回声一个img并在src中放入一个php语句？

Question

How would I format this correctly? I know I can't open a php statement inside an existing one but I'm not sure how to correct it.

echo '<img src="<?php echo dirname( get_bloginfo('stylesheet_url') ); ?>/images/isa.jpg" alt="ISA">';

Answer 1

像这样：

echo '<img src="'.dirname( get_bloginfo('stylesheet_url') ).'/images/isa.jpg" alt="ISA">';

Answer 2

像这样做

echo '<img src="'.dirname( get_bloginfo('stylesheet_url') ).'"/images/isa.jpg" alt="ISA">';

Answer 3

I prefer to use PHP's printf() and sprintf() for that. It's easier to visualize the HTML and we just need to fill it with %s and their corresponding build up.

printf(
    '<img src="%s/images/isa.jpg" alt="ISA">', // base HTML
    dirname( get_bloginfo('stylesheet_url') )  // first %s converted element
);

Answer 4

IT is better to use get_template_directory_uri() rather than get_bloginfo(). You can use following code:

echo '<img src="'.get_template_directory_uri().'/images/isa.jpg" alt="ISA">';