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Hello, I am doing a Dropdown with the database and I would like to use ajax to update a table, I confess that I have a difficulty in JS, I would like a help and/ or a north for this. Grateful
<!-- Profile Image -->
<div class="box box-primary">
<div class="box-body box-profile">
<img class="profile-user-img img-responsive img-circle" src="../dist/img/profile/'. $id .'.png" alt="User profile picture">
<h3 class="profile-username text-center">'. $nome .'</h3>
<p class="text-muted text-center">Ramal: '. $id .'</p>
<ul class="list-group list-group-unbordered">
<li class="list-group-item">
<b>Nivel de acesso: </b> <a class="pull-right">'. $nvAcess .'</a>
</li>
</ul>
<a class="pull-right image" href="#">
<i class="fa fa-edit">Editar</i>
</a>
</div>
<!-- /.box-body -->
</div>
<!-- /.box -->
<label>Users</label> </br>
<select class="form-control select2" style="width: 100%;" id="user">
<?php $conn=mysqli_connect("192.168.0.2","root","Kt6v75dX","spa") or die("ERRO AO CONECTAR");
$query = "SELECT id,nome,nivel_de_acesso,senha FROM usuarios";
$query = mysqli_query($conn,$query);
$reg = mysqli_fetch_array($query);
while($reg == 0){
}
while($reg != 0){
$id = $reg['id'];
$nome = $reg['nome'];
$nvAcess = $reg['nivel_de_acesso'];
$pass = $reg['senha'];
echo'<option value="'.$id.'">'. $nome .'</option>';
$reg = mysqli_fetch_array($query);
}
?>
</select>
How so update a table? Do you want to receive dropdown items via ajax or want to popular the table via ajax? If it is to feed the bank, what information would be, etc?
– Victor Carnaval
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