Parse error: syntax error, unexpected T_STRING in C:\xampp\htdocs\yo\index.php on line 3

Question

I am new on php and i am trying to add this code

Im finding this error please any one who can solve this for me

<?php
$myName = ‘Guest’;
$myVar = “Welcome back $myName”;
echo $myVar;
?>

This is my html

<p><?php echo $myVar; ?></p>

Thanks

It looks like you are using left and right quotes, try retyping that in a plain text editor like Notepad (Windows), TextEdit (Mac), or gedit (linux). That way it will have normal quotes, e.g. ". Note: In TextEdit you will need to press Cmd+Shift+T to make it plain text. — uınbɐɥs
– uınbɐɥs, Commented Jul 8, 2012 at 12:33
yes thanks so much.. i was using wordpad thanks so much i will get some good php editor as you said thanks so much — Deepak Kumar
– Deepak Kumar, Commented Jul 8, 2012 at 12:35
@DeepakKumar If you don't want to go all out and use an IDE, I found that TextPad is a great basic editor to work with. Allows multiple files open at once, can do syntax highlighting and has line numbering. You really don't want too much else to start with. — Fluffeh
– Fluffeh, Commented Jul 8, 2012 at 12:37

Fluffeh · Accepted Answer · 2012-07-08 12:31:02Z

4

Looks like you are using funny quotes rather than the straight ' and " types.

<?php
$myName = 'Guest';
$myVar = "Welcome back $myName";
echo $myVar;
?>

answered Jul 8, 2012 at 12:31

Fluffeh

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1 Comment

Hassan Badawi Over a year ago

for adding help ( in my case - I forgot the ( ; ) semicolon on the previous line )

mfadel · Accepted Answer · 2012-07-08 12:33:25Z

1

you need to write

$myVar = "Welcome Back $myName";

the problem is that you are using the wrong char for the quote character

answered Jul 8, 2012 at 12:33

mfadel

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Comments

sarojrana · Accepted Answer · 2017-12-01 20:40:54Z

0

<?php
    $myName = "Guest";
    $myVar = "Welcome back ".$myName;
    echo $myVar;
?>

or

<?php
    $myName = "Guest";
    $myVar = "Welcome back $myName";
    echo $myVar;
?>

This is how code works while concatenating the string.

answered Dec 1, 2017 at 20:40

sarojrana

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Comments

Anh Pham · Accepted Answer · 2017-12-16 07:46:31Z

0

Write $myVar = “Welcome back" . $myName; instead $myVar = “Welcome back $myName";

<?php
$myName = ‘Guest’;
$myVar = “Welcome back" . $myName;
echo $myVar;
?>

edited Dec 16, 2017 at 7:46

Anh Pham

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answered Dec 16, 2017 at 3:13

manoj03h

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