Dumping array into table

Question

I have an array which I would like to dump into a table, but have a new column X number of items. For example:

item 1 | item 2 | item 3 | item 4 |

item 5 | item 6 | item 7 | item 8 |

item 9 | and so on...

I can already make it add a new column after a certain number of items (in this case, 4) with this code:

$input = array('one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten'); 

$cols = 4; 

echo "<table border=\"5\" cellpadding=\"10\">"; 

for ($i=0; $i < count($input); $i++) 
{ 
echo "<tr>"; 
    for ($c=0; $c<$cols; $c++) 
    {
    $n = $i+$c;
    echo "<td>$input[$n]</td>"; 
    } 
echo "</tr>"; 
$i += $c; 
} 

echo "</table>";

But for some reason, after one column ends with 'four', the next starts with 'six'.

It's because you're resetting $i within the loop, and then the for() itself will do $i++. Yuu should do either for($i...; $i += $c) at the top or $i += $c - 1; at the bottom — Marc B
– Marc B, Commented May 24, 2011 at 20:32

Preston · Accepted Answer · 2011-05-24 20:39:10Z

3

Array functions can be quite magical:

$input = array('one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten'); 
$cols = 4; 
$table = array_chunk($input, $cols);

echo "<table border=\"5\" cellpadding=\"10\">";

foreach($table as $row) {
    echo "<tr>"; 
    foreach($row as $cell) {
        echo "<td>$cell</td>"; 
    }
    echo "</tr>"; 
}
echo "</table>";

ref: http://php.net/manual/en/function.array-chunk.php

edited May 24, 2011 at 20:39

Preston

1,0593 gold badges12 silver badges22 bronze badges

answered May 24, 2011 at 20:28

Kyle

5742 silver badges8 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

SIFE Over a year ago

I think $rows, $row, $cell will cause undefined variable message error.

Preston Over a year ago

@SIFE I think he meant $table instead of $rows. I changed it and it works fine now.

Kyle Over a year ago

@Preston Yeah, I decided I didn't like $rows, but didn't do so thoroughly, thx for the edit.

Luis Estrada · Accepted Answer · 2011-05-24 20:30:07Z

1

On your first loop $i will be incremented by $c (which will always be 3). Then the for loop will increase the $i value by one ($i++) which will make it skip the 'five'.

You either control the increment or let the for loop control it for you.

answered May 24, 2011 at 20:30

Luis Estrada

1131 silver badge5 bronze badges

Comments

Lotus Notes · Accepted Answer · 2011-05-24 20:29:10Z

0

$input = array('one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten'); 

$cols = 4; 

echo "<table border=\"5\" cellpadding=\"10\">"; 

for ($i=0; $i < count($input); $i++) 
{ 
echo "<tr>"; 
    for ($c=0; $c<$cols; $c++) 
    {
    $n = $i+$c;
    echo "<td>$input[$n]</td>"; 
    } 
echo "</tr>"; 
$i += $c - 1; 
} 

echo "</table>";

Kai's answer is a much better solution. All I did was subtract 1 from the line $i += $c.

answered May 24, 2011 at 20:29

Lotus Notes

6,3837 gold badges35 silver badges49 bronze badges

Comments

SamT · Accepted Answer · 2011-05-24 20:29:54Z

0

You're skipping every 5th because you are incrementing your counter an extra time with the for loop when you detect that you need to do a new row.

Here is another appraoch:

$ary = array('item1', 'item2', 'item3', 'item4', 'item5', 'item6', 'item7', 'item8');


echo "<table><tr>";
foreach($ary as $k => $item)
{
    if($k % 4 == 0 && $k != 0)
        echo "</tr><tr>";
    echo "<td>$item</td>";
}
echo "</tr></table>";

answered May 24, 2011 at 20:29

SamT

10.7k2 gold badges33 silver badges39 bronze badges

Comments

SIFE · Accepted Answer · 2011-05-24 20:30:26Z

0

If understand you clearly, you want to dump an array in number of rows specified by you:

for ($i=0; $i < count($input); $i++) 
{ 
echo "<tr>"; 
    for ($c=0; $c<$cols; $c++) 
    {
    echo "<td>$input[$i]</td>"; 
    } 
echo "</tr>"; 
}

answered May 24, 2011 at 20:30

SIFE

5,7537 gold badges34 silver badges48 bronze badges

Comments

Ryan · Accepted Answer · 2011-05-24 20:30:52Z

0

If you'd like to keep your original logic, change the increment of $i to $i += (c$-1) since you are incrementing $i in the loop in addition to your column width. Then also cater for empty values. This will work:

<?php
$input = array('one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine',     'ten'); 

$cols = 4; 

echo "<table border=\"5\" cellpadding=\"10\">" . PHP_EOL; 

for ($i=0; $i < count($input); $i++) 
{ 
echo "<tr>"; 
    for ($c=0; $c<$cols; $c++) 
    {
    $n = $i+$c;
    if ( $n < count($input))
        echo "<td>$input[$n]</td>" . PHP_EOL; 
    else
        echo "<td></td>" . PHP_EOL;
    } 
echo "</tr>"; 
$i += ($c-1); 
} 

echo "</table>"; 
?>

answered May 24, 2011 at 20:30

Ryan

28.4k10 gold badges58 silver badges83 bronze badges

Collectives™ on Stack Overflow

Dumping array into table

6 Answers 6

3 Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

3 Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related