How to make dropdown based query mysql in javascript?

Question

<script type="text/javascript">
var rowNum = 0;
function addRow(frm) {
 rowNum ++;
 var row = '<p id="rowNum'+rowNum+'"> Barang: ';
 row += '<select name="???">';
 row += '<option value="A1">A1</option>';
 row += '<option value="A2">A2</option>';
 row += '<option value="A3">A3</option>';
 row += '<option value="A4">A4</option>';
 row += '</select>';
 row += ' Satuan: <input type="text" size="5" name="satuan[]" value="'+frm.add_satuan.value+'"> Quantity: <input type="text" name="qty[]" value="'+frm.add_qty.value+'"> <input type="button" value="Remove" onclick="removeRow('+rowNum+');"><hr color=red></p>';

 $('#itemRows').append($(row));
  frm.add_qty.value = '';
  frm.add_nama.value = '';
         frm.add_satuan.value = '';
};
  </script>

How to make drop-down list but with query mysql to show A1, A2, dll. When submit button for drop-down can't post data anything. Can I store data with this javascript. For text input success post data.

Hi, i make dinamic row for insert data. But for Barang must from database query. Can u help for show in javascript? — deib97, Jan 19 '16 at 07:40

Matthew Herbst · Accepted Answer · 2016-01-19T07:55:51.303

DO NOT DO WHAT YOU ARE TRYING TO DO. You should never, ever, ever write queries on the front-end. You should do your absolute best to hide every detail of the server/database from the user. It is a massive security risk. Please read about SQL injection attacks for starters.

How you should do this:

Store the values of the dropsdowns in JavaScript. Let's keep them in a single object to make life easy:

Your JS:

var options = {
  A1: $("#rowNum select option[value='A1']").text(),
  A2: $("#rowNum select option[value='A2']").text(),
  A3: $("#rowNum select option[value='A3']").text(),
  A4: $("#rowNum select option[value='A4']").text()
};

// Now, send this object to your PHP via an AJAX call. Let's assume for simplicity that you will do this using jQuery:
$.ajax({
  url: 'my/php/script.php',
  data: options,
  success: function (data) { console.log('Yay, it worked!'); },
  error: function (jqXHR, textStatus, error) { console.log('crap it didn't work', jqXHR, textStatus, error); }
});

Your PHP

<?php

$options = $_REQUEST['options']
// You need to verify the options are valid (and don't have bad values) but that's a different question

// Build your query here. Your PHP is run on the server only so no one else will see it or be able to change it.

I still dont understand with new suggest and new technical. Whats key for I learn for use your suggest ? :) Thanks — deib97, Jan 19 '16 at 07:52
@user2400395 You need to go learn about these things (all easily googleable): 1) how to make an AJAX request using jQuery. 2) how to run MySQL queries from PHP — Matthew Herbst, Jan 19 '16 at 07:55
@user2400395 welcome. Please feel free to come back to SO and search for/ask other questions about what you are learning if needed. — Matthew Herbst, Jan 19 '16 at 08:01

score 0 · Answer 2 · answered Jan 19 '16 at 07:40

You can try this code with some modification. Use jQuery's each to iterate over the array value.

$('#emptyDropdown').empty();
// Parse the returned json data
//var opts = $.parseJSON(data);//Remove comment if you are using JSON

// Use jQuery's each to iterate over the opts value
$.each(opts, function(i, d) {
    // You will need to alter the below to get the right values from your data.
    $('#emptyDropdown').append('<option value="' + d.yourID + '">' + d.yourValue + '</option>');
});

How to make dropdown based query mysql in javascript?

2 Answers2