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Andrew O'Neill
2,984 PointsIssues with my php code
<?php
$name = "Mike";
?>
<?php echo $name ?>
This is my code for a PHP challenge asking me to echo the name. Unless I am mistaken this should be right as I even went back just to watch the video and copy what the instructor did. Can anyone please let me know if I have made some error and I just can't see it?
EDIT: I did put the echo name into the question just for some reason it's not being displayed on this page
Andrew O'Neill
2,984 PointsSeems like I forgot to copy the entire thing:
<?php
$name = "Mike";
?>
<?php echo $name ?>
Robert Richey
Courses Plus Student 16,352 PointsFixed markdown of php code
3 Answers
Robert Richey
Courses Plus Student 16,352 PointsAndrew,
The issue is that the challenge is expecting only 1 php code block. With 1 block, and your code, it will pass - but fails when using 2 code blocks. Just echo the name right below the variable assignment.
Cheers!
Andrew O'Neill
2,984 Pointsoh right. Thanks for that, sorry for all the trouble there
Robert Richey
Courses Plus Student 16,352 PointsNo trouble at all. Thanks for your patience!
Robert Richey
Courses Plus Student 16,352 PointsHi Andrew,
The only thing I see right now that needs fixing is that each statement needs to end with a semi-colon.
<?php echo name ?>
// needs to be
<?php echo name; ?>
If that doesn't solve the issue, can you please paste a link to the challenge you're working on?
Kind Regards
Andrew O'Neill
2,984 PointsSee in the video the instructor does say and as far as I am aware of PHP single line statements don't normally require the semi-colon but I have tried adding it just to be sure it wasn't just me being incorrect and I'm still getting nothing. I tried using the preview option with a number of outputs and it's echoing nothing.
https://teamtreehouse.com/library/php-variables here is the link
Corey Cramer
9,453 PointsOne thing to keep in mind is the PHP does not appear to output to your browser in the code challenges. It may be useful to create a workspace and test your code there to see how it acts.
Corey Cramer
9,453 PointsWhat you have done so far is assign the value of "Mike" to the variable $name. That won't immediately return any information to the browser; for that you need to use the PHP command echo followed by your variable name.
Edit: After your echo statement you need to include a semicolon to signal to the server that your line ends.
miguelcastro2
Courses Plus Student 6,573 Pointsmiguelcastro2
Courses Plus Student 6,573 PointsYour echo command is missing. You are just creating a variable $name with the value of "Mike", but you are not echoing the variable.