syntax error near unexpected token `if'

Question

I am currently trying to write a bash script which helps me step through a directory and check for .jpeg or .jpg extensions on files. I've come up with the following:

#declare $PICPATH, etc...

for file in $PICPATH
    if [ ${file -5} == ".jpeg" -o ${file -4} == ".jpg" ];
    then
        #do some exif related stuff here.
    else
        #throw some errors
    fi
done

Upon execution, bash keeps throwing a an error on the if line: "syntax error near unexpected token `if'.

I'm a completely new to scripting; what is wrong with my if statement?

Thanks.

To remove the last 5 characters of a variable use ${file::-5}, what you are doing ${file -5} won't work. — Veda
– Veda, Commented Oct 9, 2019 at 12:31

cyfur01 · Accepted Answer · 2013-01-26 21:01:15Z

12

I think you're just missing the do clause of the for loop:

#declare $PICPATH, etc...

for file in $PICPATH; do
    if [ ${file -5} == ".jpeg" -o ${file -4} == ".jpg" ];
    then
        #do some exif related stuff here.
    else
        #throw some errors
    fi
done

answered Jan 26, 2013 at 21:01

cyfur01

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2 Comments

el HO Over a year ago

I can't believe I missed that. I spent an hour trying to figure out what went wrong. Thanks a lot.

emi Over a year ago

You can put the do on a new line if you prefer it. Then, you can omit the semicolon ; before it. The same applies to the then after the if.

Gilles Quénot · Accepted Answer · 2013-01-26 21:12:09Z

2

${file -5}

is a syntax error. Maybe you mean

${file#*.}

?

Anyway, better do :

for file in $PICPATH; do
    image_type="$(file -i "$file" | awk '{print gensub(";", "", $2)}')"
    case $image_type in
        image/jpeg)
            # do something with jpg "$file"
        ;;
        image/png)
            # do something with png "$file"
        ;;
        *)
            echo >&2 "not implemented $image_type type "
            exit 1
        ;;
    esac
done

If you only want to treat jpg files, do :

for file in $PICPATH; do
    image_type="$(file -i "$file" | awk '{print gensub(";", "", $2)}')"
    if [[ $image_type == image/jpeg ]]; then
            # do something with jpg "$file"
    fi
done

edited Jan 26, 2013 at 21:12

answered Jan 26, 2013 at 21:07

Gilles Quénot

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1 Comment

el HO Over a year ago

Thanks a lot for the improved version, but I only need the case for jpeg: everything else should be ignored. Thanks a lot for the response anyways.

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