i have 2 tables
office1 > id,off1
office2 > id,off2_id,off2
i want to insert the choosen into table "saveoffice" > offices1,offices2
drop.php

<!DOCTYPE html>

<html dir="rtl"
lang="ar">
<head>
<meta charset="utf-8">
    <title>dependent dropdown list using jquery Ajax?</title>
    <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.1.1/jquery.js"></script>
    <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
    <link rel="stylesheet" href="http://demo.itsolutionstuff.com/plugin/bootstrap-3.min.css">
</head>
<body>
  <form action="savedata.php" name="myForm" method="POST"/>
<div class="container">
    <div class="panel panel-default">
      <div class="panel-heading">Select office 1 and get bellow Related office 2</div>
      <div class="panel-body">
      <div class="form-group">
                <label for="title">offices 1:</label>
                <select name="office1" id="office2" class="form-control">
                <option value="">--- choose office 1 ---</option>

                    <?php
                        require('config.php');
                        $sql = "SELECT * FROM office1";
                        $result = $con->query($sql);
                        while($row = $result->fetch_assoc()){
                            echo "<option value='".$row['id']."'>".$row['off1']."</option>";
                        }
                    ?>
                        </select>
                        </div>

                        <div class="form-group">
                        <label for="title">offices 2</label>
                        <select name="office2" id="office2" class="form-control" style="width:350px">
                        </select>

                        </div>
                      </br>
            <div class="col-md-4">
            <input type="submit" name="submit" value="Insert Data" class="btn btn-danger">
            </div>

      </div>

    </div>

</div>

<script>
$( "select[name='office1']" ).change(function () {
    var stateID = $(this).val();
    if(stateID) {
        $.ajax({
            url: "ajaxpro.php",
            dataType: 'Json',
            type:'POST',
            data: {'id':stateID},
            success: function(data) {
                $('select[name="office2"]').empty();
                $.each(data, function(key, value) {
                    $('select[name="office2"]').append('<option value="'+ key +'">'+ value +'</option>');
                });
            }
        });
    }else{
        $('select[name="office2"]').empty();
    }
});

</script>

</body>

</html>

ajaxpro.php

<?php

   require('config.php');
   $sql = "SELECT * FROM office2
         WHERE off2_id LIKE '%".$_POST['id']."%'";
   $result = $con->query($sql);

   $json = [];
   while($row = $result->fetch_assoc()){
        $json[$row['id']] = $row['off2'];
   }
   echo json_encode($json);
?>

savedata.php > it does not work

<?php
include('config.php')

$off1 = $_POST['$offices1'];
$off2 = $_POST['$offices2'];
$query = " INSERT INTO `saveoffice`(`id`, `offices1`, `offices2`) VALUES (0,".$offices1.",".$offices2.") ";

    $query_execute=mysqli_query($con,$query);
          if($query_execute)
                    {
                          header("location:drop.php?q=success");
                    }

          else
                    {
                            echo mysqli_error($con);
                            header("location:drop.php?e=error");
                    }

?>

Recommended Answers

Use prepared statement instead of direct insert to prevent from SQL injection!
You try to save into table saveoffice?
To save into tables office1 and office2 try something like this:

<?php

$link = mysqli_connect("localhost", "username", "password", "database");

$result = [];
for($i=1; $i<=2; $i++){
    $office …
Jump to Post

All 2 Replies

Use prepared statement instead of direct insert to prevent from SQL injection!
You try to save into table saveoffice?
To save into tables office1 and office2 try something like this:

<?php

$link = mysqli_connect("localhost", "username", "password", "database");

$result = [];
for($i=1; $i<=2; $i++){
    $office = filter_input(INPUT_POST, 'offices'.$i);
    $result[$i] = 0;
    if($office){
        $stmt = mysqli_prepare($link,"INSERT INTO `office$i`(`off$i`) VALUES (?)");
        mysqli_stmt_bind_param($stmt, "s", $office);
        mysqli_stmt_execute($stmt);
        $result[$i] = (string)(mysqli_stmt_affected_rows($stmt));
    }
    echo 'off'.$i.' inserted '.$result[$i].' rows<br/>';
}

?>

... or you want to crete new table "saveoffices" and migrate all data from "office1" and "office2"? Show your table structures and relations.

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge.