php json数组中的未定义错误

Question

I am getting undefined error in this code. I tried isset but it is becoming clumsy..is there any sweet way to handle this piece of code

This works well in php 5.6.30 but throwing errors in php 5.6.36

foreach($response as $key => $value){
$exp_table .= "<tr>";
foreach($fields_arr as $keys => $values){

$exp_table .= "<td>";
if($fields_arr[$keys]["json_subcategory"]==""){
$exp_table .=  $response[$key][$fields_arr[$keys]["api_subcategory"]][$fields_arr[$keys]["field"]];
} 
if($fields_arr[$keys]["api_subcategory"]==""){
$exp_table .=  $response [$key][$fields_arr[$keys]["json_subcategory"]]$fields_arr[$keys]["field"]];
}
if(($fields_arr[$keys]["json_subcategory"]=="") && ($fields_arr[$keys]["api_subcategory"]=="")){
$exp_table .=  $response [$key][$fields_arr[$keys]["field"]];
}

if(($fields_arr[$keys]["json_subcategory"]!="") && ($fields_arr[$keys]     ["api_subcategory"]!="")){
$exp_table .=  $response [$key][$fields_arr[$keys]["json_subcategory"]][$fields_arr[$keys]["api_subcategory"]][$fields_arr[$keys]["field"]];
}

Answer 1

isset will return false on null values. You may try array_key_exists instead

function checkFields($array, $key)
    {
    if (array_key_exists($key, $array))
        {
        if (is_null($array[$key]))
            return "isNull";
          else
            return "isSet";
        }
    }

foreach($response as $key => $value)
    {
    $exp_table.= "<tr>";
    foreach($fields_arr as $keys => $values)
        {
        $exp_table.= "<td>";
        if (checkFields($fields_arr, $fields_arr[$keys]["json_subcategory"]) == "isNull")
            {
            $exp_table.= $response[$key][$fields_arr[$keys]["api_subcategory"]][$fields_arr[$keys]["field"]];
            }else{
                //is not SET. else part is use only for testing(optional).
                }
        }
    }