Whats wrong with my shell script

Question

I want to take a local file and zip it at some other temp location. Following is my code which is giving error

path="/home/tft/Downloads/ie.js"
temp="/home/tft/Downloads/Temp"
name=$(basename "$path")
echo "$path"
echo "$name"
echo "$temp"
echo "$temp/$name.zip" #Its output is also weird!
zip -r -j "$temp/$name.zip" $path

Getting below output:

 /bin/bash test.sh

/home/tft/Downloads/ie.js
ie.js
/home/tft/Downloads/Temp
.zipjstft/Downloads/Temp 
        zip warning: name not matched: /home/tft/Downloads/ie.js

)zip . -i /home/tft/Downloads/ie.js -r -j /home/tft/Downloads/Temp

konsolebox · Accepted Answer · 2014-07-22 08:51:59Z

2

Your script is in DOS format. Convert it first to UNIX format:

sed -i 's|\r||' yourscript.sh

Or use dos2unix:

dos2unix yourscript.sh

The error messages you see is caused by having an extra character (carriage return \r) at the end of your values. This happens when your file's format is not UNIX but DOS since DOS' line endings is \r\n where UNIX only has \n.

answered Jul 22, 2014 at 8:51

konsolebox

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1 Comment

foobar Over a year ago

Oh great. this solved it. Never knew about it. Thanks

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