CODEWITHSUNDEEP
phphtmlsyntax-errorline-breaks
BlackCat
BlackCat

Reputation: 2044

why line break is not working in php code

I have added a line break in this piece of PHP code, but it's getting a syntax error. Does PHP not allow breaks in its code?

    <!DOCTYPE>
<html>
    <head>


    </head>
    <body>
</body>

<?php
define("Value_of_Pi", 3.14);
define ("Gravity_Value", 9.8);

echo "Pi value is".Value_of_Pi; 
<br/>
echo "Gravity Value is".Gravity_Value;
?>  
</html>

Upvotes: 0

Views: 201

Answers (5)

Claire
Claire

Reputation: 3184

<br> is html and must be echoed within php, like so:

<?php
define("Value_of_Pi", 3.14);
define ("Gravity_Value", 9.8);

echo "Pi value is".Value_of_Pi; 
echo "<br/>";
echo "Gravity Value is".Gravity_Value;
?>

You can also use \r\n in php like so:

echo "Pi value is".Value_of_Pi. "\r\n";

Upvotes: 0

Siddharth Chaudhery
Siddharth Chaudhery

Reputation: 196

You can use \r\n if you want to break line within php.

<?php
define("Value_of_Pi", 3.14);
define ("Gravity_Value", 9.8);

echo "Pi value is".Value_of_Pi. "\r\n"; 
echo "Gravity Value is".Gravity_Value;
?>

Upvotes: 2

jackeblagare
jackeblagare

Reputation: 417

You can use the code below instead:

echo '<br/>';

<br/> is an HTML tag hence the syntax error you're getting.

Upvotes: 0

cramopy
cramopy

Reputation: 3497

You need to write echo '<br/>';.

This is because you can either write HTML code or PHP within the specific sections.

You are currently in a PHP code section so you can only write PHP code there that will be valid.

<br/> is not a valid PHP code but HTML. However, because you are within <?php ?> it is not valid.

Upvotes: 0

Mureinik
Mureinik

Reputation: 311088

<br/> is not valid PHP syntax. You could either have the <br/> outside the PHP block:

<?php
define("Value_of_Pi", 3.14);
define ("Gravity_Value", 9.8);

echo "Pi value is".Value_of_Pi; 
?> 
<!-- PHP block terminated -->
<br/>
<!-- New PHP block opened: -->
<?PHP
echo "Gravity Value is".Gravity_Value;
?>

Or just echo it from PHP:

<?php
define("Value_of_Pi", 3.14);
define ("Gravity_Value", 9.8);

echo "Pi value is".Value_of_Pi; 
echo "<br/>"; # Here!
echo "Gravity Value is".Gravity_Value;
?>

Upvotes: 1

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