Error with Bash script - syntax error

Question

File: test.sh

Command: sh test.sh

Content of test.sh

   x=1
    while [ $x -le 5 ]
    do
      echo "Welcome $x times"
      x=$(( $x + 1 ))
    done

Error:

test.sh: line 6: syntax error near unexpected token `done'

test.sh: line 6: `done'

GNU bash, version 3.2.39(1)-release (x86_64-pc-linux-gnu)

RESOLVED: I tried to use "pico" instead of trasfering file via FTP and all works! thanks! — Leonardo
– Leonardo, Commented Feb 20, 2011 at 21:38

tylerl · Accepted Answer · 2011-02-20 20:28:32Z

1

It works when I run that code.

Make sure you're running with bash not sh. The $(( )) construct is relatively new.

answered Feb 20, 2011 at 20:28

tylerl

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5 Comments

Rafe Kettler Over a year ago

Works for me under sh as well. My version of sh is 4.1.

tylerl Over a year ago

@Rafe On some architectures, sh is a separate, less capable executable than bash.

Leonardo Over a year ago

i have GNU bash, version 3.2.39(1)-release (x86_64-pc-linux-gnu)

Leonardo Over a year ago

RESOLVED: I tried to use "pico" instead of trasfering file via FTP and all works! thanks!

tylerl Over a year ago

@Leonardo: Check your FTP line ending conversions. Line endings are significant to bash since a newline marks a new statement, and if the line endings are being translated incorrectly, the statements will be mixed up. You CAN explicitly terminate statements with semicolons, but you're better off just making your file transfers work properly.

Jonathan Leffler · Accepted Answer · 2011-02-20 20:40:48Z

1

Run your script with:

bash test.sh

It will work. There are two main possibilities:

When it is run as sh, bash does not necessarily recognize all the syntax that it recognizes when it is run as bash. On the Linux and MacOS X machines where I tested this, though, both sh and bash worked fine.
Alternatively, you might be on a machine running AIX, HP-UX, Solaris or similar, where /bin/sh is not the same shell as bash at all. It is more rigid Bourne shell, where the notation you used is invalid - a syntax error.

edited Feb 20, 2011 at 20:40

answered Feb 20, 2011 at 20:35

Jonathan Leffler

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1 Comment

Jonathan Leffler Over a year ago

Look hard at the code, then. Possibly, you have CRLF (DOS) line endings? Even that's a bit of a stretch, though. The Linux machine I tested on has bash 3.2.25(1) on x86/64. Look at your code with od -c; there is ample evidence that what is copied and pasted is OK. Maybe try typing it in again into test2.sh, and see if that works. It sounds silly, but at this point, your problem is most peculiar.

Navi · Accepted Answer · 2011-02-20 20:25:22Z

0

That's strange on my system I get.

Welcome 1 times
Welcome 2 times
Welcome 3 times
Welcome 4 times
Welcome 5 times

Must have something to do with whitespace.

answered Feb 20, 2011 at 20:25

Navi

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Comments

nvez · Accepted Answer · 2011-02-20 20:26:43Z

0

Say hello to semicolons. :)

x=1
while [ $x -le 5 ]
do
  echo "Welcome $x times";
  x=$(( $x + 1 ));
done

You don't need a semicolon if it's one line, however, you might want to do a loop this way:

for i in `seq 1 5`
do
  echo "Welcome $i times"
done

answered Feb 20, 2011 at 20:26

nvez

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2 Comments

Jonathan Leffler Over a year ago

The semi-colons are not relevant.

Dennis Williamson Over a year ago

seq isn't needed in Bash: for i in {1..5} or for ((i=1;i<=5;i++))

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