Adding a link into PHP printf

Question

I have a script on a buddypress set up which shows a user on the 'members area' if they have a notification, it just prints "You have new notifications (22)" or "You do not have any notifications".

Typical set up, I am a little lost on how to correctly print a link that is not going to change..

Currently the code for when they have a notification is;

if ( $instance['show_count_in_title'] ) {
    printf( "<span class='notification-count-in-title'>(%d)</span>", $count );
}

However when I add in the link to the text, it breaks, I have tried within the span and outside, but always in the (".. I am not totally clued up on PHP so maybe I am wrong to use links in this way? But I guess more so I am confused to why

When I do this;

if ( $instance['show_count_in_title'] ) {
    printf( "<a href="me/notifications"><span class='notification-count-in-title'>(%d)</span></a>", $count );
}

I get the following error:

Parse error: syntax error, unexpected 'http' (T_STRING) in /homepages/2/d676463482/htdocs/dc/wp-content/plugins/buddypress-notifications-widget/core/class-bp-notification-widget.php on line 59

I presume I am entering the a href incorrectly for PHP but I have struggle to figure out what it should be.

You can safely just switch the href to use single quotes like `'notification-count-in-title'` use. — Michael Berkowski, Aug 01 '18 at 18:10

SpotDetector · Accepted Answer · 2018-08-01T18:17:22.497

You need to escape your backslashes:

printf( "<a href=\"me/notifications\"><span class='notification-count-in-title'>(%d)</span></a>", $count );

Or continue using single slashes:

printf( "<a href='me/notifications'><span class='notification-count-in-title'>(%d)</span></a>", $count );

Furthermore, I would strongly suggest using single quotes, and using double quotes inside. (More info on this here):

printf( '<a href="me/notifications"><span class="notification-count-in-title">(%d)</span></a>', $count );

Because you are not including variables within the string (You're passing them in via printf), there's no reason to have PHP try to parse the string to look for variables.

Thanks this worked perfectly. Also super helpful to know the different ways and the preferred way to do. Thanks loads, really has helped clear this up for me. — Daniel Hobbs, Aug 01 '18 at 18:19

score 1 · Answer 2 · answered Aug 01 '18 at 18:10

Use single quotes or escape the inner double-quotes:

printf( '<a href="me/notifications"><span class="notification-count-in-title">(%d)</span></a>', $count );

... or:

printf( "<a href=\"me/notifications\"><span class='notification-count-in-title'>(%d)</span></a>", $count );

score 1 · Answer 3 · answered Aug 01 '18 at 18:12

You need to either escape

  printf( "<a href=\"me/notifications\"><span class=\"notification-count-in-title\">(%d)</span></a>", $count );

or use ''

  printf( '<a href="me/notifications"><span class="notification-count-in-title">(%d)</span></a>', $count );

Adding a link into PHP printf

3 Answers3