UPDATE at bottom of question
I'm getting the error:
Warning: mysqli_query() expects parameter 2 to be string, object given
Questions about this are incredibly common on Stack Overflow - my apologies in advance. I haven't been able to find a good answer for my specific problem. If there is a thread that addresses this, please let me know.
Here is my Ajax code:
$.ajax({
url: "get.php",
type: "post",
datatype: "json",
data:{ ajaxid: altCheck }, //this is an integer passed to MySQL statement
success: function(response){
console.log(response);
},
error: function(){
console.log("test");
}
});
get.php
<?php
$db = mysqli_connect("...", "...", "...", "...");
$value = filter_var($_REQUEST["ajaxid"], FILTER_SANITIZE_STRING);
$value = mysqli_real_escape_string($db, $value);
var_dump($value); //checking to see what $value is at this point
$sql = $db->prepare("SELECT * FROM table WHERE screeningId = ?");
$sql->bind_param("s",$value);
//THIS LINE THROWS THE ERROR
$result = mysqli_query($db, $sql);
$temp = array();
while ($row = mysqli_fetch_array($result)){
//output data
array_push($temp,$row['imageURL']);
}
echo json_encode($temp);
?>
The fourth line of code var_dump($value);
outputs string(0)
.
UPDATE: MySQLi
<?php
$db = mysqli_connect("...", "...", "...", "...");
$value = filter_var($_REQUEST["ajaxid"], FILTER_SANITIZE_STRING);
$value = mysqli_real_escape_string($db, $value);
$query = $db->prepare('SELECT * FROM table WHERE screeningId = ?');
$query->bind_param('s', $_GET[$value]);
$query->execute();
if ($result = mysqli_query($db, $query)) {
while ($url = mysqli_fetch_object($result, 'imageURL')) {
echo $url->info()."\n";
}
}
?>
Screenshot of MySQL table data columns: