jQuery Forum

live search jquery and show to the table
[4Replies]
- 19-Jul-2020 09:13 AM
- Forum: Using jQuery
hello, i'm creating live search and show it to the table

here the ajax
$("#cari").on('keyup', function(){
               var cari =  $("#cari").val();
               $.ajax({
                    type: "post",
                    url: "do_kasircari.php",
                    data : {cari: cari},
                    dataType: "text",
                    success: function(data)
                    {
                         $("#hasil").load(data);
                         console.log(cari);
                         console.log(data);
                    }
               })
          });

here is the console.log

data is right, but not showing. why? please help, thanks

cannot get data from ajax
[6Replies]
- 28-Jun-2020 03:02 AM
- Forum: Using jQuery
hello
i was creating live username check for learning

here my code
login.php
$.ajax({
                url: "cek_username.php",
                type: "post",
                data: "username="+username,
                success: function(data){
                    if(data == 0)
                    {
                        $("#idErr").html("username tersedia");
                        $("#id_daftar").css("outline-color","green");
                    }
                    else
                    {
                        $("#idErr").html("username tidak tersedia");
                        $("#id_daftar").css("outline-color", "red");
                        console.log(data);
                        error_username = true;
                    }
                },
            });

cek_username.php (code for check username avaibility on database)
<?php
     include "koneksi.php";

     $username = $_POST['username'];

     $sql= "select * from login where username = '$username' ";
     $query= mysqli_query($conn, $sql);
     $data = mysqli_num_rows($query);

     echo $data;
?>

here is the form
<div class="form-group">
     <label for="username">Username :</label>
     <input type="text" style="max-width:90%;" class="form-control" name="id_daftar"
      id="id_daftar" placeholder="Masukkan Username Anda" autocomplete="off"
      maxlength="16" value="" />
     <div class="error" id="idErr"></div>
</div>

console.log

so i created in 2 php file, 1 is simple one for my learning purpose
but after i implemented to another php for my project, it cannot get value from cek_username.php, and i dont know why
please help

thanks

live form check availability
[0Replies]
- 14-May-2020 08:10 PM
- Forum: Using jQuery
hello,
i'm newbie of jquery and ajax, so i created based on website that i read
i already created simple live username check (and others) and i works

but after i implement on to my project, it dosen't works, why?

here the code:
form code

<form name="daftar" method="post" onsubmit="return validate_daftar()">
<div class="form-group">
    <label for="username">Username :</label>
    <input type="text" style="max-width:90%;" class="form-control" name="id_daftar" id="id_daftar" placeholder="Masukkan Username Anda" maxlength="16" autocomplete="off"/>
    <div class="error" id="idErr"></div>
</div>
<div class="form-group">
    <label for="password">Password :</label>
    <input type="password" style="max-width:90%;" class="form-control" name="pass" id="pass" placeholder="Masukkan Password Anda" maxlength="16" value="" />
    <input type="checkbox" onclick=password()> Tampil Password
    <div class="error" id="passErr"></div>
</div>
<div class="form-group">
    <label for="nama">Nama :</label>
    <input type="text" style="max-width:90%" class="form-control" name="nama" id="nama" maxlength="50" placeholder="Masukkan Nama Sesuai KTP" value="" />
    <div class="error" id="namaErr"></div>
</div>
<div class="form-group">
    <label for="email">Email :</label>
    <input type="email" style="max-width:90%;" class="form-control" name="email" id="email" placeholder="Masukkan Email Anda" value="" />
    <div class="error" id="emailErr"></div>
</div>
<div class="form-group">
    <label for="nomorhp">Nomor HP:</label>
    <input type="number" style="max-width:90%;" class="form-control" placeholder="Masukkan Nomor HP Anda" name="nohp" id="nohp" maxlength="16" value="" />
    <div class="error" id="nohpErr"></div>
</div>
<button type="submit" name="daftar" value="daftar" class="btn btn-info">Daftar</button>
</div>
</form>

jquery and ajax code:
$(document).ready(function(){
    $('#id_daftar').keyup(function(){
        var id_daftar = $('#id_daftar').val();
        $.ajax({
            url: "regis_username.php",
            type: "post",
            data: "id_daftar="+id_daftar,
            dataType: "text",
            success: function(data)
            {
                if(data == 1)
                {
                    $("#idErr").html("username tersedia");
                    $("#idErr").css("outline-color", "green");
                }
                else
                {
                    $("#idErr").html("username tidak tersedia");
                    $("#idErr").css("outline-color", "red");
                    console.log(id_daftar);
                    console.log(data);
                }
            },
        });
    });

searching code: (regis_username.php)
<?php
     include "koneksi.php";

     $username = $_POST['id_daftar'];

     $sql= "select * from login where username = '$username' ";
     $query= mysqli_query($conn, $sql);
     $count = mysqli_num_rows($query);

     echo $username."-".$count;
?>

console.log

i don't know what wrong but, any help and explanation will be appreciated!

get data from combo box and store it to the session
[5Replies]
- 13-May-2020 10:20 PM
- Forum: Using jQuery
hello,
im just learning jquery and ajax via YT, and i tried to create a form with combo box, and show records from database based on combo box
after that, i want to store the data to the session
i already created the form that showing the records and already worked, but the button that i created that for store to the session, the ajax is not working.
ajax3.php (index)
<form action="cari.php" method="POST">
     <select name="kamar">
          <option value="regular single">Regular Single Rp 99.000</option>
          <option value="regular double">Regular Double Rp 120.000</option>
          <option value="family single">Family Single Rp 127.500</option>
          <option value="family double">Family Double Rp 150.000</option>
          <option value="vip single">V.I.P Single Rp 170.000</option>
          <option value="vip double">V.I.P Double Rp 200.000</option>
     </select>
     <br>
     <button type="submit" name="submit">Submit</button>
</form>
<br>
<div id="tampil"></div>
<script type="text/javascript">
$(document).ready(function(){
$('form').on('submit', function(e)
          {
e.preventDefault();
$.ajax({
type: $(this).attr('method'),
url: $(this).attr('action'),
data: $(this).serialize(),
success: function(data){
$("#tampil").html(data);
     },
});
$('.simpanData').on('submit', function(e)
{
e.preventDefault();
$.ajax({
type: "get",
url: $(this).attr('href'),
data: $(this).serialize(),
success: function(data){
$("#tampil").html(data);
},
});
});
});
});
here is cari.php (searching)
<?php
include "koneksi.php";
$jenis = $_POST['kamar'];
$sql= "select * from kamar where jkamar = '$jenis' ";
$result = mysqli_query($conn, $sql);
?>
<table>
<?php
if(mysqli_num_rows($result) > 0)
          {
while ($data = mysqli_fetch_array($result))
               {
$link_simpan = "<a class='simpanData' href='simpan2.php?nokmr=".$data['nokamar']."'>Simpan</a>";
?>
               <tr>
                    <td value="<?php echo "nokamar = $data[nokamar]"; ?>"><?php echo $data['nokamar'];?></td>
                    <td><?php echo $link_simpan ?></td>
               </tr>
<?php
               }
          }
else
          {
echo "no more data!";
          }
?>
</table>

any help will be appreciated, thanks

Response title

Attachments

jQuery Forum

Activity Trend