pull information via php ajax

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pull information via php ajax

Asked 9 years, 2 months ago

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good afternoon, I am studying ajax and I am trying to make an input pass to a file via post, but nothing happens.. I am following via Chrome firebug and I see that always the type: 'post' is in error.

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>

<script>
function ajax(nome){
    info = {"nNome" : nome};

    $.ajax({
        type: 'post';
        url: "arquivo.php";
        data: info,
    }).done(function(data) {
        data = $.parseJSON(data);
        $(".resultado span.nome").text(data.nNome);
    });
}

    $(document).ready(function(){

        $("input[name=Enviar]").click(function(){
            var nome = $("input[name=nome]").val();
        });

    });

</script>

<form>
<label>Entre com seu nome:</label>
<input type="text" name="nome"><br>
<input type="button" name="Enviar" value="Enviar">
</form>

<div class="resultado">
Olá, <span class="nome"></span>
</div>

1

The mistake are the ; inside object, you should be using commas.

– bfavaretto

2016/08/10 at 18:41
OK, I made the changes and really that was the mistake, but I still can’t display the result...

– Vinicius Nakamura

2016/08/10 at 18:49
Please show what your.php file does.

– Kayo Bruno

2016/08/10 at 18:55
<?php

echo json_encode($_POST);

?>

– Vinicius Nakamura

2016/08/10 at 19:01
data.nNome should also not be data.nome? Anyway, it will be hard to see the value change if what you return in PHP is the same as what you just sent. The .text() will trade six for half a dozen.

– bfavaretto

2016/08/10 at 19:10
then as to the date.nName it is pulling via Indice.. which comes through the info = {"nNome" : nome }; , it should appear in the div below... it’s just a test to learn and apply in other files...

– Vinicius Nakamura

2016/08/10 at 19:16

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by bfavaretto • **64,705** points · Answer 1 · 2016-08-10T19:27:24+00:00

There are several small errors in your code, try this modified version:

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>

<form>
<label>Entre com seu nome:</label>
<input type="text" name="nome"><br>
<input type="button" name="Enviar" value="Enviar">
</form>

<div class="resultado">
Olá, <span class="nome"></span>
</div>

<script>
function ajax(nome){
    var nome = ;
    var info = {"nNome" : nome};

    $.ajax({
        type: 'post',
        url: "arquivo.php",
        data: info,
    }).done(function(data) {
        data = $.parseJSON(data);
        $(".resultado span.nome").text(data.nNome);
    });
}

$(document).ready(function(){
    $("input[name=Enviar]").click(function(){
      ajax( $("input[name=nome]").val() );
    });
});
</script>