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PHP

Posted by Milton Centeno
Milton Centeno
Courses Plus Student 7,940 Points

PHP varables quiz not passing

Whats wrong with the code below? Its not passing part 2 of variables quiz.

<?php $name = "Shirts 4 Mike"; echo name; ?>

Thanks.

4 Answers

Giovanni CELESTE
Giovanni CELESTE
20,961 Points
Giovanni CELESTE
Giovanni CELESTE
20,961 Points

The answer is in the notice of the challenge :). Don't forget...

Milton Centeno
PLUS
Milton Centeno
Courses Plus Student 7,940 Points
Milton Centeno
Milton Centeno
Courses Plus Student 7,940 Points

The notice says "Bummer! Make sure you're echoing the variable you just set in the last step." Here is my code: <?php $name = "Shirts 4 Mike"; echo name; ?> What am I missing? Shouldn't "echo name;" work?

Giovanni CELESTE
Giovanni CELESTE
20,961 Points

Did you try something else? I quite understand what you were trying to do, but do you undertand every word of code you are writing?

<?php  // here you have your opening tag for php, who says to the server "hey, I am a php request!"
   $name = "Shirt 4 Mike"; // you declare your variable with a $, saying to the server "this is a variable", and you give a value with " ", saying to the server "this is a string chain". ";" is saying the server "end of the action.
echo // you say the server "display what follow me on the screen"
name; // What do you say to the server here? What does it understand? What is name?
?> Here you close your php tag, saying to your server "ok now, bye"
Milton Centeno
PLUS
Milton Centeno
Courses Plus Student 7,940 Points
Milton Centeno
Milton Centeno
Courses Plus Student 7,940 Points

Nevermind, I got it. It worked by echoing the "Shirts 4 Mike" instead of echoing the name variable.

Joseph DeFazio
Joseph DeFazio
10,599 Points

You should be echoing the variable though. Defining a variable and then echoing the value rather than the variable itself defeats the purpose of creating the variable.

Joseph DeFazio
Joseph DeFazio
10,599 Points
Joseph DeFazio
Joseph DeFazio
10,599 Points 
<?php
    $name  = "Shirt 4 Mike";
    echo $name;
?>