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If you have gnu find use -iname for case insensitive match.

otherwise, try like this for example :
find . -name '[fF]2'

to match
f2
F2

Specify the name as a parameter:
find $file_path1 -name *$file_nm* -type f -mtime +$days

Change the command to:

find $file_path1 -name "*$file_nm*" -type f -mtime +$days -exec rm -f {} \;

as it should be.

Quick and dirty, suppress ALL error messages:

set +e # on error continue
find $file_path1/*$file_nm* -type f -mtime +$days -exec rm -f {} \; 2>/dev/null
find $file_path2/*$file_nm* -type f...

Since you did not post your last given command I am going to guess and recommend that you tested first with:

find /etc/logs/*aa* -type f -atime +59 -exec ls -l {} \;

That should pick the files...

Change:
find $File_Path/*$file_nm* -type f -atime +$days -exec mv {} $Dest_Path \;to:
find $File_Path/*$file_nm* -type f -atime +$days -exec mv -f {} $Dest_Path \;

find /home/etc -type f -atime -10 -exec mv {} /other/directory \;

The suggestion hergp made shows you how to use the for loop. A simpler way to perform what this script does is:
#!/bin/ksh
Src_Dir="/home/etc"
file_nm ="ab.temp"
cd "$Src_Dir"
printf "%s\n"...

...
for File in $Src_Dir/$File_nm*
...

You may try this also...

$ cat <<EOF | awk '{gsub(/[A-Za-z._]/,x,$0);1}1'
AB_XYZA_20130930183017.log
AB_DY_XYZA_20130930183017.log
AB_GZU_20130930183017.log
EOF

20130930183017...

If I understood correctly , you mean to say copy files from /home/etc/AB*20130930* to /home/etc/ARC_20130930/

file_path="/home/etc"
Last_Day=20130930

if [ ! -d "$file_path/ARC_${Last_Day}" ]...

The dot in find . invokes the search from present working directory. The command is trying to find files under present directory and with name /home/etc/files/ABC*

If you had a file as below then...

The name argument should be a file name (or glob), not a path. . (the path) should contain the path.


find /home/etc/files -name "ABC*" | wc -l

diff returns 0 when no differences are found, 1 when there are differences, >1 (2,3,4...)
when there was an error.

Probably the best usage is to cover all possibilities by looking for all the...

there is no not command in ksh. try ! or test

Hello smile689,

could you please try this, may be it will be helpful for you.



$ cat timings_file_query.ksh


echo "Please enter the number of files present types for files like...

Using awk:
ls YDT:FILE* | awk -F: '
{
idx = $1 OFS $2
if ( A[idx] -lt $3 )
A[idx] = $3
else if ( ! ( A[idx] ) )
...

using awk:

ls YDT:FILE* | awk -F: '{a[$0]++};$3>d{d=$3}END{for(i in a){split(i,x,":");if(x[3]==d)print i}}'

Hello Smile689,

Could you please try the following code hope it wil be helpful for you.




$ cat delete_latest_file.ksh

####Script starts from here#####
value=`ls -ltr | grep...

Spaces on both side of "=" are necessary.

if [ $str1 == $str2 ]

I believe ksh88 uses single "=" syntax though not sure.

if [ $str1 = $str2 ]

Further, to avoid syntax error in case any...

The if statement has to be modified as below as per syntax

if [ $str1 == $str2 ]

Without the spaces, it considers the expression as a string and evaluates to true as it is non-zero.

What you are doing is to execute the files. You probably need something more like this:-#!/bin/ksh

for i in 19*
do
txt=`cut -c -6 $i`
echo "$txt"
doneThe cut command will get you the...

Hi,
I have the files in the following files in a folder
19996587342
19487656550
19534838736
And i need to get the first 6 characters from the abvoe files
so i used the following script
...

${file##*:} deletes the largest matching pattern (matches from beginning) until colon :

Since I put asterisks * it deletes every character before that last colon :

For further reference:...

You don't need an awk code inside for loop to do this. You can use parameter substitution:
#!/bin/ksh

for file in pay*
do
file="${file##*:}"
print "$file"
done

If you are...