retrieve mysql result to jquery - php

Question

My problem is that I can't retrieve the result of the mysql result via ajax, please help
ajax code:
$.ajax({
type: "POST",
url: "do_find_courses.php",
//data:{question_id:question_id,answer:answer},
data:{user_id:user_id}, dataType:'json',
success:function(msg) {
alert ('asdasd')
// $("#quiz_form,#demo1").addClass("hide");
// $('#result').show();
$('p').html(msg);
}
});
PHP code:
$final=array();
$sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
$row_courses = mysql_fetch_array($sql_courses);
$result=$row_courses['course_name'];
//array_push($final,$result);
//print_r($result);
echo json_encode($result);

Answer 1

Change PHP Code as below
$final = array();
$sql_courses = mysql_query("SELECT course_id, course_name FROM course") or die(mysql_error());
$row_courses = mysql_fetch_array($sql_courses);
echo json_encode($row_courses);
change php code as below:
$.ajax({
type: "POST",
url: "do_find_courses.php",
//data:{question_id:question_id,answer:answer},
data: {
user_id: user_id
},
dataType: 'json',
success: function (msg) {
$('p').html(msg.course_name);
}
});

Answer 2

Its better to send it with a key value like this:
And its better to use console.log(variable); to check variable's content
ajax code:
$.ajax({
type: "POST",
url: "do_find_courses.php",
//data:{question_id:question_id,answer:answer},
data:{user_id:user_id}, dataType:'json',
success:function(msg) {
alert ('asdasd');
console.log(msg);//You should check output of this in browser console
// $("#quiz_form,#demo1").addClass("hide");
// $('#result').show();
$('p').html(msg.cname);
}
});
PHP code:
$final=array();
$sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
$row_courses = mysql_fetch_array($sql_courses);
$result=$row_courses['course_name']; // this will have first_coures_name (an string)
$final['cname']=$result;
//print_r($result);
echo json_encode($final); //the output should be this {'cname':'first_course_name'}