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How to pass a parameter to a shell script
This is a discussion on How to pass a parameter to a shell script within the Shell scripting forums, part of the Development/Scripting category; I have 2 scripts, ftp_file.sh and sendfile.sh ftp_file.sh ~~~~~~ #!/bin/bash sftp -b /home/<instance>/script/ftp/sendfile.sh oracle@<host> sendfile.sh ~~~~~~~ #!/bin/bash cd /u01/<destination_directory> put ...
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07-23-2008, 09:49 PM
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How to pass a parameter to a shell script
I have 2 scripts, ftp_file.sh and sendfile.sh
ftp_file.sh ~~~~~~ #!/bin/bash sftp -b /home/<instance>/script/ftp/sendfile.sh oracle@<host> sendfile.sh ~~~~~~~ #!/bin/bash cd /u01/<destination_directory> put /d01/<source_directory>/xxx.out xxx.out When i run ./ftp_file.sh the sendfile.sh is executed and the xxx.out file is transferred from the source_directory to the destination_directory. This works fine. The sendfile.sh is hardcoded with xxx.out. How can i pass a parameter to sendfile in order to transfer other files to the destination_directory instead of just xxx.out? I am new to shell scripting. regards flyingrat 
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07-24-2008, 01:34 AM
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07-24-2008, 03:14 AM
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Code:
#!/bin/bash cd /u01/<destination_directory> put /d01/<source_directory>/$1.out $2.out $2 is second parameter sendfile.sh file1 file2 will transfer as Code:
put /d01/<source_directory>/file1.out file2.out
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