PHP For Loop dynamic from database

Question

I'm generating my check boxes from this code. But, I'm having difficulty with making it generate by 5 outputs then once the 5th output is displayed it should skip to the next line. Would appreciate any help.

<?php 
    $result=mysql_query("SELECT * FROM tbl_tourism_type order by type_name ");
    while($row = mysql_fetch_array($result))
    echo '<input type="checkbox" name="type" value='.$row['type_id'].'>'
            .'<label>'.$row['type_name']. '</label>'.'<br>'.'<br>';
?>

Can you explain a little more? Its already showing every next result in a new line — Hanky Panky
– Hanky Panky, Commented Jan 25, 2013 at 5:58
so put a counter in your while loop and do continue when it is == 5 — sdjuan
– sdjuan, Commented Jan 25, 2013 at 5:59
Kinda hacky but if you want a 1 liner: if (@++$i == 5) continue; — Supericy
– Supericy, Commented Jan 25, 2013 at 5:59
Yes. But, Instead of Checkbox <br> I would really like a Checkbox Checkbox Checkbox Checkbox Checkbox <br> — user1998926
– user1998926, Commented Jan 25, 2013 at 5:59

Hanky Panky · Accepted Answer · 2013-01-25 06:07:25Z

1

<?php 
$result=mysql_query("SELECT * FROM tbl_tourism_type order by type_name ");
$i=1;
while($row = mysql_fetch_array($result))
{
echo '<input type="checkbox" name="type" value='.$row['type_id'].'><label>'.$row['type_name']. '</label>';
if($i%5==0)
{
   echo "<br>";
}
$i++;
}
?>

edited Jan 25, 2013 at 6:07

answered Jan 25, 2013 at 6:02

Hanky Panky

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Comments

Rajat Garg · Accepted Answer · 2013-01-25 06:00:22Z

0

$result = mysql_query("SELECT * FROM tbl_tourism_type order by type_name ");
                $i=0;
                while($row = mysql_fetch_array($result)){
                if($i%5 == 0){
                 //skip to next line
                 //continue??
                 }
                echo '<input type="checkbox" name="type" value='.$row['type_id'].'>'.'
                <label>'.$row['type_name']. '</label>'.'<br>'.'<br>';                   
                $i++;
                }

answered Jan 25, 2013 at 6:00

Rajat Garg

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Comments

sourcecode · Accepted Answer · 2013-01-25 06:02:54Z

0

Do like this

<?php 
    $i =1;
    $result=mysql_query("SELECT * FROM tbl_tourism_type order by type_name ");
    while($row = mysql_fetch_array($result)){
    if($i == 5){
      echo "<br>";
      $i = 1;
    }
    else
    echo '<input type="checkbox" name="type" value='.$row['type_id'].'>'
            .'<label>'.$row['type_name']. '</label>';
    $i++;
}
?>

answered Jan 25, 2013 at 6:02

sourcecode

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Comments

arkascha · Accepted Answer · 2013-01-25 06:07:06Z

You question is how to skip every 5th entry and to put a linebreak instead?

Have a try like this:

<?php 
    $i=0;
    $result=mysql_query("SELECT * FROM tbl_tourism_type order by type_name ");
    while($row = mysql_fetch_array($result)) 
    {
        if (++$i % 5)
            echo sprintf ( '<input type="checkbox" name="type" value="%s">'
                          .'<label>%s</label>'."\n", 
                           htmlentities($row['type_id']), 
                           htmlentities($row['type_name']) );
        else
            echo "<br>\n";
    }
?>

In addition to iterating over all entries in the result just use a counter you increment with every iteration. Then, inside the iteration, only output the current entry if the counter modulo 5 is not zero. So if it is not a clean multiple of 5.

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