Pop-up message after submitting a form with php
Question
I'm trying to get a pop-up message saying if it was successfully submitted or not without having to go to a different page.
Now chrome gives me the pop-up message but it redirects me to a blank page after.
Here is my current code.
<?php
include "header.php";
include "conexao.php";
echo "<h1 align='center'>Pagina para alterar produtos</h1><div class='container'><hr>";
$referencia=$_GET['id'];
$sql = "SELECT * ";
$sql = $sql . " FROM tb_produto ";
$sql = $sql . " WHERE pr_codigo='".$referencia."'";
$produtos = $db->query($sql);
foreach ($produtos as $produto) {
$referencia = $produto["pr_codigo"];
$nome = $produto["pr_descricao"];
$preco = $produto["pr_preco"];
$disponivel = $produto["disponivel"];
}
echo "<h2>Referencia: ".$referencia."</h2>";
echo "<h2>Nome: ".$nome."</h2><hr>";
?>
<form action="confirmaAlterar.php">
<div class="form-group">
<label>Referencia</label>
<input class="form-control" type="text" name="referencia" value="<?php echo $referencia?>">
</div>
<div class="form-group">
<label>Nome</label>
<input class="form-control" type="text" name="nome" value="<?php echo $nome?>">
</div>
<div class="form-group">
<label>Preço</label>
<input class="form-control" type="text" name="preco" value="<?php echo $preco?>">
</div>
<button class="btn btn-primary">Alterar</button>
</form>
Here is where it submits the information of the form.
<?php
include "header.php";
include "conexao.php";
$nome=$_GET['nome'];
$referencia=$_GET['referencia'];
$preco=$_GET['preco'];
$sql="UPDATE tb_produto SET pr_descricao='".$nome;
$sql.="', pr_preco=".$preco;
$sql.= " WHERE pr_codigo='".$
try{
$comando=$db->prepare($sql);
$comando->execute();
echo "<script type='text/javascript'>alert('submitted successfully!')</script>";
header( "refresh2;Location:index.php" );
}
catch (PDOException $e){
echo "A";
}
2 answers
solution1
0 ACCPTED 2016-08-27 09:57:20
To pass values using ajax. Form:
<form id="form">
<input type="text" value="test" name="akcija">
</form>
All inputs fields values in your form will be passed.
Ajax:
jQuery(function(){
$('#form').on('submit', function (e) { //on submit function
e.preventDefault();
$.ajax({
type: 'post', //method POST
url: 'yoururl.php', //URL of page where u place query and passing values
data: $('#form').serialize(), //seriallize is passing all inputs values of form
success: function(){ //on success function
$("#input").attr("disabled", true); //example
$("#input").removeClass('btn-primary').addClass('btn-success');//example
},
});
}
});
And on the ajax page you can get values by
$akcija = $_POST['akcija']
solution2
0 2016-08-30 06:49:22
for this Problem you must use ajax method .
1- create html form and all input Required .
<form id="contactForm2" action="/your_url" method="post">
...
</form>
2- add jQuery library file in the head of html page .
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js">
</script>
...
3- add this method Under the jQuery library
<script type="text/javascript">
var frm = $('#contactForm2');
frm.submit(function (ev) {
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
data: frm.serialize(),
success: function (data) {
if(data == 'pass')
alert('ok');
else(data == 'fail')
alert('no');
}
});
ev.preventDefault();
});
</script>
4- in your_url .php file
<?php
$a = ok ;
if( $a == 'ok' ){
echo 'pass';
}else{
echo 'fail';
}
?>
this top answer is easy management form with jquery , but you need managment Complex form better use this library http://jquery.malsup.com/form/
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