Add PHP variable inside echo statement as href link address?

Question

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>';

If you switched your single `'` and double quotes `"` around this would work. — Novocaine, Sep 20 '13 at 12:38

GautamD31 · Accepted Answer · 2014-10-27T10:05:52.727

73

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>

edited Oct 27 '14 at 10:05

answered Sep 20 '13 at 12:36

GautamD31

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if you use double quotes, why use concatenation for a variable? – eis Sep 20 '13 at 12:41
1

Yup.Now I have added that too...We can use either.Thanks @eis – GautamD31 Sep 20 '13 at 12:43
1

The second suggestion here (`echo "Link";`) which is nice and simple and works perfectly, thanks. – jasonbradberry Sep 20 '13 at 12:52

score 11 · Answer 2 · answered Sep 20 '13 at 12:38

11

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.

answered Sep 20 '13 at 12:38

Davide De Santis

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score 5 · Answer 3 · edited Apr 30 '20 at 07:54

5

Basically like this,

<?php
$link = ""; // Link goes here!
print "<a href="'.$link.'">Link</a>";
?>

edited Apr 30 '20 at 07:54

BSMP

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answered Sep 20 '13 at 12:49

Ryan Paluch

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1

score 4 · Answer 4 · answered Sep 20 '13 at 12:51

4

as simple as that: echo '<a href="'.$link_address.'">Link</a>';

answered Sep 20 '13 at 12:51

Jay

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imsing · Answer 5 · 2020-12-30T11:20:23.093

0

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"

edited Dec 30 '20 at 11:20

answered Dec 30 '20 at 11:13

imsing

7
2

score 0 · Answer 6 · answered Jan 07 '21 at 01:30

0

This worked much better in my case.

HTML in PHP: <a href=".$link_address.">Link</a>

answered Jan 07 '21 at 01:30

NDi

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score 0 · Answer 7 · answered Apr 20 '21 at 18:16

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>';

score -2 · Answer 8 · edited Sep 10 '18 at 12:30

-2

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>" ;

edited Sep 10 '18 at 12:30

Hashmat

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answered Sep 14 '14 at 11:31

Alaa Moneam

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Add PHP variable inside echo statement as href link address?

8 Answers8

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