PHP variable in href in PHP

Question

I'm struggling with the syntax of this:

echo '<strong><a href="<?php echo get_post_meta( get_the_ID(), 'data_sheet', true );?>" target="_blank">Datasheet</a></strong>';

The central PHP element gets a URL from a WooCommerce custom field (I know this works ok on its own). That needs to be turned into a viable link opening a new page, and the whole sits in a PHP wrapper on the page.

Where am I going wrong?

What do you get at the moment when that code runs?

sjdaws
– sjdaws

2017-06-08 00:49:23 +00:00
Commented Jun 8, 2017 at 0:49 — sjdaws
– sjdaws, Commented Jun 8, 2017 at 0:49

Muhammad Usman · Accepted Answer · 2017-06-08 00:56:11Z

2

Try this hopefully, it will work.

echo '<strong><a href="' . get_post_meta( get_the_ID(), 'data_sheet', true ) . '" target="_blank">Datasheet</a></strong>';

answered Jun 8, 2017 at 0:56

Muhammad Usman

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1 Comment

Muhammad Usman Over a year ago

@SteveBower Glad to help ;) please marked tick the answer

Qirel · Accepted Answer · 2017-06-08 01:31:45Z

1

Try adding more to / changing the href

something like this might work

echo '<strong><a href="data_sheet.php?id='.$_GET['id'].'" target="_blank">Datasheet</a></strong>';

You'll have to adjust a little bit, but should do the trick

edited Jun 8, 2017 at 1:31

Qirel

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answered Jun 8, 2017 at 0:54

sicksixstrings

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3 Comments

Qirel Over a year ago

This isn't going to work. You're already in PHP mode, so the string should be concated. You can't echo inside echo

sicksixstrings Over a year ago

You are right, I just edited the answer to remove the PHP tags

Qirel Over a year ago

Slightly incorrect still, I corrected it for you. This will print correctly and shouldn't throw syntax errors

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