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Submitted by Tannim S (Umbreo)
over 9 years

Oops, try again. Make sure to use echo to print out your name and age! (Solved!)

My code keeps passing the console just fine, but CodeCademy throws the aforementioned error. Here’s my code. <!DOCTYPE html> <html> <head> <link type='text/css' rel='stylesheet' href='style.css'/> <title>PHP FTW!</title> </head> <body> <!-- Write your PHP code below!--> <p><?php $myName = "Umbreo"; $myAge = 16; echo $myName; echo $myAge; ?></p> </body> </html>

Answer 5421d0dd7c82ca5561000294

0 votes

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Solved! I just had to go to a different computer.

points
Submitted by Tannim S (Umbreo)
over 9 years

1 comments

Yash Naresh Sharma over 9 years

so one needs two computers to go through this course? nice!

Answer 54d1bc3386f5527698001556

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ofcourse echo on console is displaying, but its giving error in codecademy. here go my code is //php code for myname and myage.

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Submitted by Syed Rizwan
about 9 years

Answer 5589566d93767696ac000300

0 votes

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I’ve solved it by typing “echo $myName, $myAge;” instead of two separate lines.

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Submitted by Mikołaj Junik
almost 9 years