php Cannot call the function for array_map

Question

I am using php 5.2.10 i want to do the array_map on the array and i created a function for array mapping

function get_result(){
    $result = mysql_query("Select * from table");
    while($cr = mysql_fetch_array($result)){
        $b = array_map(`calc`,$cr);
        $rr_id = $cr['batch_id'].$cr['seq_id'];
  $mqrrid = '999'.$rr_id;
  $question_id = $cr['question_id'];
        foreach ($b as $k => $v){
            if(preg_match('{^Item \d+$}',$k)){
                $new_insert[] = array(
                    'r_id'=>$mqrrid,
                    'q_id' =>$q_id,
                    'c_id' =>$k,
                    'rank'=>$v
                );
            }
        }
    }
}   

function calc($n){
    foreach($n as $m=> &$x) {
        if (preg_match('{^Item \d+$}', $m)) {
            if($x == null){
                $x = $x;
            }else {
                $x = $x - 1;
            }
        }
    }   
    return $n;
}

I don't know why I cannot call the function calc in array_map.....I cannot figure out the reason..... Can anyone help me ?

original array :( actually the output after the array_map(calc,$cr) are same as follow)

array(23) {
["batch_id"]=>
string(1) "1"
["seq_id"]=>
string(1) "1"
["question_id"]=>
string(4) "2086"
["Item 1"]=>
string(1) "1"
["Item 2"]=>
string(1) "2"
["Item 3"]=>
string(1) "3"
["Item 4"]=>
string(1) "4"
["Item 5"]=>
string(1) "5"
["Item 6"]=>
NULL

what i need is : (minus the value of Item 1 to 6 by 1, if its null just leave it ~)

array(23) {
["batch_id"]=>
string(1) "1"
["seq_id"]=>
string(1) "1"
["q_id"]=>
string(4) "2086"
["Item 1"]=>
string(1) "0"
["Item 2"]=>
string(1) "1"
["Item 3"]=>
string(1) "2"
["Item 4"]=>
string(1) "3"
["Item 5"]=>
string(1) "4"
["Item 6"]=>
NULL

Finally, the result will become like this:(example of Item 1 and Item 6)

 array(4) {
["r_id"]=>
string(5) "99911"
["q_id"]=>
string(4) "2086"
["c_id"]=>
string(6) "Item 1"
["rank"]=>
string(1) "0"
}
array(4) {
["r_id"]=>
string(5) "99916"
["q_id"]=>
string(4) "2086"
["c_id"]=>
string(6) "Item 6"
["rank"]=>
string(4) NULL
}

In calc() you're returning $n o_o ? Shouldn't you return $x? By the way this $x = $x; does not make sense ... — HamZa
– HamZa, Commented May 14, 2013 at 9:15
@HamZaDzCyberDeV I want it return null , if $x is null , how to express it ? — user2210819
– user2210819, Commented May 14, 2013 at 9:19
Backticks are not valid, use quotes: array_map('calc', ...)! — deceze
– deceze ♦, Commented May 14, 2013 at 9:30

Willem Mulder · Accepted Answer · 2013-05-14 09:44:20Z

3

calc should be global, otherwise it cannot be found. Also, you should pass a string (no ` but rather enclose in ' or ").

Additionally, in general (if you used PHP 5.3), it is better to pass a function reference to the array_map function, instead of a string:

$func = function calc() { ... }
array_map($func, $cr);

edited May 14, 2013 at 9:44

answered May 14, 2013 at 9:26

Willem Mulder

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4 Comments

mpyw Over a year ago

See this: I am using php 5.2.10. Closure is supported on PHP 5.3 or later.

user2210819 Over a year ago

@Willem Mulder thank you for your response but i am using php 5.2.10

Willem Mulder Over a year ago

Yeah, so ensure that calc is global and that you use no backticks but proper quotes. Does that work?

user2210819 Over a year ago

show nothing when using proper quotes, but as what @CertaiN said may be i did have to prepare the function for array_map. If i prepared and use proper quotes, it may work. but it takes time for me to modify it.

mpyw · Accepted Answer · 2013-05-14 09:54:53Z

1

I think you don't have to prepare the function for array_map.

function get_result($link_identifier = NULL) {
    $result = mysql_query('Select * from table', $link_identifier);
    $new = array();
    while ($rows = mysql_fetch_assoc($result)) {
        $r_id = '999' . $rows['batch_id'] . $rows['seq_id'];
        foreach ($rows as $k => $v) {
            if ($v !== null && preg_match('@^Item \\d+$@', $k)) {
                $v = (string)((int)$v + 1);
            }
            $new[] = array(
                'r_id' => $r_id,
                'q_id' => $rows['question_id'],
                'c_id' => $k,
                'rank' => $v,
            );
        }
    }
    return $new;
}

edited May 14, 2013 at 9:54

answered May 14, 2013 at 9:37

mpyw

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8 Comments

user2210819 Over a year ago

It takes time to modify i will let you know if its work or not !

user2210819 Over a year ago

@^Item \\d+$@ is different from {^Item \d+$}?

mpyw Over a year ago

As a delimiter, { } and @ @ have same meanings. I prefer @ @ to { } because { and } have special meanings . [PHP Manual] php.net/manual/en/regexp.reference.repetition.php About \\d and \d, see these links: [PHP] ideone.com/FWzLF0 [C] ideone.com/qpSP2T Note: Most major languages behaves like C. Only PHP can behave oddly .

user2210819 Over a year ago

great !!!! thank you!!!!!! It works !! and I want to ask you one more question , what is the $link identifier used for?

user2210819 Over a year ago

Thank you very much. You gave me a detail explaination that help me alot. You english is good ! btw, I am not a english speaker also :P. I am chinese xp

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