Unix Shell Programming

[kbjradmin]

hi, i'm new at unix and am writing a program to classify data into data types (str, int, float); however, everytime i try to run it for debugging purposes, i get an error.

the error is:

./data_type.bash: line 72: syntax error: unexpected end of file

the program is:

#!/bin/bash
#
# data_type
# by: james brumond
#
# used for determining the data type of a string.
# returns one of the following:
#         [str | int | float]
#
# when used with the -s option, it can be used to
# change an int or float into a str.
#
# exit status codes:
#       64 - improper option use
#

helpMsg="data_type\nby:james brumond\nused to assign and determine data types.\n\
available data types for use are: [str | int | float]\n\n\
format for this command is:   data_type [variable name] [options]\n\n\
options:\n-s:type       manually sets the data_type of a variable to type\n\
-h      displays this help message\n\n\
typing this command with no options but with a valid variable name\n\
will echo the data_type of the variable."

if [ "$#" -eq 0 ]
then
        echo "$helpMsg"
        exit 0
elif [ "$#" -eq 1 ]
then
        if [ "$1" = -h ]
        then
                echo "$helpMsg"
                exit 0
        else
                # auto type set
                dataType="$?"
                if [ "$dataType" -eq 65 ]
                then
                        echo "str"
                        exit 0
                elif [ "$dataType" -eq 66 ]
                then
                        echo "int"
                        exit 0
                elif [ "$dataType" -eq 67 ]
                then
                        echo "float"
                        exit 0
                fi
        fi
elif [ "$#" -eq 2 ]
then
        if [ "$2" != -s:* ]
        then
                exit 64         # incorrect option use
        else
                # manual type set
                dataType="$?"
                if [ "$dataType" = 65 ]
                then
                        echo "str"
                        exit 0
                elif [ "$dataType" = 66 ]
                then
                        echo "int"
                        exit 0
                elif [ "$dataType" = 67 ]
                then
                        echo "float"
                        exit 0
                fi

please help!

[vol7ron]

u need two more "fi"s in there to close the outer two if statements.