Echo an img and put in a php statement inside the src?

March 28, 20232 Views

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">';

Post Views: 2

4 comments

Like this:

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

do it like this

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

Anonymous says:

March 28, 2023 at 11:31 am
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
);
```
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Anonymous says:

March 28, 2023 at 11:31 am
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">';
```
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