Why does this extremely simple PHP code not work? syntax error, unexpected T_STRING

Question

I have just signed up for Treehouse training and I have this mini challenge to complete. I can't get this to work. If the variable is set to 'vanilla' I will get a successful output. However, if I use 'cookie dough' then I get a syntax error, I think it may be because of the space between cookie and dough in the varial prehaps?

 <?php 
        $flavor = cookie dough;
        echo "<p> Your favourite flavor is ";
        echo $flavor;
        echo ".</P>";
        if ($flavor == "cookie dough") {
            echo "<p> Yeah I like cookie dough aswell! </p>";
        }
    ?>

Maybe you want tp check in php.net how to assign strings to variable. You need to quote them. — Royal Bg, Aug 20 '13 at 10:28
You need to learn how to [READ and debug](http://stackoverflow.com/questions/12769982/reference-what-does-this-error-mean-in-php) error messages. Everything you needed to solve the problem was in the error message. on the line given in the error message you have a string that php did not expect — Anigel, Aug 20 '13 at 10:29

score 2 · Accepted Answer · answered Aug 20 '13 at 10:29

2

Here you forgot ""

$flavor = cookie dough;

should be

$flavor = "cookie dough";
          ^            ^

answered Aug 20 '13 at 10:29

Yogesh Suthar

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1

+1 , Neat representation ! – Shankar Narayana Damodaran Aug 20 '13 at 10:31
1

Brilliant! Simple mistake, wont happen again! :) – ProEvilz Aug 20 '13 at 10:33
@yogesh Suthar Why does it work when I use `flavour = vanilla` and then set the conditional to check if the variable is equal to `vanilla`. Does it only have to be in quotation marks when using multiple words? – ProEvilz Aug 20 '13 at 10:37
@AshleyBrown, Exactly. You need to wrap them in quotes to treat them as a string. – Shankar Narayana Damodaran Aug 20 '13 at 11:33

score 2 · Answer 2 · answered Aug 20 '13 at 10:29

Enclose $flavor variable value with quotes.

<?php 
        $flavor = "cookie dough";
        echo "<p> Your favourite flavor is ";
        echo $flavor;
        echo ".</P>";
        if ($flavor == "cookie dough") {
            echo "<p> Yeah I like cookie dough aswell! </p>";
        }
    ?>

score 1 · Answer 3 · answered Aug 20 '13 at 10:28

1

Simply,

$flavor = cookie dough;

isn't a correct code, enclose with quotes. You can see possible ways to define strings here.

answered Aug 20 '13 at 10:28

Alma Do

36,374
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score 1 · Answer 4 · answered Aug 20 '13 at 10:29

1

$flavor = cookie dough;

should be

$flavor = 'cookie dough'; or

$flavor = "cookie dough";

To define a string you should be use "'" or '"';

answered Aug 20 '13 at 10:29

som

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score 1 · Answer 5 · answered Aug 20 '13 at 10:30

1

You forgot to enclose cookie dough in quotes:

$flavor = "cookie dough";

answered Aug 20 '13 at 10:30

Paddyd

1,850
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Why does this extremely simple PHP code not work? syntax error, unexpected T_STRING

5 Answers5