Outputting stylesheet uri in a function

Resolved iand
(@iand)

18 years, 4 months ago

For a theme specific function I need to create a path to the stylesheet directory.
The Codex Page suggests that echo '<img src="' . get_settings('stylesheet_directory') . ''; will work but as far as I can tell, of the ‘will work with 1.5+’ options, only ‘html_type’ generates anything.

At the moment I am using this within my function:
echo '<img src="' . get_settings('siteurl') . '/pathto/stylesheet/directory>'; which is not going to be good if the theme folder is renamed.

Any help/ suggestions would be appreciated, this is my first foray into coding and I am out of ideas.

Viewing 7 replies - 1 through 7 (of 7 total)

Moderator James Huff
(@macmanx)

Volunteer Moderator

18 years, 4 months ago

Try:

<?php bloginfo('stylesheet_directory'); ?>

http://codex.wordpress.org/Template_Tags/bloginfo

Les Bessant
(@lesbessant)

18 years, 4 months ago

If you want output on screen, you need bloginfo(), not get_bloginfo()

http://codex.wordpress.org/Template_Tags/bloginfo

Example:
<?php bloginfo('stylesheet_url')?> should be what you need.

EDIT: must work quicker 🙂

Thread Starter iand
(@iand)

18 years, 4 months ago

Thanks both. However, I am not after output on screen though, I am after output into a function and that will output on screen:
value of stylesheet uri -> myfunction -> screen output

Les Bessant
(@lesbessant)

18 years, 4 months ago

In which case, get_bloginfo should work nicely:

<?php $ss_directory = get_bloginfo('stylesheet_directory'); echo $ss_directory; ?> <img src="<?php echo $ss_directory?>/validcss.gif" /> <img src="<?php echo bloginfo('stylesheet_directory');?>/validcss.gif" />

All work….

Works for me.

Thread Starter iand
(@iand)

18 years, 4 months ago

The function is question is here. At the moment I am defining $imagepath by hand and using get_settings('siteurl') . ''.$imagepath.'/ as an alternative.

Les Bessant
(@lesbessant)

18 years, 4 months ago

And what happens if you use get_bloginfo(‘stylesheet_directory’) ?

Thread Starter iand
(@iand)

18 years, 4 months ago

I was just doing that – using $ss_directory = get_bloginfo('stylesheet_directory'); worked perfectly. Thanks a lot.

Viewing 7 replies - 1 through 7 (of 7 total)

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Outputting stylesheet uri in a function

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