使用Ajax无法在SQL中插入数据

Question

I'm trying to insert data in a sql table using ajax and php, but it's not working. My ajax give me the result like it works, but when i look at the table, there's not in it. Doing it without ajax works fine, so i guess my php is working ok.

Here's the code:

HTML:

<form action="servico.php?p=cadUsr" method="POST" id="frmCadUsr">
    Nome: <input type="text" maxlength="255" name="txtNome" id="txtNome"/>
    Idade: <input type="text" maxlength="3" name="txtIdade" id="txtIdade"/>

    <input type="submit" value="Enviar"/>
</form>

PHP:

    $passo = (isset($_GET['p'])) ? $_GET['p'] : "";


    switch($passo){

        case "cadUsr":
            cadUsr();
        break;

        default:
            getRetorno();
        break;
    }

    function getRetorno(){
        echo "Este texto foi escrito via PHP";
    }

    function cadUsr(){
    require("dbCon.php");
    require("mdl_usuario.php");

        $usr = $_POST["txtNome"];
        $idade = $_POST["txtIdade"];

        $resultado = usuario_cadastrar($con,$usr,$idade);

            if($resultado){
                echo "Cadastro efetuado com sucesso";
            } else {
                echo "O cadastro falhou";
            }
    }
?>

OBS: I need to pass the action of the form with the url parameter as cadUsr, so it call the function in php.

AJAX:

window.onload = function(){

        var xmlhttp;
        var frm = document.querySelector("#frmCadUsr");
        var url = frm.getAttribute("action");
        var nm = document.querySelector("#txtNome").value;
        var idade = document.querySelector("#txtIdade").value;

        frm.addEventListener("submit",function(e){
            e.preventDefault();

            try{
                if(window.XMLHttpRequest){
                    xmlhttp = new XMLHttpRequest();
                }

                xmlhttp.open("POST",url,true);
                xmlhttp.send("txtNome=" + nm + "&txtIdade="+idade + "&p=cadUsr");

                xmlhttp.onreadystatechange = function(){
                    if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
                        //alert("Deu certo");
                        console.log(xmlhttp.responseText);
                    }
                }
            } catch(err){
                alert("Ocorreu um erro.<br />"+ err);
            }
        });

}

The PHP function to insert the data:

function usuario_cadastrar($conexao,$nome,$idade){


        if($nome == "" && $idade == ""){
            return false;
        }





        $sql = sprintf("insert into usuario (nome,idade) values ('%s',%s)",$nome,$idade);

        $resultado = mysqli_query($conexao,$sql);

            return $resultado;
    }

Answer 1

I think the problem is here servico.php?p=cadUsr . You copy the action -attribute from the form with a querystring. If you cut the querystring from it, I think it will work.

The main problem is being called by Hossein:

This :

$passo = (isset($_GET['p'])) ? $_GET['p'] : "";

Will not work. You're doing a post, you can't get GET variables.

You call value on value which will result in undefined and that will put no data in your database.

 xmlhttp.send("txtNome=" + nm + "&txtIdade="+idade + "&p=cadUsr");

So remove value and add the cadUsr variable to the querystring in the send function. Update PHP to:

$passo = (isset($_POST['p'])) ? $_POST['p'] : "";

And it will work!

You can see your callback codes by adding console.log(xmlhttp.responseText); to your readystate success function.

Also you need to set the requestheader content-type to x-www-form-urlencoded when sending post.