Drupal 7 Get image field path

I would like to get the image path of a field. I'm in a node and I need the Url of image in order to put it as a background-image in inline css.
I can't find the solution. Can you help me?

I sometime use this:

$node = node_load($nid);
$img_url = $node->field_userimage['und'][0]['uri'];

<img src="<?php print image_style_url($style, $img_url) ?>" />

To get just the path of an image from it's URI:
<pre>file_create_url($node->field_image['und'][0]['uri']);</pre>

More information on file_create_url() can be found here:

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To get the path of an image created using an image style from it's URI use:
<pre>image_style_url($style, $node->field_image['und'][0]['uri']);</pre>

More information on image_style_url() can be found here:

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I am afraid, none of the solutions above are correct. They don't follow Drupal standards.

// field_video_image is the name of the image field

// using field_get_items() you can get the field values (respecting multilingual setup)
$field_video_image = field_get_items('node', $node, 'field_video_image');

// after you have the values, you can get the image URL (you can use foreach here)
$image_url = file_create_url($field_video_image[0]['uri']);

More details here:

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You also can use the [Image URL Formatter][1] module. It allows to change the field format in order to get only the URL:


![Drupal field settings][2]


Bonus: it works with Views as well:


![enter image description here][3]


[1]:

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[2]:

[3]:

to get image style url:

$field = field_get_items('node', $node, 'field_image');
$image_url = image_style_url('landingpage_slider', $field[0]['uri']);