Conditional display

Resolved

(@fabriciomanc)

How can I display a field conditionally ? (I mean, when It’s not blank, or null)

I tried using an if statement, but no luck. The FontAwesome icon appears anyway and the information didn’t because it doesn’t exist.

<?php
	if($user_info->adi_phone != '' || !empty($user_info->adi_phone) || $user_info->adi_phone != ' ') {
		echo "<span><i class='fa fa-phone-square'></i>&nbsp;&nbsp;".$user_info->adi_phone."</span>";
	}
	if($user_info->adi_skype != '' || !empty($user_info->adi_skype) || $user_info->adi_skype != ' ') {
		echo "<p><i class='fa fa-skype'></i>&nbsp;&nbsp;".$user_info->adi_skype."</p>";
	}
	?>

https://ww.wp.xz.cn/plugins/simple-user-listing/

Viewing 3 replies - 1 through 3 (of 3 total)

Plugin Author HelgaTheViking
(@helgatheviking)

10 years, 9 months ago

try var_dump($user_info->adi_phone); to see what the value is.

Plugin Author HelgaTheViking
(@helgatheviking)

10 years, 9 months ago

Both

if($user_info->description)

and

if(!empty($user_info->description))

seem to work for me.

Thread Starter fabriciomanc
(@fabriciomanc)

10 years, 9 months ago

It worked now.. Some fields where filled with ” ” (two blank spaces)
Just removed it 🙂

Viewing 3 replies - 1 through 3 (of 3 total)

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