+""+fullpath
        else:
           print '"'+fullpath+"'"+" not found"

    print ""+template % (i,blocks,blocks*512/1024)
    total_blocks+=blocks
    total_files+=i

print template % (total_files,total_blocks,total_blocks*512/1024)

You can then execute the script (after chmod +x [script_name].py) by ./script.py ft.txt and it will then use the path to the command line file as the assumed path to the files "ft.txt". You can process multiple files as well.