Creating a php script to display a random image using Mysql database [duplicate]

Question

I have a folder full of images (200), I want a php script to select 1 image every 24 hours and display it on the page (using true randomness, example not display the same image more than once until all images have been displayed)

I would like to use a mysql db for this script to store the id, image path, and the image link (example: id: 1, us.png, http://www.website.com/us.html)

Note: The link will be dynamic in nature, such as http://www.website.com/cms/united-states/

Also note: I need help with the entire script, as in the creation of the db, and the script itself as I am just now starting out...I learn best visually, I need to see the script function before I can fully grasp the concept. I appreciate all who contribute to helping learn php+mysql!

Thank you in advance! Brian

before you ask please read here stackoverflow.com/help/how-to-ask — Rendi Wahyudi Muliawan
– Rendi Wahyudi Muliawan, Commented Apr 20, 2018 at 5:51

stalin wesley · Accepted Answer · 2018-04-20 05:55:05Z

0

here is code

$folder = "images";
    $results_img_arr = array();
    if (is_dir($folder))
    {
            if ($handle = opendir($folder))
            {
                    while(($file = readdir($handle)) !== FALSE)
                    {
                        if(!in_array($file,array(".","..")))
                            $results_img_arr[] = $folder."/".$file;
                   }
             closedir($handle);
            }
    }
    $ran_img_key  = array_rand($results_img_arr);
    $dbconn = mysqli_connect("127.0.0.1", "root", "", "images");
    $img_path = $results_img_arr[$ran_img_key];
    $select_img_exist = mysqli_query($dbconn,"SELECT imagespath FROM imgfiles WHERE  imagespath = '$img_path'");
    $img_rowcount=mysqli_num_rows($select_img_exist);
        if($img_rowcount == 0 )
        {
         $insert_img = "INSERT INTO imgfiles (imagespath,views) VALUES ('$img_path',1)";
        mysqli_query($dbconn, $insert_img);
        echo "<img src='".$img_path."'>";
        }
        else{
            echo "all images are displayed";
        }

answered Apr 20, 2018 at 5:55

stalin wesley

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4 Comments

Brian Over a year ago

Hello, I forgot the most important part of the code that I was initially looking for which is to display random at midnight each day, I have a piece of code from another script I used to use, but I don't know how to incorporate it in this code: $today=getdate(); srand($today['mday']+$today['month']+$today['year']); $r=rand(0,$i-1); echo '<a href="' . $links[$imgs[$r]] . '">' . "<img src=\"flags/{$imgs[$r]}\" alt=\"Photo\" /></a>"; } That basically would display an image at 12 midnight everyday. Can you plug that into your response code to make it work?

Brian Over a year ago

Also, I believe because I am using php 7.028 your script gives me this error: Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /var/www/html/rand.php on line 21 I have resolved it by putting a @ like this: $img_rowcount=@mysqli_num_rows($select_img_exist); Is that ok, or should something else be changed? I get no errors after adding the @ however... Thank you again! Brian

Brian Over a year ago

...And how I can put a different link to each image being displayed? I will create the row in my database table for the link? Also, I believe because I am using php 7.028 your script gives me this error: Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /var/www/html/rand.php on line 21 I have resolved it by putting a @ like this: $img_rowcount=@mysqli_num_rows($select_img_exist); Is that ok, or should something else be changed? I get no errors after adding the @ however... Thank you again! Brian

Brian Over a year ago

Also, I notice that the db part is not having an effect on the script, your code only seems to be working from the folder and the db. I want to pull the image url from the db table and display it, currently it not working that way. If I comment out the db query, the image sitll display's when I comment out the SQL Query: // $select_img_exist = mysqli_query($dbconn,"SELECT image FROM test WHERE imagespath = '$img_path'"); The image still displays, it should NOT display if code is working, but it's not. Can you fix?

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