Fetch and display data from database in table format in HTML form using PHP

Question

I have done with fetching and display the data. But the problem is ,its display the data of fetched row in the same row of html table. For example : i want to display like this

                  ID | Name | Desination
                   ......................
                   1 | ABC | Developer
                   2 | PQR | Tester
                   3 | XYZ | Developer

But its showing as -

                   ID | Name | Desination
                   ......................
                    1 | ABC | Developer   2 | PQR | Tester   3 | XYZ | Developer

I have done something like this-

$sql = " SELECT candidate.cand_number,candidate.cand_fname,candidate.cand_desc FROM candidate ".$join.' where '.$condition;
$result = mysql_query($sql) or die(mysql_error());

Correct me with the display format of table.

 <div class="box-body table-responsive no-padding">
     <table class="table table-hover">
          <tr>
              <th>ID</th>
              <th>Name</th>
              <th>Designation</th>
          </tr>
          <tr>
              <?php

                   If(mysql_num_rows($result)>0)
                   {
                     while($row=mysql_fetch_array($result))
                     {  

                ?>
                  <td><?php echo $row['cand_number']; ?></td> 
                  <td><?php echo $row['cand_fname']; ?></td> 
                  <td><?php echo $row['cand_desc']; ?></td> 
                <?php

                }
                }
                 ?>

              </tr>
       </table>
    </div>

Saty · Accepted Answer · 2015-05-02 09:46:49Z

5

You need to repeat row not column

 <?php
If (mysql_num_rows($result) > 0) {
    while ($row = mysql_fetch_array($result)) {
        ?>
        <tr>
            <td><?php echo $row['cand_number']; ?></td> 
            <td><?php echo $row['cand_fname']; ?></td> 
            <td><?php echo $row['cand_desc']; ?></td> 
        </tr>
        <?php
    }
}
?>

edited May 2, 2015 at 9:46

Saty

22.5k7 gold badges35 silver badges52 bronze badges

answered May 2, 2015 at 9:05

Pranay Aher

4344 silver badges11 bronze badges

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7 Comments

vinod Over a year ago

@AbhiAhere..Thank you...it worked. How would i make the "name" as hyperlink and after on click of the link it should display the whole details of that candidate.

Pranay Aher Over a year ago

you are welcome . If you find the the answer helpful then check it as correct – abhi

Pranay Aher Over a year ago

As i have mentioned above You need to repeat row not column, all the explanation is in the line itself

vinod Over a year ago

@NarendraSisodia..ok the link too worked..but give some idea how would i display that particular candidate details when the link is clicked. i.e. it should execute another sql query on click. to display his whole details

vinod Over a year ago

...how would i display that particular candidate details when the link is clicked. i.e. it should execute another sql query on click. to display his whole details

|

Sagar · Accepted Answer · 2015-05-02 09:19:57Z

0

Use this HTML and try

<div class="box-body table-responsive no-padding">
<table class="table table-hover">
<tr>
   <th>ID</th>
   <th>Name</th>
   <th>Designation</th>
</tr>

          <?php

               If(mysql_num_rows($result)>0)
               {
                 while($row=mysql_fetch_array($result))
                 {  

            ?>
             <tr>
              <td><?php echo $row['cand_number']; ?></td> 
              <td><?php echo $row['cand_fname']; ?></td> 
              <td><?php echo $row['cand_desc']; ?></td> 
            </tr>
            <?php

            }
            }
             ?>


   </table>
</div>

answered May 2, 2015 at 9:19

Sagar

6473 silver badges14 bronze badges

Comments

Saty · Accepted Answer · 2015-05-02 09:21:30Z

0

You need to add into the while loop. Because of this your all data is printed inside first row.

<?php
If (mysql_num_rows($result) > 0) {
    while ($row = mysql_fetch_array($result)) {
        ?>
        <tr>
            <td><?php echo $row['cand_number']; ?></td> 
            <td><?php echo $row['cand_fname']; ?></td> 
            <td><?php echo $row['cand_desc']; ?></td> 
        </tr>
        <?php
    }
}
?>

edited May 2, 2015 at 9:21

Saty

22.5k7 gold badges35 silver badges52 bronze badges

answered May 2, 2015 at 9:04

Codelord

1,1202 gold badges10 silver badges21 bronze badges

6 Comments

vinod Over a year ago

@CodeLord..Thank You..it worked for me. :) How would i make the "name" as hyperlink and after on click of the link it should display the whole details of that candidate.

Codelord Over a year ago

@vinod just add anchore tag for name like <a href="#"><?php echo $row['cand_fname']; ?></a>

vinod Over a year ago

how would i display that particular candidate details when the link is clicked. i.e. it should execute another sql query on click. to display his whole details

Codelord Over a year ago

You need to create another page for this send your candidate id in url parameter. This is broad question . You need to post this question so I can give brief answer.

vinod Over a year ago

@CodeLord..wait I brief you with my code and send you the link. soon. Thank you

|

Community · Accepted Answer · 2017-05-23 12:16:54Z

0

In addition to the other answers: don't forget to sanitise $join and $condition. This to prevent SQL injection.

edited May 23, 2017 at 12:16

CommunityBot

11 silver badge

answered May 2, 2015 at 9:27

user2566212

1 Comment

vinod Over a year ago

Thank you! its just a raw code. we will implement it for sure!

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