CODEWITHSUNDEEP
php
Nexgen Technologies LLC
Nexgen Technologies LLC

Reputation: 35

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

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>

same image will be display in model and by submiting it will send flag to DB

Upvotes: 0

Views: 268

Answers (1)

user6025438
user6025438

Reputation:

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>");
}

Upvotes: 2

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