将来自数据库的图像显示为PHP中div元素的背景

Question

Currently I'm learning about PHP and database, I wrote the script below to display the image as background for my div element, but the output is actually nothing! the error is from the line:

echo "<div class=\"post\" style='background-image: url('\"<?php echo $data[Image];\"?>')'";?>

The quotes mess up! Can someone tell me how to correct this? I tried to change double quotes to single quotes, but still doesn't work at all.

This is my full script:

<div class="dashboardA">
                <?php
                    $con = mysqli_connect("localhost", "Dave", "password");
                    if (!$con){
                        die ("Could not connect to database: " . mysqli_connect_error());
                    }
                    mysqli_select_db($con, "my_blog");

                    $sql = mysqli_query($con, "select * from article");

                    while ($data=mysqli_fetch_array($sql)){
                        echo "<div class=\"post\" style='background-image: url('\"<?php echo $data[Image];\"?>')'";?>
                        <?php echo "<p>" . $data["Title"] . "</p>";
                        echo "<p>" . $data["Category"] . "</p>";
                        echo "<p>" . $data["Published"] . "</p>";
                        echo "</div>";
                    }
                ?>
            </div>

Answer 1

You are getting a little mixed up. You already have PHP opening tags, you don't need them again. Just concatenate your variable:

echo "<div class='post' style='background-image: url(\"$data[Image]\")'>";

Note: You also need to close your opening <div> tag.

Answer 2

There is a mistake in the line

echo "<div class=\"post\" style='background-image: url('\"<?php echo $data[Image];\"?>')'";?>

You are already in php, so the opening tags are wrong there.

Try:

echo "<div class=\"post\" style='background-image: url(\"" . $data[Image] . "\")'";?>

Answer 3

Try this

 while ($data=mysqli_fetch_array($sql)){
    echo "<div class=\"post\" style='background-image: url('\"<?php echo $data[Image];\"?>')'";?>
    <?php echo "<p>" . $data["Title"] . "</p>";
    echo "<p>" . $data["Category"] . "</p>";
    echo "<p>" . $data["Published"] . "</p>";
    echo "</div>";
}

Replace this

with this

while ($data=mysqli_fetch_array($sql)){
    $img = $data["Image"];?>
    <div class="post" style="background-image: url('<?php echo $img;?>')" >
    <?php echo "<p>" . $data["Title"] . "</p>";
    echo "<p>" . $data["Category"] . "</p>";
    echo "<p>" . $data["Published"] . "</p>";
    echo "</div>";
}

Answer 4

You actually did it right in the following echo commands. To combine the value of your variable and a string you use the "." that basically substitutes for "+" that is used in other languages for this purpose. Opening a new php tag is therefore not needed here.

The correct code is

echo '<div class="post" style="background-image: url(\'' . $data[Image] . '\')">';

Answer 5

在php标记之外尝试此操作。

<div class="post" style="background-image: url('<?php echo "$data[Image]"; ?>');">