使用命令行参数在Bash中检查文件是否存在

Question

How do you use a command line argument as a file path and check for file existence in Bash?

I have the simple Bash script test.sh :

#!/bin/bash

set -e
echo "arg1=$1"
if [ ! -f "$1" ]
then
    echo "File $1 does not exist."
    exit 1
fi
echo "File exists!"

and in the same directory, I have a data folder containing stuff.txt .

If I run ./test.sh data/stuff.txt I see the expected output:

arg1=data/stuff.txt
"File exists!"

However, if I call this script from a second script test2.sh , in the same directory, like:

#!/bin/bash
fn="data/stuff.txt"
./test.sh $fn

I get the mangled output:

arg1=data/stuff.txt
 does not exist

Why does the call work when I run it manually from a terminal, but not when I run it through another Bash script, even though both are receiving the same file path? What am I doing wrong?

Edit: The filename does not have spaces. Both scripts are executable. I'm running this on Ubuntu 18.04.

Answer 1

The filename was getting an extra whitespace character added to it as a result of how I was retrieving it in my second script. I didn't note this in my question, but I was retrieving the filename from folder list over SSH, like:

fn=$(ssh -t "cd /project/; ls -t data | head -n1" | head -n1)

Essentially, I wanted to get the filename of the most recent file in a directory on a remote server. Apparently, head includes the trailing newline character. I fixed it by changing it to:

fn=$(ssh -t "cd /project/; ls -t data | head -n1" | head -n1 | tr -d '\n' | tr -d '\r')

Thanks to @bigdataolddriver for hinting at the problem likely being an extra character.