basic bash programming question

[Jimb0]

I keep getting this error "line 36: syntax error: unexpected end of file", ive googled it and the only solutions i could find were not working. im still not entirely sure im running the program correctly. im using this command in my linux terminal: "bash Homework5Bash.bash"

#!/bin/bash
name=$"Joe Shmoe"
echo "$name"
var=0
while [$var -lt 101]
do
	i=var%2;  
	if [$i = 0];then
		echo "$var";
	fi
	j=var%17;
	if [$j = 0];then
		if [$i = 0];then
			echo " banana,";
		else
			echo "$var banana";
		fi
	fi
	k=var%15;
	if [$k = 0];then
		if [[$i = 0] || [$j = 0]]; then
			echo " pear";
		else
			echo "$var pear,";
		fi
	fi
	l=var%6;
	if [$l = 0];then
		echo "orange";
	fi
	var=$[$var+1];
	echo "\n";
done

Im really just looking for any way to solve that end of file error, just so I can have the program at least run.

[Salem]

> I keep getting this error "line 36: syntax error: unexpected end of file",
Well your posted code has only 33 lines, so now what?

What I do get is
$ bash foo.sh
Joe Shmoe
foo.sh: line 5: [0: command not found

[Jimb0]

yes because when i posted i deleted 2 lines worth of comments at the top containing my real name and student id

am i supposed to be naming my bash file as a ".sh" or a ".bash "?

ive only been taught 30 minutes worth of bash programming so im very new to this

EDIT: This is weird, im getting that error in my original code, yet when i copied the same exact code from what i posted im getting that same error you mentioned. and im looking at it and all that is changed is my real name to Joe Schmoe and i removed 2 comments... thats it.

[Salem]

Well this prints 100 lines of \n

#!/bin/bash
name=$"Joe Shmoe"
echo "$name"
var=0
while [ $var -lt 101 ]
do
    i=var%2;
    if [ $i = 0 ]; then
        echo "$var";
    fi
    j=var%17;
    if [ $j = 0 ]; then
        if [ $i = 0 ]; then
            echo " banana,";
        else
            echo "$var banana";
        fi
    fi
    k=var%15;
    if [ $k = 0 ]; then
        if [ [ $i = 0 ] || [ $j = 0 ] ]; then
            echo " pear";
        else
            echo "$var pear,";
        fi
    fi
    l=var%6;
    if [ $l = 0 ]; then
        echo "orange";
    fi
    var=$[ $var+1 ];
    echo "\n";
done

Note that shell scripts are a bit more sensitive to white space (compared to a C program).
Look at all the if statements, and then look at your while statement. Use -lt type operators.

[Jimb0]

ahh i see thank you so much i almost have this working:

#!/bin/bash
name=$"Joe Shmoe"
echo "$name"
var=0
while [ $var -lt 101 ]
do
    i=$[ $var%2];
    if [ $i -eq 0 ]; then
        echo "$var";
    fi
    j=$[ $var%17];
    if [ $j -eq 0 ]; then
        if [ $i -eq 0 ]; then
            echo " banana,";
        else
            echo "$var banana";
        fi
    fi
    k=$[ $var%15 ];
    if [ $k -eq 0 ]; then
        if [ [ [ $i -eq 0 ] || [ $j -eq 0 ] ] ]; then
            echo " pear";
        else
            echo "$var pear";
        fi
    fi
    l=$[ $var%6 ];
    if [ $l -eq 0 ]; then
        echo " orange";
    fi
    var=$[ $var+1 ];
    echo "\n";
done

everything is printing out correctly except im just getting some weird errors with the multiple conditional statement on line 21.

EDIT: nm i got it completely working just discovered the "-o"

thank you so much...again.