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0 points
Submitted by Justina
over 7 years

its not accepting my answer!

Answer 528eb581548c35d675000339

1 vote

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252 points
Submitted by 2JZGTE
over 7 years

Answer 52b5d118282ae303e4002006

0 votes

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$myAge not $myage

347 points
Submitted by Joakim Kjellander
over 7 years

Answer 52b6070a8c1cccffc3002ad2

0 votes

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Same for $myName

…and take away the first ; in your code to see if you can proceed on to the next lesson.

:)

415 points
Submitted by hp6721
over 7 years

Answer 53f21228282ae3360b00014f

0 votes

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I had the same problem and couldn’t get through it until I typed in the answer exactly as it was in the hint. ?> had to be one the same line and everything.

105 points
Submitted by Adam Evertsson
over 6 years

Answer 54380ee97c82ca1d420006cd

0 votes

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you have to use echo like this: echo $myName = “your name”; echo $myAge = “your age”;

1107 points
Submitted by rocket quiah
over 6 years

1 comments

Christopher mlalazi almost 6 years

delete echo its not supposed to print at this stage

Answer 5478b2a89c4e9dc553007b3b

0 votes

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hi i am getting the same problem this is my code

157 points
Submitted by gingerkid3102000
over 6 years

1 comments

Jordan M Tschetter about 6 years

don’t wrap numbers in quotes

Answer 554cda7d51b8874df90000bb

0 votes

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it took a while for me to figure this out but i did it like this

119 points
Submitted by BradfordPA
about 6 years

1 comments

Venkat K K over 5 years

I think you missed out the ‘

Answer 560a6e3c3e0ec8b6ac000045

0 votes

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This works for me

<!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>
            echo $myName = "Djamal";
            echo $myAge = 54;
        </p>   
    </body>
</html>
1234 points
Submitted by DAGIROV Djamal
over 5 years