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So is there any way to go about this?? Am really getting tired :(

Nope....Intel Core Duo

Sorry was too desperate , have been trying for hours together.....

Error message with rpm -i gcc-xxx.rpm



warning: gcc-4.3.2-7.ppc64.rpm: Header V3 DSA signature: NOKEY, key ID 4ebfc273...

Hi guys am using Open Suse 10 and I misplaced the DVD somewhere...i dont have any pre existing gcc compilers. How do i install gcc now? ./configure isnt working, make install isnt working. Please...

Fine then, I guess there is no easy way out of this one, most probably will go for the sorting of du -k and then depending on length, will add a multiplying factor (like 1/1024 for e.g). Thanks for...

Well I've discovered another problem with what you've said...now suppose I have a folder with size 1.1 M (indeed i do!) then I guess I cant use this logic.

Well I did have the idea of using sed to replace all G's with six zeros, M's with 3 zeros but the problem is in converting them back...suppose a file has 1000K i dont want it to appear as 1KK after...

I have the following file. Its an output from a du command with certain conditions attached to it. I used du -ah as I need the 1st column to look human readable. sort -nr is not giving me the output...

Well as a programmer I have no problems in counting off digits, but damn these end users who need to be spoonfed :p

Well if that's the only solution, its a little more complicated than a k, If I do a division by 1024 once, i need to append K, twice, by M, and thrice, by G and so on.

That's why I was looking...

Well now that I've got my result. I want to try out something here. Suppose I want the file size (column 1 of my output) to be printed in human readable form, How do I go about it??

performing...

I think I can give myself the Idiot of the week award :mad:

The solution is



read fs

du -a <folder> | nawk '$1 > fs {print $0}' fs=$fs




Sorry Zaxxon for the trouble :( . Thanks...

I've got the output you've shown. But the real problem starts when you give some variable instead of 20.

du -a . | head -20

output :

2 ./.profile
2 ./local.cshrc
2 ./local.login
2 ./local.profile
2 ./abat.ksh
22 ...

Sadly it isnt working even now....and one thing is for sure, it has got nothing to do with du. Because the output i posted was a sample one, not the complete form of du and still the code you gave...

I am not him and he is not me :) . I've removed that k factor by removing the -h option from du but it hasnt brought me any closer to finding the solution.

I did try that but unfortunately the output still shows all the files present and does not perform the comparison.

28270 .
28170 ./OMautomation
27926 ./OMautomation/pqms
26096 ./OMautomation/pqms/EDM_Batch20080907.log
1744 ./OMautomation/pqms/REDX20080907.log
242 ./OMautomation/pvd
52 ...

Hi

I've tried to use a different approach by first sorting the list that du gives me. Is it now possible to use this sorted list to find all entries which are greater than a particular value??
...

I havent been able to find the suitable version of du and zaxxon's code isnt giving me the result :( .


Also, using

awk -v var = ${filesize} just isnt working for integers even after...

I've tried using -ak as well as the code that you've posted.


Both are giving the same result : displaying all the files and subfolders present in the directory mentioned.

Pludi

my version of unix doesnt support -b option on du..


Zaxxon

I've tried what you've told...

Code ( after read filesize) :

typeset -i filesize = ${filesize}
awk -v filesize =...

Hi I'm trying to write a script which takes an integer as input and checks for files with size > input and displays it on screen.

The code I've tried :

echo 'Enter the min file size expected'...