how i can print the variable as it not the value of the variable

Question

In this code

<?php print "value of a is $a" ?>

I need to print this statement as it how?

Answer 1

If you want to print $a and not the a variable's value, there are two ways:

1. Escape the $: print "value of a is \$a";

2. Use single quotes instead

Everything in single quotes, including $something and special chars (newline \n, Unicode chars \uxxxx) are treated literally.

print 'value of a is $a';

Answer 2

Append with a dot, escape double-quoted strings with a backlash, or use single quotes and obviate the need for escaping.

Answer 3

If you want to include the variable in the string you must use single quotes like so

<? php print 'value of a is $a' ?>

Hope this helps.

Answer 4

PHP performs variable parsing in double quoted strings,

You can use single quoted strings.

So depending on want you really want, you can do

print 'value of a is $a';

which literally print the string 'value of a is $a' (which does not make much sense).

Or you can use string concatenation to print the variable and it's value:

print 'value of $a is ' . $a;

which will print 'value of $a is 42' (if $a = 42;).

Answer 5

If you want to print $a and not the a variable's value, there are two ways:

1. Escape the $

<?php print "value of a is \$a"; ?>

2. Use single quotes instead

Everything in single quotes, including $something and special chars (newline \n, Unicode chars \uxxxx) are treated literally.

<?php print 'value of a is $a'; ?>

Answer 6

You could either use single quotes or escape $a within double quotes with a \.

Either:

<?php print 'value of a is $a' ?>

Or:

<?php print "value of a is \$a" ?>

Answer 7

use single quote if you don't want to print the value of variable

<? php print 'value of a is $a' ?>