foreach loop in database table column with multiple images

Question

<?php  
    $images= $uploading['front_image'];
    $whileimg= explode(",", $images);
    foreach ($whileimg as $key => $image) {
        echo  $uploading["front_image"];          
    }
?>

In database I have my images saved in one column(2,3,4 it depends). now when I want them all to show in web ,Instead of images I get the path of the photos, as are saved in database. thank you

Hey @newuser - I have edited your question to improve the formatting of the code and to make it more readable — Alon Eitan
– Alon Eitan, Commented Aug 31, 2017 at 19:33
You need to make it an image take and echo the path in the img tag — clearshot66
– clearshot66, Commented Aug 31, 2017 at 19:34
There is another way to do it, just create a new table and join it. But @clearshot66 have the answer for your case — Muhammad Ibnuh
– Muhammad Ibnuh, Commented Aug 31, 2017 at 19:35

Ravinder Reddy · Accepted Answer · 2017-08-31 19:35:56Z

1

To display images, you have use img tag

foreach ($whileimg as $key => $image) {
        echo  "<img src='".$image."' alt='image'>";   
    }

answered Aug 31, 2017 at 19:35

Ravinder Reddy

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1 Comment

Ravinder Reddy Over a year ago

$whileimg might have only one element. Print it and see elements in it.

clearshot66 · Accepted Answer · 2017-08-31 19:35:57Z

1

You need to make it an image take and echo the path in the img is what it sounds like you're getting:

 echo "<img src='".$image."' />

instead of

echo  $uploading["front_image"];

So:

<?php  
    $images= $uploading['front_image'];
    $whileimg= explode(",", $images);
    foreach ($whileimg as $key => $image) {
        echo "<img src='".$image."' />          
    }
?>

answered Aug 31, 2017 at 19:35

clearshot66

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1 Comment

clearshot66 Over a year ago

Well check your array of images because foreach don't terminate until a break is issued or the array ends

Deepansh Sachdeva · Accepted Answer · 2017-08-31 19:43:13Z

0

Assuming you are fetching comma separated paths of images using the following line:

$images= $uploading['front_image'];

You just need to use the following loop after creating array of paths:

foreach ($whileimg as $key => $image) {
    echo "<img src='$image'>";
}

answered Aug 31, 2017 at 19:43

Deepansh Sachdeva

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