How to get Find command work with a variable containing "*"?

[6656]

Hello guys,

I have a problem to pass a variable ($optn) containing quoted wild card to FIND command. for exsample.

#$1=name 
#$2=*.log

if [  "$1" == "name" ]; then
    optn="-name '$2'"
else
    optn="-mtim +$2"
fi 
find . -type f $optn -ls

Please give your advise to make it work.
Thanks in advance.

 

[feherke]

Hi

find . -type f [red]"[/red]$optn[red]"[/red] -ls

Feherke.

http://rootshell.be/~feherke/

 

[columb]

Er...
There are a number of bugs in your code, but I'll leave those as an excercise.
However the code

$2="*.log"
find . -name "$2"

will work. Note the double quotes.

Ceci n'est pas une signature
Columb Healy

 

[PHV]

to make it work
what is not working ?
computer crash ? any error message ? unexpected behaviour ? ... ?

To trace the execution of your script use the set -x command.

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[6656]

Thanks for your speedy response.

to feherke, the "$optn" doesn't work. error msg

find: bad option -name '*.log'
find: path-list predicate-list

to columb, option -name or -mtime is determined by if...then..fi.

Below is the expected result return to be excuted:

find . -type f -name "*.log"
or
find . -type f -mtime +100

Thanks again.

 

[columb]

6656

Use PHV's suggestion and run the script with the -x parameter. Either

ksh -x myscript.ksh

or amened the first line of your script to

#!/bin/ksh -x

Then, if you post the output we can see what's going on. feherke's suggestion should work, it does on my system. As to your

to columb, option -name or -mtime is determined by if...then..fi.

Below is the expected result return to be excuted:

find . -type f -name "*.log"
or
find . -type f -mtime +100

That's not what your code snippet does!

Ceci n'est pas une signature
Columb Healy

 

[6656]

Here is full code:

 #!/bin/ksh -x
 #$1 - name of option
 #$2 - pattern of filename or age of file

optn=`echo "$1"| tr "[A-Z]" "[a-z]"`
if [ "$1" == "name" ]; then
    optn="-name '$2'"
else
    optn="-mtime +$2"
fi
find . -type f $optn -ls

output:
~/temp>rmf.sh name *.txt
+ rmf.sh name eli.txt ftpfile.txt section.txt test_syb.txt user.txt

There is no problem with " rmf.sh mtime 300" command

 

[PHV]

Call your script like this:
rmf.sh name "*.txt"

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[6656]

Call your script like this:
rmf.sh name "*.txt"

output:
~/temp>rmf.sh name "*.txt"
+ rmf.sh name *.txt
+ + tr [A-Z] [a-z]
+ echo name
optn=name
+ [ name == name ]
+ optn=-name '*.txt'
+ find . -type f -name '*.txt' -ls

In above '*.txt', single quote is treated as part of tring, but not quote function.

 

[PHV]

Even if you replace this:
optn="-name '$2'"
with this ?
optn="-name $2"

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[6656]

with this ?
optn="-name $2"

output:

~/temp>rmf.sh name "*.txt"
+ rmf.sh name *.txt
+ + tr [A-Z] [a-z]
+ echo name
optn=name
+ [ name == name ]
+ optn=-name *.txt
+ find . -type f -name eli.txt ftpfile.txt section.txt test_syb.txt user.txt -ls
find: bad option ftpfile.txt
find: path-list predicate-list

If it can be resulted in quotes around the filenames at final execute time, such as,
find . -type f -name "eli.txt ftpfile.txt section.txt"

 

[kHz]

What are you trying to do? Above you have $2=*.log then above you have optn=*.txt. What happened to *.log?

Anyway, as first mentioned there are too many problems with the code and it wouldn't run anyway.

 

[6656]

To kHz:

The issue is how to fix the code to pass the paremater value with '*' in the script.

Something is weird that it causes the syntax error when assigning a value '*' or '*.log' to a parameter, but '*.txt', '*.abc', '*.log.*' or 'a*.log' is OK. Any idea why?

Thanks again

 

[columb]

If the script fails when you pass '*' or '*.log' but not with other patterns then this nearly always means that you have a number of *.log files in your working directory
Consider the code

find . -name *.log

If there are no files called something.log then this will be passed to the shell unaltered and work as expected. Now, suppose that we have two files one.log and two.log, in this case what will be passed to the shell is

find . -name one.log two.log

which is a syntax error.

The problem you now mention is slightly different to how we understood the orignal post. Anyway the script

find . -name "$1"

works for me when called using

./testscript "*.log"

Note all the double quotes, both within and without the script.

Ceci n'est pas une signature
Columb Healy

 

[columb]

Going through your code line by line

#!/bin/ksh -x
 #$1 - name of option
 #$2 - pattern of filename or age of file

no problems so far

optn=`echo "$1"| tr "[A-Z]" "[a-z]"`

You've set 'optn' to $1 translated to lower case, but thenyou never use this. However you then test $1 against 'name' I assume you want to test $1, translated to lower case against 'name'
# so

if [ "$1" == "name" ]; then

should be

if [ "$optn" == "name" ]; then

    optn="-name '$2'"

Well, er, '$2', because of the single quotes, is uninterpreted, and you want optn to be '-name *.log' so try

  optn="-name \"$2\""

Note the escaped double quotes

else
   optn="-mtime +$2"
fi

Not to much wrong here. You haven't checked that $2 is numeric but we'll live with that.
Finally

find . -type f $optn -ls

Whilst this produces the right command, for some reason it doesnt seem to work. I resolved this by changing

  optn="-name \"$2\""

to

  optn="find . -type f -name \"$2\" -ls"

and the final line to

ksh -c "$optn"

Putting it all together you get

#!/bin/ksh -x
 #$1 - name of option
 #$2 - pattern of filename or age of file
optn=`echo $1 | tr "[A-Z]" "[a-z]"`
if [ "$optn" = "name" ]
then
  optn="find . -type f -name \"$2\" -ls"
else
  optn="find . -type f -mtime +$2"
fi
ksh -c "$optn"

Which works for me.

Ceci n'est pas une signature
Columb Healy

 

[6656]

Thanks a lot! Columb,

I think i'm going to take the codeing method as below.

if [ "$optn" = "name" ]
then
  find . -type f -name "$2" -ls
elif [ ...condition... ]; then
 ... more....
else
  find . -type f -mtime +$2 -ls
fi

It just seems more typing than the way in my original approach.
Thanks all again.