Array Undefined index error (notice) in PHP

Question

I have this function:

function coin_matrix($test, $revs) {
    $coin = array();

    for ($i = 0; $i < count($test); $i++) {
        foreach ($revs as $j => $rev) {
            foreach ($revs as $k => $rev) {
            if ($j != $k && 
                $test[$i][$j] != null && 
                $test[$i][$k] != null) {

                $coin[$test[$i][$j]][$test[$i][$k]] += 1 / ($some_var - 1);
                }
            }
        }
    }
    return $coin;
}

where

$test = array(
array('3'=>'1','5'=>'1'),
array('3'=>'2','5'=>'2'),
array('3'=>'1','5'=>'2'),
array('3'=>'1','5'=>'1'));

and

$revs = array('3'=>'A','5'=>'B');

the problem is that when I run it, it returns these errors (notices):

Notice: Undefined index: 1 at line 10

Notice: Undefined index: 2 at line 10

Notice: Undefined index: 1 at line 10

which is this line: $coin[$test[$i][$j]][$test[$i][$k]] += 1 / ($some_var - 1);

The problem is that at the end the function returns the correct matrix (array) and if I test to see if $coin[$test[$i][$j]][$test[$i][$k]] exists, then it doesn't return it anymore.

Any suggestion is greatly appreciated!

Thanks!

Powerlord · Accepted Answer · 2010-04-20 15:58:21Z

5

$coin[$test[$i][$j]][$test[$i][$k]] += 1 / ($some_var - 1);

The warning is being generated by the +=. += needs to look up the element before adding to it, and you haven't initialized any of the elements in $coin the first time you access them.

answered Apr 20, 2010 at 15:58

Powerlord

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Comments

Ro Yo Mi · Accepted Answer · 2013-07-06 05:02:50Z

3

You can/should test to make sure that $coin[$test[$i][$j]][$test[$i][$k]] is set before incrementing the value. This shouldn't change the functionality of your code, but will make the notices go away (which is good practice).

function coin_matrix($test, $revs) {
    $coin = array();

    for ($i = 0; $i < count($test); $i++) {
        foreach ($revs as $j => $rev) {
            foreach ($revs as $k => $rev) {
            if ($j != $k && 
                $test[$i][$j] != null && 
                $test[$i][$k] != null) {

                    // new tests go here
                    if(!isset(coin[$test[$i][$j]])) 
                    {
                        coin[$test[$i][$j]] = array(); 
                    }
                    if(!isset(coin[$test[$i][$j]][$test[$i][$k]])) 
                    {
                        coin[$test[$i][$j]][$test[$i][$k]] = 0; 
                    }

                    $coin[$test[$i][$j]][$test[$i][$k]] += 1 / ($some_var - 1);
                }
            }
        }
    }
    return $coin;
}

edited Jul 6, 2013 at 5:02

Ro Yo Mi

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answered Apr 20, 2010 at 15:52

Scott Saunders

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3 Comments

Alex Over a year ago

if i do that then the function doesn't work any more. i should have mentioned that at the end it returns the right matrix (array)

Scott Saunders Over a year ago

How does it break? Are you sure you only set the variable if it was not set?

Scott Saunders Over a year ago

I've extended the code so you can see the context. Glad to help.

coolnalu · Accepted Answer · 2010-04-20 15:59:05Z

I think the problem is you are trying to use $coin as a two dimensional array.

if you want it to be two dimensional, $coin has to be an array of arrays.

function coin_matrix($test, $revs) {

    $coin = array();

    for ($i = 0; $i < count($test); $i++) {
        foreach ($revs as $j => $rev) {
            foreach ($revs as $k => $rev) {
            if ($j != $k && 
                $test[$i][$j] != null && 
                $test[$i][$k] != null) {
                // add this.
                if ($coin[$test[$i][$j]] == null){
                    $coin[$test[$i][$j]] = array();
                }
                // end
                $coin[$test[$i][$j]][$test[$i][$k]] += 1 / ($some_var - 1);
                }
            }
        }
    }
    return $coin;
}

Ali Demirci · Accepted Answer · 2010-04-20 15:50:37Z

0

I don't understand much but I can suggest you to use

function coin_matrix($test, $revs) {
    $coin = array();

    for ($i = 0; $i < count($test); $i++) {
        foreach ($revs as $j => $rev) {
            foreach ($revs as $k => $rev) {
            if (($j != $k) && 
                ($test[$i][$j] != null) && 
                ($test[$i][$k] != null)) {

                $coin[$test[$i][$j]][$test[$i][$k]] += 1 / ($some_var - 1);
                }
            }
        }
    }
    return $coin;
}

answered Apr 20, 2010 at 15:50

Ali Demirci

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Comments

cwallenpoole · Accepted Answer · 2010-04-20 16:17:33Z

Have you thought about swapping out the for loop with a foreach loop? Eg:

foreach( $tests as $i => $test )

This has the benefit of having a value in $test[ $i ]. Then, in your $test[ $i ][ $j ] == null block, place this:

        if ($j != $k && 
            // I suspect that this will cause errors too.
            // Do yourself a favor and add this sanity check.
            isset( $test[$j] ) && $test[$j] != null && 
            isset( $test[$k] ) && $test[$k] != null) {

                $currentK = $test[$k];
                $currentJ = $test[$j];
                // Use debug lines if setting things directly won't work
                if( !isset( $coin[ $currentJ ] ) || !is_array( $coin[ $currentJ ] ) )
                {
                    // $coin[ $currentJ ] = array();
                    die( "<pre>$currentK $currentJ \n" .  print_r( $coin ) );
                }
                $currentCoin =& $coin[ $currentJ ];
                if( !isset( $currentCoin [ $currentK ] ) || 
                    !is_array( $currentCoin [ $currentK ] ) )
                {
                    // Just curious, but when doing these checks before,
                    // did you remember to assign a numeric value here?
                    // 
                    // $currentCoin[ $currentK ] = 0;
                    die( "<pre>$currentK $currentJ \n" .  print_r( $coin ) );
                }
                $currentCoin[ $currentK ] += 1 / ($some_var - 1);
            }
        }

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