Appending file content to a String variable in loop within a bash file is not working

Question

I am passing a text file to a bash while running. Text file has contents I want to supply to a java program as argument. Text file has each content in a new line. The contents print fine within the loop but I need to create a concatenated string with all the contents to pass to java program and appending to a string variable in loop is not working. This is how the program looks like:

#!/bin/bash
args=""
for var in $(cat payments.txt)
do 
  echo "Line:$var"
  args+="$var "
done
echo "$args"

It prints:

Line: str1
Line:str2
 str2  // args should have appended values of each line but it got only last line

File looks like:

str1
str2

Can anyone suggests what I am doing wrong here?

Thanks

Answer 1

Edit: the issue was due to \r\n line endings.

for var in $(cat payments.txt) is a nice example of useless use of cat. Prefer a while loop:

#!/bin/bash
args=""
while IFS= read -r var; do
  args+="$var "
done < payments.txt
echo "$args"

But instead of looping, which is not very efficient with bash, you could as well use a bash array:

$ declare -a args=($(< payments.txt))
$ echo "${args[@]}"
str1 str2

"${args[@]}" expands as separate words. Use "${args[*]}" to expand as a single word . If your line endings are \r\n (Windows) instead of \n (recent macOS, GNU/Linux), the \r will interfere. To remove the \r before printing:

$ echo "${args[@]%$'\r'}"

Answer 2

Your first echo is printing out the combination and not storing it in a new variable. Try:

#!/bin/bash
args=""
for var in $(cat payments.txt)
do 
  echo = "Line:$var" # this line prints but doesn't alter $var
  args+="Line:$var2 " #add Line: in here
done
echo "$args"