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For loop in bash shell


 
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# 1  
For loop in bash shell

Hi,

I am using a for loop to manipulate files data_1.txt through data_100.txt. The for-loop is set up like this:

Code:
for i in {1..100}; do cut -f1 data_$i.txt > output$i.txt

I get the following error message when I run the code:

cannot open `data.txt' for reading: No such file or directory

The code seems to run fine though if I take out the "_" sign in the filenames and rename them data1.txt through data1000.txt. Do you know why I'm getting the error and how do I go about fixing it?
# 2  
Just enclose the file name variable in double quotes as below:

Code:
for i in {1..100}; do cut -f1 "data_$i.txt" > "output$i.txt"

This User Gave Thanks to hiten.r.chauhan For This Post:
# 3  
This does not seem to fix the issue.
# 4  
Missing double quotes is definitely not the reason for that error.

Verify if you are running this script inside the directory where you indeed have these files.

Also perform following changes:
Code:
#!/bin/bash
for i in {1..100}
do
        [ -f "data_$i.txt" ] && cut -d' ' -f1 "data_$i.txt" > "output$i.txt"
done

# 5  
Adding quotes won't make any difference for this code snippet. There is no way that data_$i.txt will ever expand to data.txt, although it might appear that way if $i contained a backspace character. (But, of course i won't contain a backspace in the loop you listed.)

Please show us the rest of the script or at least add set -xv to the start of your script to figure out what command in your script is actually failing.
 

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