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PHP PHP Basics (Retired) PHP Conditionals & Loops Conditionals Challenge

Posted by Henry Nduka
Henry Nduka
1,500 Points

Please help <?php $name = "Mike"; if($name == "Mike") $info = "Hi,I am Mike!"; ?> where have I gone wrong

<?php $name = "Mike"; if($name == "Mike") $info = "Hi,I am Mike!"; ?>

Please where have I gone wrong

index.php 
<?php
$name = "Mike";
  if($name == "Mike")
  $info = "Hi,I am Mike!";
?>

3 Answers

Nejc Vukovic
.a{fill-rule:evenodd;}techdegree seal-36
Nejc Vukovic
Full Stack JavaScript Techdegree Graduate 51,574 Points

interesting. use ' instead of " .

<?php
  $name = 'Mike';
if ($name == 'Mike')
  echo 'Hi, I am Mike!';
?>
Henry Nduka
Henry Nduka
1,500 Points

Thank you it works

Nejc Vukovic
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Nejc Vukovic
Full Stack JavaScript Techdegree Graduate 51,574 Points
Nejc Vukovic
.a{fill-rule:evenodd;}techdegree seal-36
Nejc Vukovic
Full Stack JavaScript Techdegree Graduate 51,574 Points

Hi Henry.

The challange states that you have to "echo" the string 'Hello I'm Mike!' - > so you have to echo it :)

<?php
$name = "Mike";
  if($name == "Mike")
    echo  "Hi,I am Mike!";
?>

You wrote the string to another variable, but if you want to give an output the you have to use echo.

Happy Coding!!!

nWEBd

Henry Nduka
Henry Nduka
1,500 Points

Still doesn't work that way

Nejc Vukovic
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Nejc Vukovic
Full Stack JavaScript Techdegree Graduate 51,574 Points
Nejc Vukovic
.a{fill-rule:evenodd;}techdegree seal-36
Nejc Vukovic
Full Stack JavaScript Techdegree Graduate 51,574 Points

You are welcome.

I only find it strange that the challange so strict is that a single quote must be used.

Because single or double quotes represent a string literal. ( You only have to be careful when you combine them together in a string, but there you just need a char break so that the string is correctly taken).

Regards.