PHP String for HTML style

Question

I want to show a png file in background of a div. I have this CSS style for put it in the correct place. I'v tested directly in my HTML and the style works properly. I't doesn't work wen I use php variable.

<?php 

    if ($RScoluna1['nova'] == 1){
        $style = "background-image:url(layout/nova.png); background-repeat:no-repeat; background-position: 220px 0;";
    } else {
        $style = "";
    }


    ?>
    <div class="<?php echo $row_RScoluna1['class_cor']; ?>" style="<?php echo $style; ?>">

are you sure that $RScoluna1['nova'] sometimes holds 1 as a value? — Masiorama
– Masiorama, Commented May 6, 2015 at 10:09
Make sure the relative path to the image is working and the div has enough width and heigth to display the image. — Jordy
– Jordy, Commented May 6, 2015 at 10:12
I'v checked. If you copy background-image:url(layout/nova.png); background-repeat:no-repeat; background-position: 220px 0; and past it on style="" it works — CMartins
– CMartins, Commented May 6, 2015 at 10:20

Masiorama · Accepted Answer · 2015-05-06 10:13:27Z

1

If you copy pasted

background-image:url(layout/nova.png)

from a css file, then you should fix the path accordingly to the position of the script.

In example:

background-image:url("../css/layout/nova.png");

I hope it helps.

answered May 6, 2015 at 10:13

Masiorama

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1 Comment

CMartins Over a year ago

My bad. if ($RScoluna1['nova'] == 1){ should be if ($row_RScoluna1['nova'] == 1){

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