Error outputting img html tag to php echo source

Question

I have searched the site for an answer to my question, however the solutions have been unable to solve my problem.

My Problem is: I have tried to add a php echo element inside of an html tag. However it is throwing off an error as seen in the screenshot. I have tried various ways to adjust this code unsuccessfully.

I should mention that it is a simple MVC application.

screenshot

Thanks in advance for any advice.

You are printing a string not an html tag ! where is the '<>' ? — Omis Brown, May 26 '17 at 12:07
Stack allows you to post code. perhaps you did not get the memo — Rotimi, May 26 '17 at 12:08

The Truth · Answer 1 · 2017-05-26T12:20:09.447

1

You were concating wrong. Be careful with ' and ".

echo '<img src= " '. $page->image . ' " <br>';

The spaces between " are here just for readability.

And btw you can't execute PHP code inside html.

edited May 26 '17 at 12:20

answered May 26 '17 at 12:07

The Truth

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You should do the tag like this – Omis Brown May 26 '17 at 12:08
Ok, thank you for the explanation, I will research further on Concating. Thanks for solution as well. Appreciate it. – JediMaster Orion May 26 '17 at 12:10

score 1 · Answer 2 · answered May 26 '17 at 12:07

1

You can't separate your string like that:

echo 'Image' 'img src='...;
            ^

If you want to concatenate everything, use . or just put it all on the same string:

echo 'Image <img src="'. $yourVariable.'"/>'; ...

answered May 26 '17 at 12:07

rbock

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Thank you I was thinking about the concatenating using the ., appreciate the advice. – JediMaster Orion May 26 '17 at 12:10
@JediMasterOrion glad I could help :) – rbock May 26 '17 at 12:11

score 0 · Answer 3 · answered May 26 '17 at 12:50

0

Your ' and " were in the wrong spots. Also, not that it's related to your PHP erorr but you were also missing the < /> around your image.

echo 'Image: <img src="' . $page->image . '" alt=""/></br>';

answered May 26 '17 at 12:50

Blueline

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Error outputting img html tag to php echo source

3 Answers3