How to save result of a query in variable? - php

Question

I'm trying to save result of a MYSQL query in variable. I know that results save as an array but I don't know how to assign result to variable.
I could echo result but I couldn't assign it to variable.
here is my code with error:
require('db.php');
$id =$_REQUEST['id'];
$name ="SELECT name FROM table2 WHERE id='".$id."'";
$name_result = $con->query($name);
while($row = $name_result->fetch_assoc()) {
$name=$row['name'];
}
$ins_query=" insert into table1 (`id`,`name`) values ('$id','$name')";
$stat=mysqli_query($con,$ins_query) or die(" error".mysql_error());
Id gets from user as html input field and will save in table1.
Name exists in table2 and I want to select it where id in table1 is equals with id in table2 then insert it in table1.

Answer 1

$name ="SELECT name FROM table2 WHERE id='$id'";
As you said that we can store variables as arrays
$array = array();
while($row = mysql_fetch_object($name))
{
$array[] = $row;
}
So after that $array[] will have the value that you want but if you want to have single variable instead of an array you can convert it by using equivalent. I hope that this one helps you.

Answer 2

Try this code:
require('db.php');
$id = $_REQUEST['id'];
//modification
$query = $this->db->prepare("SELECT name FROM table2 WHERE id=?");
$query->bind_param('s', $id);
$query->execute();
//assuming you are retrieving only 1 value
$result = $query->get_result->fetch_assoc();
if ($result) {
//value stored...use as you wish!
$name = $result["name"];
} //you can add your else
$this->db is your db connection