Adding a variable to PHP script causes another line to create an error

Question

Quick question... I'm using this code to communicate with my flash application:

<?php
// Only run this script if the sendRequest is from our flash application
if ($_POST['sendRequest'] == "parse") {
// Print the var to flash
print "var1=The user's name is Harry";

}

?>

Now this works fine but as soon as I add a variable:

<?php
// Only run this script if the sendRequest is from our flash application
if ($_POST['sendRequest'] == "parse") {
$uname = "Name"
// Print the var to flash
print "var1=The user's name is Harry";

}

?>

I get an error stating:

Parse error: syntax error, unexpected T_PRINT in /home/a4935911/public_html/usersOnline.php on line 7

Where line 7 is my print statement. Why is this happening??? Please can someone help. PHP is driving me crazy...

Funk Forty Niner · Accepted Answer · 2013-10-26 16:46:41Z

2

Missing semi-colon in $uname = "Name"

change it to $uname = "Name";

Rewrite:

<?php
// Only run this script if the sendRequest is from our flash application
if ($_POST['sendRequest'] == "parse") {
$uname = "Name";
// Print the var to flash
print "var1=The user's name is Harry";

}

?>

Added note: A semi-colon can be omitted if it's the last line of code with nothing else that will be executed/included thereafter.

edited Oct 26, 2013 at 16:46

answered Oct 26, 2013 at 16:38

Funk Forty Niner

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1 Comment

Thomas Hupkens Over a year ago

Exactly. When in doubt about a PHP error, always look at the last statement before the line at which PHP say there is a problem…

hexacyanide · Accepted Answer · 2013-10-26 17:01:17Z

1

PHP requires that you end lines with a semicolon unless it is the last line. This is because if the semicolon is omitted, it will be evaluated as the same expression as the next line. Change this:

$uname = "Name"

To this:

$uname = "Name";

edited Oct 26, 2013 at 17:01

answered Oct 26, 2013 at 16:38

hexacyanide

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8 Comments

user2867893 Over a year ago

I think it's obvious that I'm new to PHP. Why is this causing a problem with the print function though? I think that's what confused me the most.

hexacyanide Over a year ago

I'm pretty sure it's because the two lines are being evaluated as one.

user2867893 Over a year ago

Oh, that makes sense. I think I'm a bit over my head with PHP. Trying to use it before learning it. Thanks for your help everyone :)

Funk Forty Niner Over a year ago

@user2867893 You're welcome. Added note (from my answer) A semi-colon can be omitted if it's the last line of code with nothing else that will be executed/included thereafter.

user2867893 Over a year ago

I'm not that interested in PHP at the moment but I just need a bit of it to setup a project I'm working on. You both seem really knowledgable about PHP so do you think either of you could just give me a private hand sorting out some issues I have (I imagine all simple like this) so I don't have to ask too many questions on the forum. Thoughts?

|

Matt Clark · Accepted Answer · 2013-10-26 16:39:04Z

0

Missing a semicollen.

$uname = "name";

answered Oct 26, 2013 at 16:39

Matt Clark

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Comments

Funk Forty Niner · Accepted Answer · 2013-10-26 16:41:36Z

0

write it like this just add ; at the end

$uname = "Name";

edited Oct 26, 2013 at 16:41

Funk Forty Niner

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answered Oct 26, 2013 at 16:38

basil.shine

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Comments

user2092317 · Accepted Answer · 2013-10-26 16:43:09Z

0

You need to end each statement with a semicolon ;

This error most often means you forgot one of these

' ; ' " ) ( [ ]

Change

$uname = "Name"

with

$uname = "Name";

answered Oct 26, 2013 at 16:43

user2092317

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