Variable class names

Question

I want my link class to be a variable (I am using the filtering system Isotope).

    $categories = get_categories($args);

    foreach($categories as $category) {
        $name = '.'.$category->name;
        echo '<a href="#" data-filter=$name>'; //$name does not work here 
        echo $category->name; 
        echo '</a>';
        echo ' / ';
}

My $name is only displayed as $name in the browser and not the category name I want. Outside the data-filter echoing $name gives all the categories as expected. Is there a way to solve this problem with putting the $name in the data-filter? If the answer is too difficult, please point me to the right direction of what I should do to fix this problem myself please.

Thanks!

tlenss: I didnt know the quotes were the problem so I did not search for that problem name. — Folthan, May 26 '14 at 08:42

score 1 · Accepted Answer · answered May 26 '14 at 08:19

1

Strings in single quotes do not do variable expansion. You have to change it to something like this:

echo '<a href="#" data-filter='.$name.'>';

An alternative without the need for concatenation would be:

echo "<a href=\"#\" data-filter=$name>";

Or, more elegant in my eyes:

echo sprintf('<a href="#" data-filter=%s>', $name);

Also I guess you have to add double quotes around that data-filter attributes content...

answered May 26 '14 at 08:19

arkascha

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Thanks for the clarification about the quotes, I had no idea but I shall remember this now. also the \ before and after the # was necessary like you said. Used your 2nd example. – Folthan May 26 '14 at 08:41

Variable class names

1 Answers1