the right way of displaying an array in php

Question

So, I have this type of array

$fields = array('result'=>234, '1'=>1, '2'=>2, '3'=>4, ....'20'=>5);

and I have o display it in a table, but the 'result' key has to be the last column. Until now I've used this approach, but the problem arrives as soon as i need to add other keys besides 'results', to the end of the table.

foreach($fields as $key=>$value) {
  if(strcmp('key', 'result') != 0)
    echo "<td>$value</td>";
  }
echo "<td>$fields['result']</td>";

I assume that is has to be a better approach. What do you think?

Jan-Henk · Accepted Answer · 2011-10-26 10:25:35Z

3

I assume you know the names of the keys you want to display at the end, so you can do something like the following:

$result = $fields['result'];
$otherResult = $fields ['other_result'];
unset($fields['result'], $fields['other_result']);

foreach($fields as $key=>$value) {
    echo "<td>$value</td>";
}

echo "<td>$result</td>";
echo "<td>$otherResult</td>";

This is only useful to do for a small number of keys.

answered Oct 26, 2011 at 10:25

Jan-Henk

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PeeHaa · Accepted Answer · 2011-10-26 10:29:48Z

2

// array of columns you want to add at the end of the table in sequence
// I used an array of columns and an exrta foreach loop to be able to support possible future more columns to be added
$last_cols = array('result'=>null, 'other'=>null);

foreach($fields as $key=>$value) {
    // check if current column is one of the cols to be added at the end
    if (in_array($key, array_keys($last_cols))) {
        $last_cols[$key] = $value;
        continue;
    }

    echo '<td>'.$fields['result'].'</td>';
}
// loop through last columns
foreach($last_cols as $col) {
    echo '<td>'.$col.'</td>'
}

edited Oct 26, 2011 at 10:29

answered Oct 26, 2011 at 10:24

PeeHaa

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Comments

hsz · Accepted Answer · 2011-10-26 10:31:50Z

1

There is faster way:

$result = $fields['result'];
$otherResult = $fields['otherResult'];

unset($fields['result'], $fields['otherResult']);
foreach($fields as $key=>$value) {
  echo '<td>' . $value . '</td>';
}

echo '<td>' . $result . '</td>';
echo '<td>' . $otherResult . '</td>';

You don't check key every time.

However if you really want - do it in following way:

if( $key == 'result' )

edited Oct 26, 2011 at 10:31

answered Oct 26, 2011 at 10:23

hsz

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2 Comments

PeeHaa Over a year ago

Where is the extra key OP requested at the end of the table?

hsz Over a year ago

Huh - I've noticed that @Jan-Henk has an identical answer. o0

FloydThreepwood · Accepted Answer · 2011-10-26 10:25:46Z

1

In this case your Datastructure is not well formed. Why dont you use

$result = 123;

? Or a simple object containing $result and $values like so:

class MyDataContainer {
   public $result = 0;
   public $values = array();
}

? I would recommend this, since it would be a lot cleaner.

answered Oct 26, 2011 at 10:25

FloydThreepwood

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1 Comment

FloydThreepwood Over a year ago

If you have completely different types of values, they should be structured correctly. since array(1,2,3,4) is a list of numbers and array('result' => 999) is a single scalar result they should not be mixed in one structure in the first place.

hakre · Accepted Answer · 2011-10-26 11:08:55Z

0

Make the first element the last element in an array (as you don't use the keys in your question):

$fields = array('result'=>234, '1'=>1, '2'=>2, '3'=>4, ....'20'=>5);

$fields[] = array_shift($fields);

Demo

answered Oct 26, 2011 at 11:08

hakre

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