I'm trying to use a PHP variable to add a href value for a link in an echo statement.
Here's a simplified version of the code I want to use. I know that I can't just add the variable into the echo statement, but I can't seem to find an example anywhere that works.
$link_address = '#';
echo '<a href="$link_address">Link</a>';
Try like
HTML in PHP :
echo "<a href='".$link_address."'>Link</a>";
Or even you can try like
echo "<a href='$link_address'>Link</a>";
Or you can use PHP in HTML like
PHP in HTML :
<a href="<?php echo $link_address;?>"> Link </a>
echo "<a href='$link_address'>Link</a>";) which is nice and simple and works perfectly, thanks.you can either use
echo '<a href="'.$link_address.'">Link</a>';
or
echo "<a href=\"$link_address\">Link</a>';
if you use double quotes you can insert the variable into the string and it will be parsed.
Basically like this,
<?php
$link = ""; // Link goes here!
print "<a href="'.$link.'">Link</a>";
?>
as simple as that: echo '<a href="'.$link_address.'">Link</a>';
You can use one and more echo statement inside href
<a href="profile.php?usr=<?php echo $_SESSION['firstname']."&email=". $_SESSION['email']; ?> ">Link</a>
link : "/profile.php?usr=firstname&email=email"
This worked much better in my case.
HTML in PHP: <a href=".$link_address.">Link</a>
The safest way to generate links in PHP is to use the built-in function http_build_query(). This function is very easy to use and takes an array as an argument.
To create a dynamic link simply echo out the result of http_build_query() like so:
$data = [
'id' => $id,
'name' => $name
];
echo '<a href="index.php?'.http_build_query($data).'">Link</a>';
If you want to print in the tabular form with, then you can use this:
echo "<tr> <td><h3> ".$cat['id']."</h3></td><td><h3> ".$cat['title']."<h3></</td><td> <h3>".$cat['desc']."</h3></td><td><h3> ".$cat['process']."%"."<a href='taskUpdate.php' >Update</a>"."</h3></td></tr>" ;
'and double quotes"around this would work.