I have the below if statement. It gives me syntax error when I have both the csv file and the zip file in the folder. But it works when I have the csv file alone or the zip file alone or no files exist.
I'm not sure what causes the syntax error in the command below:
if [ -f ctf_error_report_??_????????????.csv -o -f ctf_error_report_??_????????????.???.zip ]; then echo "Successful" else echo "Problem" fi exit 0 Any insight regarding this?
This is a dangerous way to check for files. If you happen to have more than one match in that directory then it won't work because each file will become a separate parameter, which won't match the syntax of the text. A better (although not foolproof) way to do it is like this:
set ctf_error_report_??_????????????.csv csv_file=$1 set ctf_error_report_??_????????????.???.zip zip_file=$1 if [ -f "$csv_file" -o -f "$zip_file" ] then echo "Successful" else echo "Problem" fi 
set ...?.csv; n_csvfiles=$#; set ...?.???.zip; n_zipfiles=$#; if [ $n_csvfiles -gt 0 -0 $nzipfiles -gt 0 ]; ...
I suspect you need this to test the success of some file transfer or report generation script (am I close?). The way I would do it (may not be the best, but it should work) is this:
#!/bin/bash FILE_FOUND=0 ls ./ctf_error_report_??_????????????.csv &>/dev/null if [ $? -eq 0 ]; then FILE_FOUND=1 fi ls ./ctf_error_report_??_????????????.???.zip &>/dev/null if [ $? -eq 0 ]; then FILE_FOUND=1 fi if [ $FILE_FOUND -eq 1 ]; then echo "Successful" else echo "Problem" fi This would return successful the moment the file pattern you're looking for is found in the directory of the script. You could change ./ for the full path of the files of course...
ksh -x your-scriptto see what gets executed. The chances are that the file name expansion expressions yield more than one file name, as the other people have diagnosed in their answers.