How to display images(.jpg) only?

Question

Array ( [0] => assets/image/man.jpg [1] => assets/image/violin.jpg [2] => assets/image/test.txt )

The data from data base is like above.It contain images and txt.how can i display only images.

You could update your query so only image files are returned — brombeer, Dec 31 '18 at 08:06

score 1 · Answer 1 · answered Dec 31 '18 at 08:56

$ar = ['assets/image/man.jpg','assets/image/violin.jpg','assets/image/test.txt'];
 $allowed = ['jpg']; //your image extensions
 $img_ar = [];
 foreach($ar as $img){
     $ext = pathinfo($img,PATHINFO_EXTENSION);
     if(in_array($ext,$allowed)){
         $img_ar[] = $img;
     }

 }
 print_r($img_ar);

score 0 · Answer 2 · edited Dec 31 '18 at 09:14

0

Why do you check it when you want write it on page?
You can:
1. split assets/image/man.jpg with '/'
2. get last one,
3. split last one with '.'
4. get extension and if it was 'jpg' write it to page.

edited Dec 31 '18 at 09:14

PrakashG

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answered Dec 31 '18 at 07:50

Bahman Shafiei

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score 0 · Answer 3 · answered Dec 31 '18 at 07:51

$array= Array ( [0] => assets/image/man.jpg [1] => assets/image/violin.jpg [2] => assets/image/test.txt )
$m_array = preg_grep('/^.jpg\s.*/', $array);

$m_array contains matched elements of array.

For more detail have look at this thread search a php array for partial string match

Danish Ali · Answer 4 · 2018-12-31T08:12:32.127

0

Use foreach loop and get an extension of the file and display.

foreach($array_result as $result){

    //$array_result is array data
    //condition is checking the file that if it is an image or not
    $allowed =  array('gif','png' ,'jpg');
    $filename = $result;
    $ext = pathinfo($filename, PATHINFO_EXTENSION);
    if(in_array($ext, $allowed) ) {
        echo '<img src="'.$result.'" alt="" /> ';
    }
 }

edited Dec 31 '18 at 08:12

answered Dec 31 '18 at 07:56

Danish Ali

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This will fail on `fakefile-jpg.txt` for example. – Rotimi Dec 31 '18 at 08:03
`fakefile-jpg.txt` will be a text file not an image – Danish Ali Dec 31 '18 at 08:03
Yes and your if statement would mistake it for an image. You need to get the proper file extension in the loop the right way – Rotimi Dec 31 '18 at 08:04
You are making a mistake by checking if the word `jpg` is present in the text. – Rotimi Dec 31 '18 at 08:05
@Akintunde-Rotimi I've made changes to my answer – Danish Ali Dec 31 '18 at 08:05

Sherif Salah · Answer 5 · 2018-12-31T08:21:57.273

0

This should do the trick:

$images = array();
$images_exts = array('.jpg','.png');
foreach($db_array as $key => $value)
{ 
    foreach($images_exts as $ext)
    {
        if (strpos($value,$ext))
        {
            $images[] = $value;
        }
    }
}

Here is an example https://3v4l.org/8W3Be

And here is another way, whatever you like the most:

$images = array();
$images_exts = array('jpg','png');
foreach($input as $value)
{
    if(in_array(@end(explode('.', $value)), $images_exts))
    {
        $images[] = $value;
    }
}

Here is an example https://3v4l.org/b0njd

edited Dec 31 '18 at 08:21

answered Dec 31 '18 at 08:03

Sherif Salah

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This will fail on `fakefile-jpg.txt` for example. You should check for the extension the right way. All you are doing is checking for the text, `jpg` not the extension – Rotimi Dec 31 '18 at 08:06
@Akintunde-Rotimi what about this? – Sherif Salah Dec 31 '18 at 08:12
Added another way that won't fail on `-jpg` or even `.jpg` if it's a part of the filename itself :D @Akintunde-Rotimi – Sherif Salah Dec 31 '18 at 08:23

score 0 · Answer 6 · answered Dec 31 '18 at 08:09

For this you can directly filter it when you are querying like

field like '%.jpg'

If you don't want to do that and manipulate the array you can use array_filter like,

$array= Array ('assets/image/man.jpg', 'assets/image/violin.jpg', 'assets/image/test.txt');

$output = array_filter($array, function($arr) {
        if (strpos($arr, '.jpg') == true){
                return $arr;
        }
});

$output array contains only the entries which having the .jpg string.

Here am using strpos to check .jpg exists or not.

you maybe use substr($str, -4) == '.jpg' to check the last 4characters.

score 0 · Answer 7 · answered Dec 31 '18 at 08:11

0

If you are using PHP 5+ (which I hope you are on 7.0+), use SplFileInfo() class

$spl  = new SplFileInfo($fileName); 
if ($spl->getExtension() == 'jpg') {
//image
}

answered Dec 31 '18 at 08:11

Rotimi

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score 0 · Answer 8 · answered Dec 31 '18 at 09:19

<?php
  $error = array();
  $file_extArr = explode(".", $file_name);
  $file_extEnd = end($file_extArr);
  $file_ext = strtolower($file_extEnd);

  $validateImage = array("png", "jpg", "jpeg", "gif");

  if (!in_array($file_ext, $validateImage)) {
    $error[] = "wrong format image";
  }

  if (!empty($error)) {
   return;
  }
?>

score 0 · Answer 9 · answered Jan 03 '19 at 09:43

    <?php 
    $data = Array ( 
       'assets/image/man.jpg ',
       'assets/image/violin.jpg ',
       'assets/image/test.txt ',
    );

    $arrDara = array();
    foreach ($data as $value) {
        $fileName = explode('/', $value);
        $arrDara[] = end($fileName);
    }

    print_r($arrDara);
?>

Just loop your array and explode every sting. the last index is what all you need.

How to display images(.jpg) only?

9 Answers9