Redirect page by passing "Answer" from ajax to a div

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I know this is a strange question, and it may seem pointless, but I need an answer, even if it’s for impossível. I do a search in my indexand need to pass the result in the same to another page positioning in a div. I did some research and tried some alternatives, but without success and as I said, I don’t even know if that’s possible.

What I already have is this, the return of a research, I consult response and already visualise the content desired.

$(function() {      
    $("#frmBusca").validate({
        submitHandler: function(form) {
            var data = $(form).serialize();             

            $.ajax({
                type: 'POST',
                url: 'pBuscaGenerica.php',
                data: data,
                dataType: 'html',

                success: function(response) {

                    window.location.replace("outra_pag.php");

                    // EXIBINDO O RETORNO DAS INFORMAÇÕES   
                     $("#msgRamais").html(response);
                    // RESETANDO A FUNÇÃO
                     _toggle();                 

                },
                error: function(xhr, ajaxOptions, thrownError) {
                    console.log(xhr, ajaxOptions, thrownError);
                    $("#msgRamais").html('×ATENÇÃO! Ocorreu um erro ao tentar obter o ramal. Contate o suporte técnico.');
                }
            });
            return false;
        }
    });
});

I need to redirect to a particular div, I will not post the attempts because they were unproductive and would not add anything to the post.

php ajax

  • Want to put the result (Replay) inside a div of another page? That’s it?

  • Hello @Miguel, just that.

  • Not yet? I see that you have updated the code... So I see you want to redirect to another page when you receive the answer?

  • Hello @Miguel, what I need is a little complex, I need to move Replay to the next page and show it in a div, but I don’t know if this is possible.

  • Of course it is, follow the steps of my reply. Also read the notes. Want to redirect to the other pag in the ajax function succeso?

  • redirect and pass the content of 'Answer'.

  • Edited answer. I think that’s it... Read it carefully

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3 answers

2

I won’t post codes, but the idea would be:

Perform the search via AJAX. Store in one variable and pass the other page via PHP POST.

  • #Shura16 already answered the new rss post and I know what boring to do kkkkk

2


From what I understand is what you want (the way I would do):

JS:

$.ajax({
            type: 'POST',
            url: 'pBuscaGenerica.php',
            data: data,
            dataType: 'html',

            success: function(response) {
                window.location.replace("outra_pag.php");
            },
            error: function(xhr, ajaxOptions, thrownError) {
                console.log(xhr, ajaxOptions, thrownError);
                $("#msgRamais").html('×ATENÇÃO! Ocorreu um erro ao tentar obter o ramal. Contate o suporte técnico.');
            }
        });

On the server side, php, stored the information you wanted to pass to the other page, this when you call ajax:

pBuscaGenerica.php:

session_start();
...
$_SESSION['ajax_result'] = 'isto vai ser exibido dentro de uma div numa outra página que não a que fez a chamada';
...

other_pag.php:

session_start();
...
if(!isset($_SESSION['ajax_result'])) {
    // fazer uma coisa qualquer caso não haja essa variável, não tenha armazenado a informação
}
else {
    echo '<div>' .$_SESSION['ajax_result']. '</div>';
}

Some notes:

  1. Your mistake is what you’re doing $("#msgRamais").html(response); but this is useless, do not need this on this page (because it will be redirected to another, window.location.replace("outra_pag.php");)

  2. In case you need to have the variable $_SESSION['ajax_result'] set to view the other_pag.php within the if(!isset($_SESSION['ajax_result'])) {... put the redirect... In this case you don’t need to have the else {... which is down... Make sure you have this redirect at the top of the page before any print (output).

  • Hello @Miguel, what is the value of this variable? $_SESSION['ajax_result']; Where are you assigning value to it?

  • It is whatever is presented in the div, your @adventistapr answer . Set this in pBuscaGenerica.php when you call ajax

2

@adventistapr, So the example below on Ajax + PHP

File (index.html):

<html>
<head>
    <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
</head>

<body>
    <form id="ajax_form">
        <p>Texto:</p>
        <p><textarea name="texto" rows="10" cols="30" placeholder="Descreva um comentario" /></textarea></p>
        <p><input type="submit" name="enviar"/></p>
    </form>
    <hr><p>Ajax:</p>
    <textarea id="log" rows="20" cols="70"></textarea>
</body>

<script>
$(document).ready(function(){
    $('#ajax_form').submit(function(){
        var dados = $(this).serialize();
        $.ajax({
            type: "POST",
            url: "log.php",
            data: dados,
            success: function(testlog) {
                $('textarea#log').text(testlog);
            }
        });
        return false;
    });
});
</script>

</html>

File (log.php):

<?php
if (!empty($_POST)) {
    echo $_POST['texto'];
}
?>

  • I didn’t understand very well @Kingrider, I need to pass the content of 'Answer' to the next page, as was your example I have to perform a new rescue?

  • 1

    This is an example of code, to register and exit the reply give success without leaving on the success page.

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