Basic picture insert

[anarchy3200]

Hi

I have been searching everywhere but have not been able to find a way to just show a single picture with php code.

eg. in HTML it is: <img src="picture.gif">

what is it with php

thanks for any help

[Nightfire]

It's no different, just put it in print or echo eg

<?php
echo '<img src="picture.gif">';
?>

But it is really best to escape from php when using html, makes things much easier for you

<?php
//php code here
?>
<img src="picture.gif">
<?php
// more php code
?>

[anarchy3200]

What i am trying to do is
script:

<?php

if ($_GET["name"] == comp)
   echo "TEXT"; \\Make the picture appear with this text


   elseif ($_GET["name"] == lap)
   echo "TEXT";
   else 
   echo "wrong";
?>

but i get an error when i try to

[Nightfire]

<?php

if ($_GET["name"] == comp)
echo "TEXT";

//Put the picture here
echo '<img src="picture.gif">';

elseif ($_GET["name"] == lap)
echo "TEXT";
else
echo "wrong";
?>

[anarchy3200]

When i do that though i get an error:

Parse error: parse error, unexpected T_ELSEIF in C:\apache\minixampp\htdocs\xampp\form\action.php on line 21

which is the line with the 'elseif' statement in this.

[Nightfire]

It would be best to use curly braces in the code, like this. I changed the code to how you wanted it too.

<?php

if ($_GET["name"] == comp){
    echo "TEXT <img src=\"picture.gif\">";
}
elseif ($_GET["name"] == lap){
    echo "TEXT";
}
else{
    echo "wrong";
}
?>

As the html for the image contains double quotes and the echo is also using dowuble quotes, you need to use backslashes in the html like I did above, or else you'll get a parse error

[anarchy3200]

Thanks for all your help - it is working now.