Parsing Array And Object json Via PHP - Invalid argument supplied - php

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?

Answer 1

you also try is_array before your for each loop
if (is_array($jason_array))
{
foreach ($jason_array as $data) {
{
...
}
}