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How to display images on PHP page from my database? – Phpmyadmin

I’m trying to display images and relevant data based on what the user enters in a form(IMG 1). Everything is working but I need the table to show the actual image instead of URL.

In my database, I’ve got the image data stored as URL but I want to show as an image on the PHP page (IMG 3).

IMG – 1: What User Enters

enter image description here

IMG – 2: Result of Search
enter image description here

IMG – 3: Set up on phpMyAdmin
enter image description here

So in the table (IMG 2), I want the actual image to show up instead of link

Here is the most important part of the code that does the searching and displays the data.

    $title = trim($_POST["PhotoTitle"]);
    $FromDate = trim($_POST["FromDate"]);
    $ToDate = trim($_POST["ToDate"]);
    $keywords = trim($_POST["keywords"]);

    $query ="SELECT * FROM `photos` WHERE `Photo-Title` like '%".$title."%' AND `Keywords` like '%".$keywords."%' AND `Date-of-Photo` BETWEEN '$FromDate' AND '$ToDate'";

    while ($row = mysqli_fetch_assoc($result)) {
        echo "<tr>n ";
        echo "<td>",$row["Photo-Title"],"</td>n";
        echo "<td>",$row["Description"],"</td>n";
        echo "<td>",$row["Date-of-Photo"],"</td>n";
        echo "<td>",$row["Keywords"],"</td>n";
        echo "<td>",$row["Reference-to-S3"],"</td>n";
        echo "</tr>n ";
    }

I can provide any other information if needed.

How do I solve this problem?

Tags:

2

Answers


  1. 0

    First: Your code is highly insecure. It is vulnerable to sql injections and XSS.

    Regarding your Question: You need to output an html tag with the image url as the src attribute.

    Like this:

    echo "<td><img src='" . htmlspecialchars($row["Reference-to-S3"]) . "' /></td>n";
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  2. 0

    Just edit your while to this piece of code :

    while ($row = mysqli_fetch_assoc($result)) {
        echo "<tr>n ";
        echo "<td>",$row["Photo-Title"],"</td>n";
        echo "<td>",$row["Description"],"</td>n";
        echo "<td>",$row["Date-of-Photo"],"</td>n";
        echo "<td>",$row["Keywords"],"</td>n";
        echo "<td><img src='" . $row["Reference-to-S3"] . "' alt='" . $row["Photo-Title"] . "' /></td>n";
        echo "</tr>n ";
    }

    what it does is displaying a URL as an image using <img /> tag of HTML. Also, alt adds a title to the image which if you hover image, you can see the value of this attribute.

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