Parsing Array And Object json Via PHP - Invalid argument supplied

Question

I'm Parsing This json Array and I Want to Take type Object and Put That in New Column type2, and This is one Row of My json Rows, Why I Get This Warning for Some Rows? Warning: Invalid argument supplied for foreach() in C:\wamp64\www\json\json.php on line 18

[{"id":"26","answer":[{"option":"3","text":"HIGH"}],"type":"3"},
{"id":"30","answer":[{"option":"3","text":"LOW"}],"type":"3"},
{"id":"31","answer":[{"option":"3","text":"LOW"}],"type":"3"}]

And This is My Code:

<?php
$con=mysqli_connect("localhost","root","","array");
mysqli_set_charset($con,"utf8");

// Check connection
if (mysqli_connect_errno()){
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

$sql="SELECT `survey_answers`,us_id FROM `user_survey_start`";
if ($result=mysqli_query($con,$sql)){
    while ($row = mysqli_fetch_row($result)){
        $json = $row[0];
        if(!is_null($json)){
            $jason_array = json_decode($json,true);
            // type2
            $type = array();
            foreach ($jason_array as $data) {
                if (array_key_exists('type', $data)) {
                    // Now we will only use it if it actually exists
                    $type[] = $data['type'];
                }
            }         
            // lets check first your $types variable has value or not?
            if(!empty($type)) {
             $types= implode(',',$type); /// implode yes if you got values
            } 
            else { 
                $types = ''; //blank if not have any values
            }
            $sql2="update user_survey_start set type2='$types' where us_id=".$row[1];//run update sql
            echo $sql2."<br>";
            mysqli_query($con,$sql2);
        }
    }
}
mysqli_close($con);
?>

That is Strang, Why Some Row Has Output And Some Rows Hasn't Any Output, Those Json Type Are Same. I Find The Problem, Because Some json entered, I Mean. This One Has Warning: Invalid argument supplied for foreach()

[{"id":"26","answer":[{"option":"4","text":"Hello
"}],"type":"3"}]

And This One is Okey

[{"id":"26","answer":[{"option":"4","text":"Hello"}],"type":"3"}]

How Can I Fix The Problem?

Dipali Sakle Systematix · Accepted Answer · 2017-02-22 08:53:06Z

1

you also try is_array before your for each loop

if (is_array($jason_array))
{
     foreach ($jason_array as $data) {
    {
        ...
    }
}

answered Feb 22, 2017 at 8:53

Dipali Sakle Systematix

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7 Comments

Saeed Heidarizarei Over a year ago

Yessss,Solved, Appriciate It.

Saeed Heidarizarei Over a year ago

I Haven't Any Error But in Output I Haven't Value For Those Rows Type Too, I Mean , I Have update user_survey_start set type2='' where us_id=267593

Dipali Sakle Systematix Over a year ago

i think your variable $jason_array is not a array that'why not come any output

Saeed Heidarizarei Over a year ago

I Don't Know Why, Because My Json Type is Like Above, Two Same Row but Defference Value, One Retrun Output and Other One Have Invalid argument supplied

Dipali Sakle Systematix Over a year ago

print your varibale $jason_array first then check what is return

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