MySQL Query with PHP and Send Data to JavaScript -

edit:

i have number input id on it. in mysql database, have table ids , list of options separated commas id.

i'll take 1 of database records example: id "ham_and_cheese" , options "mayonnaise, salad cream, chutney, no dressing". person selects 1 on number input id "ham_and_cheese".

then, 1 drop down appear options: "mayonnaise, salad cream, chutney, no dressing" (each own option). if choose 2, 2 dropdowns appear, 3 3 appear, etc.

i new ajax great if me out.

my new code:

...  <td><input type='number' min='0' max='5' name='ham_and_cheese_amount' /></td>  ...  <td id='ham_and_cheese_dressing'></td>  ...  function changedropdown(name, amount){      //gets id.     var newname = name.replace("_amount", "");      //gets div drop down go in.     var el2 = document.getelementbyid(newname + "_dressing");      //if number 0 selected in number input.     if(amount == 0){         el2.innerhtml = "";     }      //if number 1 selected in number input.     if(amount == 1){         var html = "<select>";         var id = newname;         $.ajax({             url: 'updatedropdown.php',             datatype: 'json',             data: {'option_id':id},             type: 'get',             success: function(r){                 for(i = 0; < r.length; i++){                     // use r[i].tolowercase().replace(" ", "_") convert "salad cream" id "salad_cream", etc.                     html += "<option value='" + r[i].tolowercase().replace(" ", "_") + "'>" + r[i] + "</option>";                 }                 html += "</select>";                 el2.innerhtml = html;             }         });     }  } 

my updatedropdown.php:

<?php      include "/home/pi/config.php";     $id = $_get['option_id'];     //i know connect works because use same code other queries.     $connect = mysqli_connect("localhost", $name, $pass, "items");     if(mysqli_connect_errno()){         return "failed connect products database!";     }     $result = mysqli_query($connect, "select options products id='$id' limit 1");     $resultarray = explode(", ", mysqli_fetch_row($result)[0]);      return json_encode($resultarray);  ?> 

ajax how information between back-end , front-end.

$.ajax({                                                                                                     url: "/url/of/php/page/that/echos/json/results",                         data: "post variables send",                         type: "post",                         error: function(xmlhttprequest, textstatus, errorthrown)  {                             alert("an error has occurred making request: " + errorthrown);                         },                         success: function(){                             //do stuff here                         }                     }); 

just build php page echos out results of query json_encoded, nothing else sent ui.