Question

I have a custom field in WordPress called "thumb-url" which contains the exact location of an image. I want to only display the image if "thumb-url" contains the location of the image.

I m start with an if statement that echoes photo exists if there s a value in the "thumb-url" custom field, otherwise it does nothing.

<div class="excerpt">
<?php
$key =  thumb-url ;
$themeta = get_post_meta($post->ID, $key, TRUE);
if($themeta !=   ) {
echo  photo exists ;
}
?>

Now, here s the code that I really want the above if statement to echo if there s a value in "thumb-url":

<img alt="<?php the_title() ?>" src="<?php if ( function_exists( get_custom_field_value ) ){ get_custom_field_value( thumb-url , true); } ?>" align="absmiddle" height="62" width="62" class="writtenpostimg" />

How do I get that ↑ inside the echo part of the if statement?

Much appreciated...

Answer 1

Assuming you re echoing it to the page for some sort of copy/paste instructions:

<div class="excerpt">
<?php
$key =  thumb-url ;
$themeta = get_post_meta($post->ID, $key, TRUE);
if($themeta !=   ) {
    echo htmlspecialchars( <img alt="<?php the_title() ?>" src="<?php if ( function_exists( get_custom_field_value ) ){ get_custom_field_value( thumb-url , true); } ?>" align="absmiddle" height="62" width="62" class="writtenpostimg" /> );
}

?>

Answer 2

This escapes the code... The other comment actually prints out "<img...>", so it depends on what you re trying to do.

You can remove the PHP tags and use conditional expressions:

<div class="excerpt">
<?php
$key =  thumb-url ;
$themeta = get_post_meta($post->ID, $key, TRUE);
if($themeta !=   ) {
echo  <img alt=" .the_title(). " src=" .(function_exists( get_custom_field_value )?get_custom_field_value( thumb-url , true):  ). " align="absmiddle" height="62" width="62" class="writtenpostimg" /> ;

It might be easier to understand if you actually use a variable for it:

$url = ""
if(function_exists( get_custom_field_value ))
  $url = get_custom_field_value( thumb-url , true);
echo  <img alt=" .the_title(). " src=" .$url. " align="absmiddle" height="62" width="62" class="writtenpostimg" /> ;

友情链接