Populate an html table columwise with mysql data - php

Question

I have a MySQL query that returns data using PHP.
My problem is that I need to populate my html table (with 3 columns) with the data returned by the query but the data should be populated Columnwise.
Like first the first column should be populated .. then the second and finally the third one.
Also I would like to know that is it possible to do so for an unlimited set of data?
Following the screen shot of the desired layout

Answer 1

you can use while.
<table>
$sql=mysql_query("select * from table");
while($s=mysql_fetch_array($sql))
{
$x=$s["x"];
<tr>
<td ><?php echo $x; ?></td>
<td ><?php echo $y; ?></td>
<td ><?php echo $z; ?></td>
</tr>
}
</table>

Answer 2

guybennet's answer is correct but if you want it to work for an unlimited amount of columns you could do this: (I also threw in some column headers for readability)
echo '<table>';
$counter = 0;
$result = mysql_query("select * from table");
while($row = mysql_fetch_array($result)) {
$counter++;
if($counter == 1) {
echo '<tr>';
foreach($row as $key => $val) {
echo "<th>$key</th>";
}
echo '</tr>';
}
echo '<tr>';
foreach($row as $key => $val) {
echo "<td>$val</td>";
}
echo '</tr>';
}
echo '</table>';
Also of course you should use mysqli or PDO I'm just showing a quick example.

Answer 3

If you don't care about how it's organized, you can try something like this:
echo "<table><tr>";
$i=0;
while($row = mysql_fetch_array($sql))
{
echo "<td>".$row[0]."</td>\n";
$i++;
if($i==3)
{
echo "</tr>\n<tr>";
$i=0;
}
}
echo "</tr></table>";
Otherwise, I'd suggest putting it all into an array and then putting it into the table.

Answer 4

$data = array();
$result = mysql_query("SELECT * FROM your_table");
while ($row = mysql_fetch_array($result)) {
$data[] = $row;
}
$itemsAmount = count($data);
$ceilAmount = ($itemsAmount - $itemsAmount % 3) / 3;
$lastAmount = $itemsAmount % 3;
$firstArray = array_slice($data, 0, $itemsAmount);
$secondArray = array_slice($data, 0, $itemsAmount*2);
$thirdArray = array_slice($data, 0, $lastAmount);
$output = "<table>";
foreach ($data as $key => $value) {
$output .= "<tr>";
$output .= "<td>" . $firstArray[$key][0] . "</td>";
$output .= "<td>" . $secondArray[$key][0] . "</td>";
if (empty($thirdArray[$key])) {
$str = '';
} else {
$str = $thirdArray[$key][0];
}
$output .= "<td>" . $str . "</td>";
$output .= "</tr>";
}
$output .= "</table>";
echo $output;

Answer 5

You need to check all the results returned, then print them as you won't print in order:
First get an array with all the results
<?php
$sql= mysql_query('SELECT * FROM table');
$num= mysql_affected_rows($sql);
$items = array();
$i=0;
while($item=mysql_fetch_array($sql)){
$items[++$i]=$item['data'];
}
Now start printing
int $num_rows;
$num_rows=$num%3;
echo '<table>';
for ($i=0;$i<$num_rows;$i++){
echo '<tr>';
for ($j=0;$j<2,$j++){
$n=$i+1+($j*$num_rows);
if($items[$n]!==null)
echo '<td>'.$items[$n].'</td>';
else
echo '<td></td>';
}echo '</tr>';
}echo'</table>';
?>