CODEWITHSUNDEEP
phparraysjson
user5443928
user5443928

Reputation:

Can not check if the json key value is not present using PHP

I need some help.I need to check weather the key value is present inside the json array or not using PHP but its throwing some warning message and no value is coming. I am explaining my code below.

$mainArr=array(array("type"=>1,"name"=>"hello"),array("type"=>1,"name"=>"hii"));
//echo json_encode($mainArr);
foreach ($mainArr as $v) {
    if(!in_array(1, $v['type'])){
        $result['primary'][]=array();
    }else{
        $result['primary'][]=$v;
    }
    if(!in_array(2, $v['type'])){
        $result['secondary'][]=array();
    }else{
        $result['secondary'][]=$v;
    }
}
echo json_encode($result);

Here I need to check if type==2 is not present inside that array it should return blank array but its throwing the below message.

Warning: in_array() expects parameter 2 to be array, integer given in /opt/lampp/htdocs/test/arrchk.php on line 5

Warning: in_array() expects parameter 2 to be array, integer given in /opt/lampp/htdocs/test/arrchk.php on line 10

Warning: in_array() expects parameter 2 to be array, integer given in /opt/lampp/htdocs/test/arrchk.php on line 5

Warning: in_array() expects parameter 2 to be array, integer given in /opt/lampp/htdocs/test/arrchk.php on line 10
{"primary":[[],[]],"secondary":[[],[]]}

Please help me to resolve this issue.

Upvotes: 0

Views: 113

Answers (4)

Bojan Radaković
Bojan Radaković

Reputation: 440

Use array_search() function. For example:

foreach ($mainArr as $v) {
    if (!array_search('1', $v)) {
        $result['primary'][]=array();
    } else {...}

Upvotes: 0

Kimberlee Ho
Kimberlee Ho

Reputation: 487

$mainArr=array(array("type"=>1,"name"=>"hello"),array("type"=>1,"name"=>"hii"));
//echo json_encode($mainArr);
foreach ($mainArr as $v => $values) {
    if(!in_array(1, $values['type'])){
        $result['primary'][]=array();
}else{
    $result['primary'][]=$values;
}
if(!in_array(2, $values['type'])){
    $result['secondary'][]=array();
}else{
    $result['secondary'][]=$values;
}
}
echo json_encode($result);

Upvotes: 0

mehfatitem
mehfatitem

Reputation: 147

You search element in an array element not an array.

<?php
$mainArr = array(
    array(
        "type" => 1,
        "name" => "hello"
    ),
    array(
        "type" => 1,
        "name" => "hii"
    )
);
//echo json_encode($mainArr);
foreach ($mainArr as $v) {
    if (!in_array(1, $v)) {
        $result['primary'][] = array();
    } else {
        $result['primary'][] = $v;
    }
    if (!in_array(2, $v)) {
        $result['secondary'][] = array();
    } else {
        $result['secondary'][] = $v;
    }
}
echo json_encode($result);

?>

Upvotes: 0

Mcsky
Mcsky

Reputation: 1445

You are using the PHP's in_array function on a scalar value.

$v['type'] is an integer, a simple comparaison like 

1 === $v['type']

will do the job

Upvotes: 1

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