PHP, resource id# instead of actual string - php

Question

I have run across a problem during my query service to add a row in an online database in PHP. The addition of the row works just fine. I get user id and book id from the url and fetch the names of the book and the user to put into the row which i add to my third and last table.
When I get the names, put them in an array, json encode it and then echo it, it works. But when I put them in the row it prints resource id#3 and resource id#4 instead of the names.
Any ideas?
Here is my service:
<?php
$con = mysql_connect("localhost","root","root");
$userid=$_GET['uid'];
$id = $_GET['bookid'];
$type = $_GET['type'];
$zero = '0';
$one = '1';
$date = date("Y-m-d");
$arr = array();
if (!$con)
{
die('Could not connect: ' . mysql_error());
echo "error connection";
}
mysql_select_db("Jineel_lib",$con) or die("Could not select database");
$bkName = mysql_query("SELECT Name from books where ID='".$id."'");
$userName = mysql_query("SELECT Name from people WHERE User_ID='".$userid."'");
while($obj = mysql_fetch_object($userName))
{
$arr[] = $obj;
}
echo json_encode($arr);
if($type == 'borrow')
{
$query="UPDATE books set Availablity = '".$zero."' where ID= '".$id."' ";
mysql_query($query) or die (" borrow operation failed due to query 1");
$query1="INSERT into borrowed (BookID, BookName, BorrowerID, BorrowedName, DateBorrowed, Extended, Returned) values('".$id."','".$bkName."','".$userid."','".$userName."','".$date."','".$zer‌​o."','".$zero."')";
mysql_query($query1) or die (" borrow operation failed to due query 2");
echo "borrow success";
}
else if($type=='return')
{
$query="UPDATE books set Availablity = '".$one."' where ID= '".$id."' ";
mysql_query($query) or die (" return operation failed");
$query1="UPDATE borrowed set Returned = '".$one."' where BookID= '".$id."' ";
mysql_query($query1) or die (" return operation failed 1");
echo "return success";
}
else
echo "invalid parameters";
?>
THANK YOU IN ADVANCE

Answer 1

You don't actually retrieve the userName value here:
$userName = mysql_query("SELECT Name...
$userName is just the result resource object returned from the query. You do use mysql_fetch_object later on, which is appropriate, but then you try to use the actual result resource in your insert query:
$query1="INSERT into borrowed ...
It gets converted to the string you see. Instead, you need to use $obj->Name (you fetch the result into $obj, and presumably there is only one result). If there is more than one possible result, you will have to do that in a loop.
Listen to all of the comments on your question.