$variable in style=“background-image: url:( );”

Question

I have a question for using a variable as IconPath for setting an background-image: url()

<div style="background-image: url($iconPath)"></div>

I have no idea how I can set the url for the background-image with a variable?

I tried with "" , '' , \\"

Has anyone an idea?

Answer 1

PHP is a server side language, so if you want anything to be visible to DOM (ie to your HTML), you will have to print it to DOM.This is achieved through echo

<div style="background-image: url('<?php echo $iconPath ?>')"></div>

Any variable value if needed to be printed, just echo it and it'll be visible to DOM.

Answer 2

You can echo the PHP variables to get its value in HTML. So try like this

<div style="background-image: url('<?php echo $iconPath;?>')"></div>

Answer 3

You could use

<div style="background-image: url(<?php echo $iconPath; ?>)"></div>

Or if it is supported (short tags)

<div style="background-image: url(<?= $iconPath; ?>)"></div>

Answer 4

You must echo the variable with PHP. Try like this:

<?php
 $iconPath = "/yourPath/image.jpg";
?>

<div style="background-image: url(<?php echo $iconPath; ?> )"></div>

Answer 5

如果您的服务器端语言是PHP，则可以使用以下代码：

<div style="background-image: url('<?php echo $iconPath?>')"></div>

Answer 6

假设背景图片是“ picture.jpg”，则需要输入以下php代码：

echo "<div style='background-image:url(&#039;picture.jpg&#039;)'">;