Parse error with php function

Question

I've been going through w3schools trying to get to grips with php. I've done a little Java and C++ in the past but nothing web side. I can't figure out why I'm getting a parse error

 <!DOCTYPE html>
<html>
<body>

<?php

// calling a variable can be done with $variable or . $Variable .
// php is loosely typed - you do not have to tell it that a var is a int or string etc


$color = "red"; //variables are case sensitive

echo "My car is " . $color . "<br>";


//scopes for vars are either local, global or static. Local are declared inside a function and global are declared outside a function

function myTest()
{
    $x = 5; //local scope
    echo "<p> Variable X inside function is: $x</p>;
}


?>

</body>
</html>

This returns a "Parse error: syntax error, unexpected end of file in C:\inetpub\wwwroot\phpinfo.php on line 28"

This runs fine with the function removed. What am I missing?

Get a good IDE/Editor with syntax checking. It will go a long way to saving you headaches. — aynber, Mar 15 '18 at 13:52

score -2 · Accepted Answer · edited Mar 15 '18 at 13:58

-2

You missed the closing double quote (") on your echo line.

Change this line:

echo "<p> Variable X inside function is: $x</p>;

To this:

echo "<p> Variable X inside function is: $x</p>";

Also use good syntax editor. as @aynber suggest.

edited Mar 15 '18 at 13:58

JAAulde

19,250
5
52
63

answered Mar 15 '18 at 13:53

Hamza Zafeer

2,360
13
30
42

I feel like an idiot. Thank you! – C Mills Mar 15 '18 at 13:56

Parse error with php function

1 Answers1