Insert function name with php

Question

I'm trying to do a simple task, insert a second function name under "onmouseover" but probably something is wrong with the syntax.

echo '<div onmouseover="changeText(); <?php if($title==""){echo changeViz();}; ?> ">';

I probably need to escape some quotes and add some, but i can't figure out the correct solution. Thanks

Nothing it's working. Let me give you the complete code...:

echo '<div class="frame" onmouseover="changeText(\''.$text[$i].'\'); <?php if($title[$i]==""){echo changeViz();}; ?>">';

Community · Accepted Answer · 2023-11-17 20:42:18Z

4

You are nesting <?php ?> inside existing php code, which is a syntax error. Instead, concatenate in the javascript function changeViz() as a quoted string.

This version uses a ternary operator to duplicate the if() statement you had originally.

echo '<div onmouseover="changeText(); ' . ($title == '' ? 'changeViz();' : '') .  '">';

The ternary operation here will concatenate changeViz(); onto the echo string if $title == "", or otherwise just concatenate on an empty string.

Update after seeing full code:

You have the quote escaping correct in the first part.

echo '<div class="frame" onmouseover="changeText(\'' . $text[$i] . '\'); ' . ($title == '' ? 'changeViz();' : '') .  '">';

edited Nov 17, 2023 at 20:42

CommunityBot

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answered Jun 21, 2012 at 13:24

Michael Berkowski

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5 Comments

Michael Berkowski Over a year ago

@RyanB I think the quotes are all correct, since the final " closes the onmouseover. In any case, I changed them to single quotes on the PHP side to be clear.

Ryan B Over a year ago

wouldn't the first " after title == end the pair for the mouseover?

Michael Berkowski Over a year ago

@RyanB It's a different string, in a different context. Inside the single quotes, the opening " doesnt need escaping and isn't parsed as an open string as far as PHP is concerned. The pair of double quotes after $title (which I thought I changed to single) are a separate string as far as PHP is concerned as well, and won't close any opened in the previous string.

Ryan B Over a year ago

It might have been while you were updating

Michael Berkowski Over a year ago

@chitoiudaniel Sorry, looks like the ternary statement needs to be enclosed in (). Edited.

devsnd · Accepted Answer · 2012-06-21 13:29:20Z

1

You can make your code much more readable if you do not try to do everything in one line:

$onmouseover_action = "changeText(); ";
if($title==""){
    $onmouseover_action .= "changeViz(); ";
}
echo '<div onmouseover="'.$onmouseover_action.'">';

It makes your code easier to maintain, and you have less need to comment it, because it describes itself quite much better.

answered Jun 21, 2012 at 13:29

devsnd

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Comments

Bogdan Burym · Accepted Answer · 2012-08-16 10:08:08Z

0

Try this:

echo '<div class="frame" onmouseover="changeText(\''.$text[$i].'\'); '. ($title[$i]=="")?'changeViz();':'').'">';

answered Aug 16, 2012 at 10:08

Bogdan Burym

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