php multidimensional array if loop

Question

I have a multidimensional array like this

$array['value'][1][1]

Now i would like to implement if loop like this

if ($value = $array['value'][1][1]) {
echo "It works";
}

Now it works if i assign the values like [1][1],[2][1].

Is it possible to compare the whole array.

I mean if the array looks like

array[value][1][1],array[value][2][1],..........,array[value][n][1]

It works should be echoed.

I tried like this.

if ($value = $array['value'][][]) {
echo "It works";
}

But its not working. Can anyone give me the correct syntax?

what's the problem with using a for loop iteratively through the array? — cjds
– cjds, Commented Dec 14, 2012 at 17:03
can you give a little more information on what you're trying to accomplish? i'd love to help, but i'm not quite understanding the scope of what you're trying to accomplish. are you trying to ensure the array has the proper number of dimensions, or trying to iterate over each entry in each array and if present, print "It Works"? — Joshua Burns
– Joshua Burns, Commented Dec 14, 2012 at 17:07
@JoshuaBurns I have some text fields. Many of them can be duplicated by clicking 'add another' button. There is a field called amazon product. Near that form field i would like to display a link button to search amazon products. This link should be displayed only when the array looks like forms[amazon][*][1]. I hope you get what i'm trying to do. I tried the for loop solution. But it works only for first field. — Giri
– Giri, Commented Dec 14, 2012 at 17:26

cjds · Accepted Answer · 2012-12-14 18:00:14Z

2

I'm not sure if this is what you're looking for but you could try this function $value is 1 in your case

function($array,$value)
    foreach($array['amazon'] as $val){
        if($value != $val[1])return FALSE;
    }
    return TRUE;
}

The function runs through $array['amazon'][*] and checks condition for each value. If found FALSE for any it returns

edited Dec 14, 2012 at 18:00

answered Dec 14, 2012 at 17:06

cjds

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1 Comment

Giri Over a year ago

Hey carl, Thanks for the answer. I have some text fields. Many of them can be duplicated by clicking 'add another' button. There is a field called amazon product. Near that form field i would like to display a link button to search amazon products. This link should be displayed only when the array looks like forms[amazon][*][1]. I hope you get what i'm trying to do. I tried the for loop solution. But it works only for first field.

Joshua Burns · Accepted Answer · 2012-12-14 17:51:41Z

1

Based on your comments of what you're trying to accomplish, I think the following may solve your dilemma. Let me know if I'm going the wrong direction with this or have any additional questions/concerns.

<?php
$forms = array(
    'amazon' => array(
        0 => array(
            1 => 111,
        ),
        1 => array(
            1 => 222,
        ),
        2 => array(
            1 => 333
        )
    )
);

$value = 333;

$it_works = False;
foreach($forms['amazon'] as $array) {
    if($array[1] == $value) {
        $it_works = True;
        break;
    }
}

if($it_works === True) {
    print 'It Works!';
}
?>

edited Dec 14, 2012 at 17:51

answered Dec 14, 2012 at 17:31

Joshua Burns

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6 Comments

Giri Over a year ago

Hello mate, I'm not sure how to use for each loop here. Actually I have a function display_field($field); I should pass the value to display the input field. So actually when passing the value like form[amazon][1][1] it displays the field. Is it possible to use regex? I mean like this form[amazon][regex goes here][1]

Joshua Burns Over a year ago

if i'm understanding correctly, you want to see if any of the arrays contain an array with a key "1". if this is the case, you can't use regular expressions to access multiple keys at the same time. hence the foreach to check each value.

Giri Over a year ago

No. Actually I would like to use regex to check value like 1,2,3,..etc.

Giri Over a year ago

If ($value = form[amazon][regex matches number][1]) { echo "link";}

Giri Over a year ago

Here is the full function code. I need to display the button after line 72

|

bob dope · Accepted Answer · 2012-12-14 17:07:53Z

0

You can try this:

$itWorks = true;

for( $a = 0; $a < sizeof( $array[value] ); $a ++ ){
    if ( $value != $array[value][$a][1] ){
        $itWorks = false;
        break;
    }
}

if( $itWorks ){
    echo "It works";
}

Of course you must replace [value] with a legitimate value

answered Dec 14, 2012 at 17:07

bob dope

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Comments

user0103 · Accepted Answer · 2012-12-14 17:14:35Z

0

function check() {
    for($i = 0; $i < count($array[$value]); $i++) {
        if($value != $array[$value][$i][1])
            return FALSE;
    }
    echo 'It works!';
}

edited Dec 14, 2012 at 17:14

answered Dec 14, 2012 at 17:05

user0103

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1 Comment

cjds Over a year ago

return to what? Set the false to a variable

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