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Can you sort the files by it creation date? old files should be older,
can you place here an example to the files names (old+new)

Igal

Hi,

Please specify what have you done to solve this by yourself , it would avoid people being suspicious :-)
there are many ways to find it, try this one:
grep -o -c 'aaa' <FileName>; grep -o -c...

Enjoy...

#!/bin/bash

numOfRecords=0
prevDay=`head -1 log.txt| awk '{print $1}'`
startTime=`head -1 log.txt| awk '{print $2}'`
endTime=$startTime
while read data
do
curDay=`echo...

Ok thanks
Ill look on that

Hi
You right I should give u that b4
its 2.4.21-32

The cache is on 4308:

total used free shared buffers cached
Mem: 5954 5935 18 0 210 4308

the bad symtomps is...

Hi,

I having problem with my linux machine
it have 6Gb physical memory and somehow it always almost coming to the bottom neck and than it start writing to the swap memory
you can see that there...

for any case the paramter u looking for should be:

LimitRequestBody <Size of limit in bytes>

You should put it in your /config/httpd_<service>.conf file and restart apache

When you use:

for x in 1 2 3 4

It takes the first param and set it to x,
after that it runs into 2nd param and so on till it finish read all params.
At yours case u wrote

for i in [1..9] ...

Hi,

Your problem is that you are not running any loop ....., you just set the string [1..9] into i and print it to the screen,
For what you want to do u need the following:

for (( i = 0 ; ...

Ok, I take the line u wrote and try to run it on my server,
U actually not enter anything to var1 and thats why the if fails,

Try to do it like that:

var1=`awk -F" " '/filter/ {print $3}'...

Why you cant do the if command?
are u getting an error? if so what is the error?
did you tried it like that:

if [ $var1 -eq 5 ]; then
echo $var1
fi

If it still not working
try to put echo...

Try using the 'tr' option - this is replacing any string to anything u want
at your case you need to replace <SPACE>" with "
the syntax should be somthing like: | tr ' ' '"'
this will replace...

sed "s/ *$//g" p2.txt - this part is looking into p2.txt file and select one line from it,
the line number will be the number that replace the $ sign,

| grep -vi "[rows|row] selected" | grep -v...

Type top, you will get somthing like:
Look at the bold lines, you can see that on that machine u have 2G mem and 4 CPU

20:43:35 up 8 days, 22:32, 1 user, load average: 0.00, 0.00, 0.00
65...