PHP HTML Tables

Question

I'm trying to display

$row["name"]

through the form of a HTML Table, like this:

echo "<html>
<table class="."table".">
  <tr>
    <td>".$row["name"]."</td>
  </tr>
</table>
</html>";

I'm getting the row variable from here, too:

$query = "SELECT * FROM servers WHERE public = 1";

if ($result = $db->query($query)) {

while ($row = $result->fetch_assoc()) {
    printf ("Server name: %s", $row["name"].'<br>');
    printf ("Hostname: %s", $row["host"].'<br>');
    printf ("Port: %s", $row["port"].'<br>');
    printf ("Player count: %s", $row["players"].'<br>');
    printf ("Server Status: %s", $row["status"].'<br>');
    printf ("Last pinged: %s", $row["last_ping"].'<br>');
    printf ("Current ms: %s", $row["ms"].'<br>');

}

$result->free();
}

It does successfully display the information printed, but doesn't seem to be able to put it into a table.

https://i.sstatic.net/R78cg.png

@HarryDenley if you look at the imgur link, it's still creating the table, just not displaying the $row["name"]. I've tried multiple ways of it, it's having none of it. — Shepley
– Shepley, Commented Dec 20, 2013 at 17:21
<table class="."table"."> ? – Try <table class=\"table\"> or <table class='table'> More than likely the issue. — Funk Forty Niner
– Funk Forty Niner, Commented Dec 20, 2013 at 17:21
Have you tried var_dump($row) before you output the HTML, to check the contents of the $row variable? — JAL
– JAL, Commented Dec 20, 2013 at 17:22
Yeah, the table row there would be <table class=table>. That's not valid html since attributes must be quoted — JAL
– JAL, Commented Dec 20, 2013 at 17:25

Lloyd Banks · Accepted Answer · 2013-12-20 17:32:28Z

1

You never actually assign any of the row values in the WHILE loop to a variable. If the SQL statement returns more than 1 row, you need to build an array in your WHILE loop to use for later.

Define your array first

$myarray = array();

Then assign the object returned from the SQL statement to the array

while($row = $result->fetch_assoc()){
    $myarray[] = $row;
    //your other printf statements if you want to keep them
}

Now you have a populated $myarray where the first level key is the row number. Since you are probably building a dynamic table (one whose size depends on the amount of information returned), you'll also need to incorporate PHP into the process.

<table class = "table">
    <?php
        for($i = 0; $i < sizeof($myarray); $i++){
             echo "<tr>";
                 echo "<td>";
                      echo $myarray[$i]["name"];
                 echo "</td>";
             echo "</tr>";
        } 
    ?>
</table>

In the above, you run a for loop on every item in $myarray and for each row returned from your original database pull, a new table row and corresponding table column are created.

answered Dec 20, 2013 at 17:32

Lloyd Banks

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2 Comments

Shepley Over a year ago

Managed to get it working, really need to refresh my PHP, not done it for so long. :L Thanks for the help, greatly appreciated. :)

Funk Forty Niner Over a year ago

~ Another Happy Ending ~

Kyle Goslan · Accepted Answer · 2013-12-20 17:21:33Z

1

Have you tried taking the whole table out of the php echo statement?

So having it as html then just echoing the row inside the table

<td><?php echo $row["name"] ?></td>

answered Dec 20, 2013 at 17:21

Kyle Goslan

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Comments

Funk Forty Niner · Accepted Answer · 2013-12-20 17:32:37Z

0

you have to put the table row <tr></tr> inside the while loop and use separate <td> columns for each data field.

Sample code:

echo '<table>';
while{$row}
{
 echo '<tr>
       <td>'.$row['name'].'</td>
       </tr>';
}
echo '</table>';

edited Dec 20, 2013 at 17:32

Funk Forty Niner

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answered Dec 20, 2013 at 17:31

user3080343

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YRM · Accepted Answer · 2013-12-20 17:37:59Z

0

this:

echo "<html>
<table class="."table".">
  <tr>
    <td>".$row["name"]."</td>
  </tr>
</table>
</html>";

will echo this:

<html>
<table class=table>
  <tr>
    <td>name</td>
  </tr>
</table>
</html>

thats not valid html

as suggested by Kyle Goslan you should try it like this:

<html>
<table class="table">
  <tr>
    <td><?php echo $row['name']; ?></td>
  </tr>
</table>
</html>

this is how i would do it:

<html>
<table class="table">
   <tr>
      <th>Name</th>
   </tr>
   <?php
$query = "SELECT * FROM servers WHERE public = 1";

if ($result = $db->query($query)) {

    while ($row = $result->fetch_assoc()) { ?>
        <tr>
           <td>
              <?php echo $row['name']; ?>
           </td>
       </tr>
   <?php }
}

$result->free();
?>

edited Dec 20, 2013 at 17:37

answered Dec 20, 2013 at 17:26

YRM

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2 Comments

Funk Forty Niner Over a year ago

I was just about to make a comment about that. I wasn't going to put in an actual answer, because questions like these tend to open up both a Pandora's box and the proverbial "can of worms".

Shepley Over a year ago

Yeah, i've previously tried that, but it's done nothing. The table is displaying, but blank. It's not getting the variable, I don't understand why.

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