PHP variable variables getting array value

Question

I'm building a condition check where I'm passing in the variable and a value to check. In this case the variable is an array value, but I can't get it to return it correctly

//happens somewhere else
$specialFeature['option1']="on";
$specialFeature['option2']="on";
$specialFeature['option3']="off";

//what I'm trying to do
#query a db
$row = $result->fetch_array()
#results for purpose of demo
#$row['var'] = "specialFeature['option2']";
#$row['val'] = "on";
if($$row['var'] == $row['val']){
    //what i'm expecting
    echo "OK";
}

My issue is $$var is always null. What am I doing wrong? Is this possible?

why not you try like this simply call as $var = "$ARRAY['myKey3']"; — Vignesh Pichamani, Dec 11 '14 at 06:28

RST · Answer 1 · 2014-12-11T08:16:26.727

//happens somewhere else
$ARRAY['myKey1']=1;
$ARRAY['myKey2']=2;
$ARRAY['myKey3']=3;

//what I'm trying to do
$var = "ARRAY['myKey3']";
$val = 3;
if(${$var} == $val){
    //what i'm expecting
    echo "OK";
}

From PHP.net: In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.

Edit: The array-index seems to be the problem.

To split the variable name you could use something like

preg_match('/(.*)\[\'(.*)\'\]/', $row['var'], $matches);

score 1 · Answer 2 · answered Dec 11 '14 at 06:42

1

With $$var you get following: ${"ARRAY['myKey3']"}, so it is treated as a variable with name ARRAY['myKey3'] (Which doesn't exist, though you can create it with $$var = 'new value', but it will differ from $ARRAY['myKey3'] as they will be 2 different variables). Probably you are looking for if ($ARRAY['myKey3'] == $val) ?

answered Dec 11 '14 at 06:42

Zudwa

1,360
10
13

Thanks for the explanation. This put me on the right path. – KPK Dec 11 '14 at 07:21

score 1 · Answer 3 · answered Dec 11 '14 at 06:45

1

$ARRAY['myKey1']=1;
$ARRAY['myKey2']=2;
$ARRAY['myKey3']=3;

//what I'm trying to do
$var = "ARRAY";//pass variable name or array name
$val = 3;
if((${$var}['myKey3']) == $val){
    //what i'm expecting
    echo "OK";
}

answered Dec 11 '14 at 06:45

user3510665

218
3
13

score 0 · Answer 4 · edited Dec 11 '14 at 06:53

0

Try this:

if(${$var} == $val){
    //what i'm expecting
    echo "OK";
}

edited Dec 11 '14 at 06:53

Andrey Korneyev

26,353
15
70
71

answered Dec 11 '14 at 06:33

Mysteryos

5,581
2
32
52

Tried that with no success – KPK Dec 11 '14 at 06:57

score 0 · Answer 5 · answered Dec 11 '14 at 06:35

0

Is that what you are looking for?

    $var = $ARRAY['myKey3'];
    $val = 3;
     if($var == $val){
     //what i'm expecting
     echo "OK";
           }

answered Dec 11 '14 at 06:35

monirz

534
5
14

And yes you are doing something wrong. You are assigning a value to the condition block, $$ for assigning a value that is equal to original variable but condition is just for checking the value its true or false, for in this instance $$var is NULL so the condition is not true that's why its returning nothing. Just see what happened with this code var_dump($$var); – monirz Dec 11 '14 at 06:47

score 0 · Accepted Answer · answered Dec 11 '14 at 07:18

I'm not sure if what I'm doing is "correct", but I got the result I was looking for by doing this:

//happens somewhere else
$specialFeature['option1']="on";
$specialFeature['option2']="on";
$specialFeature['option3']="off";

//what I'm trying to do
#query a db
$row = $result->fetch_array()
#results for purpose of demo
#$row['var'] = "specialFeature['option2']";
#$row['val'] = "on";

$var = $row['var'];

if (strpos($var,'[') !== false) {
    $varA = str_split($var,strpos($var,'['));
    $varA[1] = substr($varA[1],1,-1);
    if (strpos($varA[1],"'") !== false) {
        $varA[1] = substr($varA[1],1,-1);
    }
}

if(${$varA[0]}[$varA[1]] == $row['val']){
    //what i'm expecting
    echo "OK";
}

PHP variable variables getting array value

6 Answers6