AJAX sends information to PHP but fails to INSERT only one

Question

AJAX sends information to PHP but fails to INSERT only one

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#1 by (4 votes)

3

See this code for me:

HTML:

<script type="text/javascript" src="js/jquery-2.1.1.min.js"></script>
<script type="text/javascript" src="js/criaEvento.js"></script>
<form method="post">
<table width="420" align="center">
    <tr>
        <td>Nome</td>
        <td colspan="3"><input class="form-control" type="text" name="nomeEvento" placeholder="exemplo: Festa no ap" size="30"></td>
    </tr>
    <tr>
        <td height="10"></td>
    </tr>
    <tr>
        <td>Detalhes
        <td colspan="3"><textarea class="form-control" name="maisInfos" placeholder="informais adicionais" rows="3" cols="33"></textarea></td>
    </tr>
    <tr>
        <td height="10"></td>
    </tr>
    <tr>
        <td>Onde</td>
        <td colspan="3"><input class="form-control" type="text" name="onde" placeholder="local" size="30"></td>
    </tr>
    <tr>
        <td height="10"></td>
    </tr>
    <tr>
        <td>Quando</td>
        <td><input id="dia" type="date" name="dia"></td>
        <td>Horário</td>
        <td><input id="hora" type="time" name="hora"></td>
    </tr>
    <tr>
        <td height="10"></td>
    </tr>
    <tr>
        <td></td>
        <td colspan="3"><input type="submit" class="btn btn-lg btn-primary btn-block" id="salvar" name="salvar" value="SALVAR"></td>
    </tr>
</table>
</form>

JavaScript:

(document).ready( function(){

    $("#salvar").click( function(){

        var dataString = $("form").serialize();

        $.ajax({
            url: 'php/evento.php',
            type: 'POST',
            dataType: 'json',
            data: 'data='+dataString,

            success: function(data){

            }

        });

    });
})

PHP:

<?php
include_once 'con.php';

$nomeEvento = $_POST['nomeEvento'];
$maisInfos = $_POST['maisInfos'];
$onde = $_POST['onde'];
$dia = $_POST['dia'];
$hora = $_POST['hora'];

$qryInsert = mysql_query("INSERT INTO evento VALUES(NULL, '$nomeEvento', '$maisInfos', '$onde', '$dia', '$hora')");

It passes the data to PHP but in INSERT to DB, only one of the data is NOT saved, the rest goes ...

php mysql jquery ajax

asked by anonymous 06.02.2015 / 21:22

1 answer

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score 4 · Accepted Answer

The problem is that you are sending the parameters incorrectly in the ajax call.

data: 'data='+dataString,

In this line you end up generating a new data called data where the value ends up being data=nomeEvento=[valor preenchido] , ie your first item in dataString ends up becoming the value of data and is lost. >

The .serialize() function already generates everything you need to send the data via POST. You just have to do this:

$.ajax({
    url: 'php/evento.php',
    type: 'POST',
    dataType: 'json',
    data: dataString, // passa diretamente a string contendo os dados do post

    success: function(data){

    }
});