Mixing PHP and html basic

Question

In a var i want to mix php and html/ inline css, I think i'm almost there with the code below but it keeps failing.

$bgThumb = 'style="background-image:url(' . echo $url . ');"';

echo performs output immediately. it is NOT a function and has no return value, so you cannot assign echoed data to a variable. — Marc B
– Marc B, Commented Jan 29, 2014 at 17:11

Stéphane Bruckert · Accepted Answer · 2014-01-29 17:10:02Z

2

$bgThumb = 'style="background-image:url(' . $url . ');"';

answered Jan 29, 2014 at 17:10

Stéphane Bruckert

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Comments

aksu · Accepted Answer · 2014-01-29 17:10:28Z

1

No need for echo to concatenate variables:

$bgThumb = 'style="background-image:url(' . $url . ');"';

echo just displays content in the webpage, it can't be used in the variable.

answered Jan 29, 2014 at 17:10

aksu

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1 Comment

Patrick Moore Over a year ago

connect = 'concatenate'

Jérôme · Accepted Answer · 2014-01-29 17:11:26Z

0

You can't have an echo in a variable!!!

$bgThumb = 'style="background-image:url(' . $url . ');"';

then

echo $bgThumb;

answered Jan 29, 2014 at 17:11

Jérôme

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Comments

Daniel F · Accepted Answer · 2017-07-03 10:03:20Z

0

no need to echo

$bgThumb = 'style="background-image:url(' . $url . ');"';

edited Jul 3, 2017 at 10:03

Daniel F

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answered Jul 3, 2017 at 9:39

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