PHP variables

Anon

Hi All,

I am new to this php scripting.I did a very simple script with variables.When I ran that php script from, the system is not recognizing the variables.It is giving a syntax error.With out variables it is working fine. One more thing is the system is not even recognizing functions(like phpinfo()).Here is the code.

<title>This is my first PHP</title>
<body>
$name="Bobby"
$age=3
<? echo "The name of the person is" $name; ?>
<? echo "The name of the person is" $age ;?>

</body>

I am getting the following error.

Parse error: parse error, expecting ','' or';'' in c:\program files\apache group\apache\htdocs\detail.php on line 6

If anybody has seen the same problem.Please help me.

TXS.
sankar

daarius

First of all the variables needs to be declared inside the <?php ?> tags. remember all the PHP codes needs to go inside these tags.

and secondly i think you best to go to php.net and read more on syntax of php. you have error in above echo statement. e.g. to add up the string you need to put "." (period or decimal) character. so, e.g. you will do echo "My name is ".$name;

and also at the end of the line you need to put ; character to terminate the statement.

re-write the above code to:
<?php
$name "Bobby";
$age = 3;
echo "My name is ".$name;
echo "My age is ".$age;
?>

jc94062

Why use ".$variable?
Its easier just to put it inside the quotes

<title>This is my first PHP</title>
<body>

<?php
$name = "Bobby";
$age = "3";
echo "The name of the person is $name";
echo "<br>";
echo "The age of the person is $age";
?>

</body>

which can of course be compressed

<title>This is my first PHP</title>
<body>
<?
$name="Bobby";
$age=3;
echo "The name of the person is $name<br>The age of the person is $age";
?>
</body>

But the above is probably easier to read.

Anon

It is working fine.

Thanks a lot to Daarius and James.

Anon

Yeah, you just need to throw a ; on the end of each line so that PHP knows that command line is complete. Should work fine then

-Kevin