I am not an expert of php, I developed a small service which query a mysql db.
However I developed with php 5.4, and then discovered that my web hosting plan has 5.2.6, so I am having few problems with some undefined function.
Specifically, in this case, how can I solve the mysqli_stmt_get_result undefined function available on > 5.3 ? Here is the code:
$stmt = mysqli_prepare($con,$db_query);
if($stmt) {
mysqli_stmt_bind_param($stmt,'ss',$after,$lang);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt); // <-- getting undefined error here !!!
$updated = array();
$deleted = array();
while($row = mysqli_fetch_assoc($result)) {
if($row['status']==1) {
array_push($updated,$row);
} else {
$cardName=$row['cardName'];
$cardStatus=$row['status'];
$cardId=$row['cardId'];
$language=$row['language'];
array_push($deleted,array(
'cardName'=>$cardName,
'status'=>$cardStatus,
'cardId'=>$cardId,
'language'=>$language
)
);
}
}
$response = array(
'cards'=>array(
'updated'=>$updated,
'deleted'=>$deleted
)
);
$json = json_encode($response);
mysqli_close($con);
echo $json;
}
The point is that I am using a prepared statement, due to my lack of php knowledge I found no other way of resolving the issue without rewriting the whole script.
I thought that some of you may have a simple and easy solution.