I want to show date (dd-mm-yyyy) format

Question

In my table currently shows 2019-01-21 this format and i want to show 21-01-2019 like this

I am trying like this

<table>
<tr>
 <th>DOB</th>
</tr>

 <?php foreach($employees as $post){?>
 <tr>
 <td><?php echo $post->user_date_of_birth;?></td>
 </tr>
 <?php }?>
</table>

use date() function with strtotime for formatting date. eg- date(d-m-YY, strtotime($post->user_date_of_birth)) — Gajanan Kolpuke, Mar 05 '19 at 05:52
on a personal matter, use yyyy-mm-dd or dd.mm.yyyy. please. there are too many non-standardised datum formats out there, help keep their count low. — Franz Gleichmann, Mar 05 '19 at 06:02

score 1 · Answer 1 · answered Mar 05 '19 at 05:53

1

    Please set format in sql like this.

    SELECT FORMAT(FieldName,'dd-MM-yyyy')

answered Mar 05 '19 at 05:53

Hemang A

1,012
1
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16

score 1 · Accepted Answer · edited Mar 05 '19 at 07:23

1

Okay, so let's assume that the user_date_of_birth has a value of 2019-01-21

Let's use your code:

<table>
    <tr>
      <th>DOB</th>
    </tr>

    <?php foreach($employees as $post){?>
    <tr>
      <td><?php echo date("d-m-Y", strtotime($post->user_date_of_birth));?></td>
    </tr>
    <?php }?>
</table>

Hope this helps!

edited Mar 05 '19 at 07:23

Smithiam

162
1
16

answered Mar 05 '19 at 05:54

Roshan

786
1
9
20

Eiji · Answer 3 · 2019-03-05T05:54:24.213

0

 echo date('d-m-y', strtotime($post->user_date_of_birth));

edited Mar 05 '19 at 05:54

answered Mar 05 '19 at 05:52

Eiji

454
6
15

'd-m-Y' should be the format – Devi A Mar 05 '19 at 05:53
Sorry my bad ^^ – Eiji Mar 05 '19 at 05:55

Ankur Tiwari · Answer 4 · 2019-03-05T05:59:49.640

0

You can use the following code

<?php
    $yourDate = "2019-01-21";
    echo date("d-m-Y", strtotime($yourDate) );
?>

Output: 21-01-2019

Then your final code will be like this

<table>
<tr>
 <th>DOB</th>
</tr>

 <?php foreach($employees as $post){?>
 <tr>
 <td><?php echo date("d-m-Y", strtotime($post->user_date_of_birth));?></td>
 </tr>
 <?php }?>
</table>

edited Mar 05 '19 at 05:59

answered Mar 05 '19 at 05:52

Ankur Tiwari

2,762
2
23
40

CHATHURA · Answer 5 · 2019-03-05T05:57:38.350

0

<td><?php echo date('d-m-Y', strtotime($post->user_date_of_birth));?></td>

edited Mar 05 '19 at 05:57

answered Mar 05 '19 at 05:53

CHATHURA

106
7

1

one `)` is missed in your answer – Danish Ali Mar 05 '19 at 05:56
Please add some explanation to your answer such that others can learn from it – Nico Haase Mar 05 '19 at 06:33

score 0 · Answer 6 · answered Mar 05 '19 at 05:53

0

Try this

<td><?php echo date('d-m-Y', strtotime($post->user_date_of_birth));?></td>

Read more date() and strtotime()

answered Mar 05 '19 at 05:53

Danish Ali

2,354
3
15
26

score 0 · Answer 7 · answered Mar 05 '19 at 06:03

0

<table>
<tr>
 <th>DOB</th>
</tr>

 <?php foreach($employees as $post){ 
 $datetime=$post->user_date_of_birth;
        $timeformat = date("d-m-Y",strtotime($datetime));   ?>
 <tr>
 <td><?php echo $timeformat; ?></td>
 </tr>
 <?php }?>
</table>

answered Mar 05 '19 at 06:03

Maruthi Prasad

170
1
12

score 0 · Answer 8 · edited Mar 05 '19 at 06:48

0

you should to manage this in programming side, like `$origDate = "2018-04-20";

$newDate = date("d-m-Y", strtotime($origDate));
echo $newDate;`

edited Mar 05 '19 at 06:48

Mohammad Malek

624
1
7
17

answered Mar 05 '19 at 06:03

Amit Vekaria

1
2

I want to show date (dd-mm-yyyy) format

8 Answers8