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Search: Posts Made By: ch33ry
2,236
Posted By bartus11
I think abc has to be member of sftp group to...
I think abc has to be member of sftp group to make it work.
7,000
Posted By dude2cool
oops sorry :), try the below statement. I did not...
oops sorry :), try the below statement. I did not look close enough before posting.

if [ -f $path/*.log -o -f $path/*.out ]
2,415
Posted By Corona688
find never prints nonexistent files. If find...
find never prints nonexistent files. If find finds no files at all, it prints exactly nothing.

If the while read loop reads nothing the very first time, the code inside it won't run even once.
...
2,771
Posted By alister
The value stored in $? at the time of the...
The value stored in $? at the time of the if-statement's test is the exit status of the echo statement. You need to store $? in another variable, so that you can print it and then test it.
...
2,771
Posted By Corona688
Any command you run changes the value of $?, echo...
Any command you run changes the value of $?, echo included.
2,415
Posted By Corona688
Since you're on Linux, just add -maxdepth 1...
Since you're on Linux, just add -maxdepth 1 before the '('
2,242
Posted By Corona688
I think you got it right in the first place; ...
I think you got it right in the first place; looks more like moving than removal to me :)

The -exec ls -1 is redundant, find will print the files one by one by default when not given -exec.

ls...
1,048
Posted By agama
If I read your question correctly, you have files...
If I read your question correctly, you have files in the subdirectories that you do want to be deleted, but don't want to try to remove the directories themselves. If that is correct then try:

...
2,787
Posted By yazu
Yes, because DINFA_DOMAINS_FILE is not a process,...
Yes, because DINFA_DOMAINS_FILE is not a process, just a part of its name. You can use
pgrep DINFA_DOMAINS_FILE
if you have it or
ps -ef | grep DINFA_DOMAINS_FILE | grep -v grep | awk '{print $2}'
1,358
Posted By yazu
echo 'A/B/C 10' | awk '{print "var1 = " $1; print...
echo 'A/B/C 10' | awk '{print "var1 = " $1; print "var2 = " $2}'
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