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
– Vignesh Pichamani, Commented Dec 11, 2014 at 6:28

Zudwa · Accepted Answer · 2014-12-11 06:42:51Z

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, 2014 at 6:42

Zudwa

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KPK Over a year ago

Thanks for the explanation. This put me on the right path.

user3510665 · Accepted Answer · 2014-12-11 06:45:53Z

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, 2014 at 6:45

user3510665

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Comments

RST · Accepted Answer · 2014-12-11 08:16:26Z

//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);

monirz · Accepted Answer · 2014-12-11 06:35:26Z

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, 2014 at 6:35

monirz

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1 Comment

monirz Over a year ago

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);

Andrey Korneyev · Accepted Answer · 2014-12-11 06:53:53Z

0

Try this:

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

edited Dec 11, 2014 at 6:53

Andrey Korneyev

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answered Dec 11, 2014 at 6:33

Mysteryos

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1 Comment

KPK Over a year ago

Tried that with no success

KPK · Accepted Answer · 2014-12-11 07:18:30Z

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";
}

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