html下拉下拉选择值未在MYSQL中插入

Question

i have two drop downs. first drop down populate from database. second drop down populated from database based on selected value of first drop down.

 $(document).ready(function() { $("#c").change(function() { var c1 = $('#c :selected').text(); if(c1 != "") { $.ajax({ url:'getstatw.php', data:{c:c1}, type:'POST', success:function(response) { var resp = $.trim(response); $("#c").html(resp); } }); } else { $("#c").html("<option value=''>Select state</option>"); } }); }); 

  <form id = "world" method="post" action="insert.php"> <select name="country" id="c" style = "width:200px" class="btn btn-primary dropdown-toggle" ;> <option>country</option> <?php $sql = "select DISTINCT country from table1"; $res = mysqli_query($con, $sql); if(mysqli_num_rows($res) > 0) { while($row = mysqli_fetch_object($res)) { echo "<option value='".$row->id."'>".$row->c."</option>"; } } ?> </select> <br><br> <label for="s" >State</label> <select name="State" id="s" style = "width:200px " ; class="btn btn-primary dropdown-toggle";><option>Select state</option></select><br><br> <button id = "sub" type="submit" class="btn btn-primary" disabled>Submit</button> </form> 

insert.php

 $con=mysqli_connect("localhost","root","","world"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } // escape variables for security //home tab $c = mysqli_real_escape_string($con, $_POST['country']); $s = mysqli_real_escape_string($con, $_POST['state']); //query for table_mainast $sql1="INSERT INTO table1 (Country, State) VALUES ('$c', '$s',)"; //query for table_dataast if (!mysqli_query($con,$sql1)) { die('Error: ' . mysqli_error($con)); } echo "1 record added"; mysqli_close($con); 

getstate.php

 <?php $con=mysqli_connect("localhost","root","","test"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } if(isset($_POST['c'])) { $sql = "select DISTINCT `State` from `table2` where `Country`='".mysqli_real_escape_string($con, $_POST['c'])."'"; $res = mysqli_query($con, $sql); if(mysqli_num_rows($res) > 0) { echo "<option value=''>------- Select --------</option>"; while($row = mysqli_fetch_object($res)) { echo "<option value='".$row->id."'>".$row->c."</option>"; } } } else { header('location: ./'); } ?> 

i have tried almost all solution given on net. but do not understand my data is not inserted into mysql database. How to insert HTML select value as text in MySQL via PHP PHP Drop down list selected value not inserted in the database

Answer 1

You need to concantenate them properly and also you don't you put , after last column in the query

Change following

$sql1="INSERT INTO table1 (Country, State)
VALUES ('$c', '$s',)";

to

$sql1="INSERT INTO table1 (country, state)
VALUES ('".$c."', '".$s."')";

Answer 2

if want to add options to the second drop down you need to use append not html and your using same ID ie #c to write the response in ajax success, change it to second drop down ID ie #s

you can try this:

$(document).ready(function() {
  $("#c").change(function() {
    var c1 = $('#c :selected').text();
    if(c1 != "") {
      $.ajax({
        url:'getstatw.php',
        data:{c:c1},
        type:'POST',
        success:function(response) {
          var resp = $.trim(response);
          $("#s").append(resp);
        }
      });
    } else {
      $("#c").append("<option selected value=''>Select state</option>");
    }
  });
});

then remove , in insert query.

$sql1="INSERT INTO table1 (Country, State)
VALUES ('$c', '$s')";