whats worng in my code ????

Dexter6996 · Oct 31, 2015

<?php

$abc = 98 ;

$xyz = 89;

?>

<? php
echo '$abc' ;

?>

Stiletto · Oct 31, 2015

I have no idea, but I know what's worng in your title...

iBrazilian · Oct 31, 2015

$abc = "98";
$xyz = "89";

echo $abc;

mantic0re · Nov 1, 2015

Hi,

Simple questions such as these are perfectly answered in other sources, such as w3schools.

ScarfaceMontana · Nov 1, 2015

Here is your mistake:

echo '$abc' ;

This is how its done:
echo $abc ;

Hawkster · Nov 1, 2015

Quoting so OP can see what you put

ScarfaceMontana said:
Here is your mistake:

echo '$abc' ;

This is how its done:
echo $abc ;

Shirko · Nov 1, 2015

You cannot do that. You should either use:

echo "$abc"; // note the double quotes
echo $abc; // just the variable

botsource · Nov 4, 2015

Pay attention to that style of coding when you are creating sql queries. (sql injection)

member8200 · Nov 18, 2015

Dexter6996 said:
<?php

$abc = 98 ;

$xyz = 89;

?>

<? php
echo '$abc' ;

?>

<?php
$abc = 98 ;
$xyz = 89;

echo $abc;
?>
here's how it's done. when you put '$abc' you are stating a string and not the variable.

revproxy · Nov 21, 2015

Its must be "$abc" or "{$abc}"... not '$abc'

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