$ output里面的php echo语句

Question

I have the following php code:

$skizzar_masonry_item_width = $masonry_item_width;
$skizzar_masonry_item_padding = $masonry_item_padding;
$skizzar_double_width_size = $masonry_item_width*2 +$masonry_item_padding;

$output .= '<style>.skizzar_masonry_entry.skizzar_ma_double, .skizzar_masonry_entry.skizzar_ma_double img {width:'.$skizzar_double_width_size.'}</style>';

return $output;

For some reason though, the value of $skizzar_double_width_size is not being added into the $output - is there a way to echo a value in an output variable?

Answer 1

As @Rizier123 mentioned, ensure you initialise any string variables before trying to append to them.

$var = ''; $var .= 'I appended'; $var .= ' a string!';

I would also like to strongly discourage you from using inline styles as well as generating them with inline PHP. Things get very messy very quickly.

Answer 2

In a situation like this you need to check that all the variables you are using in the calculation are valid before you panic.

So try

echo 'before I use these values they contain<br>';
echo '$masonry_item_width = ' . $masonry_item_width . '<br>';
echo '$masonry_item_padding = ' . $masonry_item_padding . '<br>';


$skizzar_masonry_item_width = $masonry_item_width;
$skizzar_masonry_item_padding = $masonry_item_padding;
$skizzar_double_width_size = $masonry_item_width*2 +$masonry_item_padding;

echo 'after moving the fields to an unnecessary intemediary field<br>';
echo '$skizzar_masonry_item_width = ' . $skizzar_masonry_item_width . '<br>';
echo '$skizzar_masonry_item_padding = ' . $skizzar_masonry_item_padding . '<br>';
echo '$skizzar_double_width_size = ' . $skizzar_double_width_size . '<br>';


$output .= '<style>.skizzar_masonry_entry.skizzar_ma_double, .skizzar_masonry_entry.skizzar_ma_double img {width:'.$skizzar_double_width_size.'}</style>';

echo $output;

This should identify which fields are causing you problems.

Also while testing always run with display_errors = On It saves so much time in the long run.