Can we can pass the filename as variable in the shell script.
Sending the filename as a parameter file, the shell script takes the filename, needs to replace the string containing the filename with the variable in the shell script.

EX: test1.sh is the shell script and takes file1.csv as parameter file. Then file1.csv has to be replaced in the test1.ctl file where the control file contains the infile option taking as test1.csv

From the command line execute
test1.sh file1.csv

#Take the file1.csv file and replace it in test1.ctl

#test1.ctl
load data
infile 'testing.csv'
......
......

Here the testing.csv has to be replaced by 'test1.csv' in the test1.ctl file. Hope the scenario is clear and any solution is greatly appreciated. Thanks.

What do you mean by replacing the file.. do you want to copy or rename it..?

I receive different files with different filenames daily. These files needs to be loaded using sqlldr utility.
The filename in the control file is set to 'testing.csv'. Instead of changing the control file everyday, I am trying to set the filename as variable through a shell script and pass that filename as variable to replace the string in control file from testing.csv to the variable which is nothing but the filename.

so you simply want to do something like this:
Correct?

Actually the control file contains the file information as shown. The control file needs to be modified.

#test1.ctl
load data
infile 'testing.csv'
............



In the above test1.ctl, we need to replace testing.csv to the filename that we pass as a parameter from the shell script.

Why? The sqlldr "data=" command line parameter overrides the INFILE parameter. Also, you can simply remove it from the file altogether using vi.

But if you are really want to do it the way that you described, you can use sed.

or more specifically:

Thank you. It works for me.