how to display images from folder while displaying image by name using php?

Question

I have two folders, 1)- 23feb2016 and 2)- 23feb2016. I want to get one image from folder1 and one from folder2 , the image name must be unique, like userid etc.

This is my html code:

<div class="col-md-12">
    <div class="col-md-6">
        <Image src="../images/registered/20115.jpg" style="width: 100%; height: 100%"></Image> Registered
    </div>
    <div class="col-md-6">
        <Image src="../images/registered/20115.jpg" style="width: 100%; height: 100%"></Image> Attendance
    </div>
 </div>

Use date('Y-m-d-H-i-s') . '_' . uniqid() for unique image name — Ravi Hirani
– Ravi Hirani, Commented Mar 7, 2016 at 7:07

score 2 · Accepted Answer · 2016-03-07 07:54:53Z

2

You can use the following approach to find the unique files.

//path to directory to scan
$directory1 = "./folder1/";
$directory2 = "./folder2/";

//get all Image files with a .jpg extension.
$jpgFolder1 = glob($directory1 . "*.jpg");
$jpgFolder2 = glob($directory2 . "*.jpg");

$UniqeImage[];

if(($jpgFolder1 != null) && ($jpgFolder2 != null))
{
    $UniqeImage = array_unique(array_intersect($jpgFolder1, $jpgFolder2));
}

Now the Array Variable $UniqeImage[] contains the Unique file names using this you can construct the HTML

foreach($UniqeImage as $img)
{
    echo("<div class='col-md-12'>");
    echo("    <div class='col-md-6'>");
    echo("        <Image src='../images/Folder1/" . $img . ".jpg' style='width: 100%; height: 100%'></Image> Registered");
    echo("    </div>");
    echo("    <div class='col-md-6'>");
    echo("        <Image src='../images/Folder2/" . $img . ".jpg' style='width: 100%; height: 100%'></Image> Attendance");
    echo("    </div>");
    echo("</div>");
}

edited Mar 7, 2016 at 7:54

answered Mar 7, 2016 at 7:32

user6025438

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3 Comments

Nexgen Technologies LLC Over a year ago

Thanks for help the folders must be dynamically changing as per date .. means each date have own folder of images and the name of folder will be the date and image name will be userid.jpg like this .

user6025438 Over a year ago

Happy with your reply. The answer really fulfilled your requirement means, then mark it as your Answer.

Nexgen Technologies LLC Over a year ago

it displays all the images while i have to check for the image with folder name and image with transiction ,

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