CODEWITHSUNDEEP
phpjqueryjquery-pluginsdatatables
Shane
Shane

Reputation: 753

Data table error

Im using a datatable to show records from a DB, using PHP.

See my code below;

<table class="table datatable-basic">
  <thead>
    <tr>
      <th> ID </th>
      <th> Name</th>
      <th>Address </th>
      <th class="text-center">Actions</th>
    </tr>
  </thead>
  <tbody>

  <?
  $result9023=mysql_query("SELECT * FROM hr_locations")or die('ERROR 315' );
  $num_rows = mysql_num_rows($result9023);

  for ($i = 1; $i <= mysql_num_rows($result9023); $i++) {
    $row = mysql_fetch_array($result9023);
    $location_id = $row ['id'];
    $name = $row ['location'];
    $address = $row ['address'];

    echo "         <tr>
        <td>$location_id</td>
        <td>$name</td>
        <td>$address</td>
        <td class='text-center'>Edit</td>
      </tr>
    "; } ?>
  </tbody>  
</table>

The table is showing correctly, with the data populated as it should however I am getting the below error when the page loads.

Error

I have looked at https://datatables.net/tn/4 however it's not making much sense?

Upvotes: 1

Views: 355

Answers (1)

Jay Blanchard
Jay Blanchard

Reputation: 34416

Since most MySQL array start with 0 do this:

for ($i = 0; $i < $num_rows; $i++) {

Set $i = 0 then state that $i must be less than the number of rows (if you have 4 you will get 0,1,2,3 (4 rows, properly indexed). Instead of counting the rows again, just use the variable you already set for the count.

You're actually probably doing a little too much with your code. Instead of using an unnecessary for loop, just use a while:

while($row = mysql_fetch_array($result9023)){ 
    $location_id = $row ['id'];
    $name = $row ['location'];
    $address = $row ['address'];

    echo "         <tr>
        <td>$location_id</td>
        <td>$name</td>
        <td>$address</td>
        <td class='text-center'>Edit</td>
      </tr>
    "; } ?>

This way you're making sure to catch each row returned from your query without possibly duplicating or skipping rows as you might do when using a for loop along with mysql_fetch_array() as you're trying to do.

Upvotes: 2

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