How to create dynamic table using the results - php

Question

$.ajax({
type: "post",
url: "<?php echo site_url(); ?>/controller_d/login/admin_search_user",
cache: false,
data: $('#docContainer1').serialize(),
dataType:"JSON", //<----here
success: function(json){
var str= "<table><tr>";
$.each(json.query,function(i,v){
alert(v.uID); //gives U0016
alert(v.name); //gives saman
str+="<td>"+v.uID+"</td>";
str+="<td>"+v.name+"</td>";
})
str+="</tr></table">;
$("body").append(str);
}
I want to create a dynamic table using json object values it must create a table but this is not working it will says v.ID not define

Answer 1

row += "$('<td>').append("+dataString[i]+")";
or
var row = $('<tr>');
for(var i = 0; i < n; i++) {
row.append($('<td>').html(dataString[i]));
}
$('#results').append(row);

Answer 2

create a table on the layout (on html code) then inside this create a <tbody> tag. assign an ID, then on the success function of your jquery, use $("#*tbody ID*").html to assign the value on the tbody.
i've got this code on one of my projects. maybe you can use this as a reference (tblApproved is the id of the tbody):
function updateApprovedTable(){
// retrieve Unit Record
$.ajax({
type:'post',
url:'php/requests.php',
data:{mode:"getApproved"},
success:function(data){
$('#tblApproved').html(data);
}//success
});//ajax
}

Answer 3

can you please used this :
$.ajax({
type: "post",
url: "<?php echo site_url(); ?>/controller_d/login/admin_search_user",
cache: false,
data: $('#docContainer1').serialize(),
success: function(data){
$("body").append(data);
});
Also write your code for table create into this url "/controller_d/login/admin_search_user"
PHP Code
echo "<table>";
echo "<tr><td>content</td></tr>";
echo "<tr><td>content</td></tr>";
echo "<tr><td>content</td></tr>";
echo "</table>";