If statement set variable but syntax error

Question

I'm getting a syntax error with this statement

$catlist = if (function_exists('put_cat_icons')) {
put_cat_icons( wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id.'&echo=0'));
} else {
wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id);
}

If I remove $catlist = I get no errors. What's the problem in the code? Can I set a variable like this?

KingCrunch · Accepted Answer · 2012-08-01 09:51:30Z

1

if as value of an assigment is invalid. Don#t know, where you get this from. However, you can use the ternary operator here

$catlist = function_exists('put_cat_icons')
    ? put_cat_icons( wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id.'&echo=0'))
    : wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id);

Also, when you look at your code, you may realized, that the wp_list_categories()-call is the same in both cases

$catlist = wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id);
if (function_exists('put_cat_icons')) $catlist = put_cat_icons($catlist);

answered Aug 1, 2012 at 9:51

KingCrunch

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Comments

Dale · Accepted Answer · 2012-08-01 09:51:12Z

0

I suggest rearranging your code a little

if (function_exists('put_cat_icons')) {
    $catlist = put_cat_icons( wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id.'&echo=0'));
} else {
    $catlist = wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id);
}

answered Aug 1, 2012 at 9:51

Dale

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Comments

Mitya · Accepted Answer · 2012-08-01 09:51:23Z

0

You cannot assign to the result of a condition - they must be freestanding. Possibly you meant:

if (function_exists('put_cat_icons')) {
    $catlist = put_cat_icons( wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id.'&echo=0'));
} else {
    $catlist = wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id);
}

You may have got confused between ordinary and ternary conditions, the latter of which do allow assignment.

$something = 1 < 2 ? 'foo' : 'bar';

answered Aug 1, 2012 at 9:51

Mitya

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Comments

Christofer Eliasson · Accepted Answer · 2012-08-01 09:51:51Z

0

From the looks of it, you are mixing an ordinary if statement with the ternary operator which should look something like this $someVar = (condition) ? "TRUE VALUE" : "FALSE VALUE";

Using the ternary operator in your case would look something like this:

$catlist = (function_exists('put_cat_icons')) ? 
    put_cat_icons( wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id.'&echo=0')) : 
    wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id);

answered Aug 1, 2012 at 9:51

Christofer Eliasson

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Comments

Dru · Accepted Answer · 2012-08-01 09:52:15Z

0

I dont this that is the correct way to set a variable. Are your trying to do this

$catlist = (function_exists('put_cat_icons')) ? put_cat_icons( wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id.'&echo=0')):
wp_list_categories('echo=0&orderby=id&show_count=1&title_li=&child_of=' . $cat_id);

answered Aug 1, 2012 at 9:52

Dru

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