Question

下午好。

我的问题是当我用jquery发送数据时。

我点击按钮但没有收到回复。我想我的问题是：

$.post("car.php",
            {
              action: "cargar_stock",
              cajas_dulce_de_leche: CajaDulceDeLeche,
              cajas_fruta: CajaFruta
            },
            function(data){
              alert("Data: "+data);
            }
        );

register.php：

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script type="text/javascript">
function cargar_carga() {

    var CajaDulceDeLeche = document.getElementById("cajaDl").value;
    var CajaFruta = document.getElementById("cajaFru").value;


    if ((CajaFruta != "") || (CajaDulceDeLeche != "")){
        $.post("car.php",
            {
              action: "cargar_stock",
              cajas_dulce_de_leche: CajaDulceDeLeche,
              cajas_fruta: CajaFruta
            },
            function(data){
              alert("Data: "+data);
            }
        );
    }   
    else{
        alert("Debe ingresar al menos algun valor");
    }
}

<label for="cajaDl">Dulce de Leche: </label>
<input type="text" id="cajaDl" />

<label for="cajaFru">Fruta: </label>
<input type="text" id="cajaFru" /><br/>
<input type="button" id="botonRegCaja" value="Registrar" onclick="cargar_carga()"/>

这里我从register.php

中获取数据

car.php：

<?php
include_once('helpers.php');
require_once("../some_libraries/config/db.php");

$port=3306;
$socket="";

$con = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME, $port, $socket)
    or die ('Could not connect to the database server' . mysqli_connect_error());

if (isset($_POST['action']) && ($_POST['action'] != "")) {
    $action = $_POST['action'];
    if ($action == "cargar_stock") {

        if (isset($_POST['cajas_dulce_de_leche']) && ($_POST['cajas_dulce_de_leche'] != "") &&
            isset($_POST['cajas_fruta']) && ($_POST['cajas_fruta'] != "")) {

            $boxs_to_stock_caramel = $_POST['cajas_dulce_de_leche'];
            $boxs_to_stock_fruit = $_POST['cajas_fruta'];

            $sql_stock = "SELECT `fruta`, `dulce` FROM `stock`";
            $result_query_stock = mysqli_query($con, $sql_stock);

            if ($result_query_stock) {

                $row_result = mysqli_fetch_array($result_query_stock);

                $row_dulce_stock = $row_result['dulce'];
                $row_fruta_stock = $row_result['fruta'];

                $total_boxes_caramel = $boxs_to_stock_caramel + $row_dulce_stock;
                $total_boxes_fruit = $boxs_to_stock_fruit + $row_fruta_stock;

                $sql_update_stock = '"UPDATE `stock` SET `fruta`='.$total_boxes_fruit.',
                                                            `dulce`='.$total_boxes_caramel.' WHERE id=1"';

                $result_sql_update_stock = mysqli_query($con, $sql_update_stock);

                if ($result_sql_update_stock) {
                    $result_gral = "Stock actualizado";
                }
                else{
                    $result_gral = "Problema con la consulta, stock no actualizado";
                }
            }
            else{
                $result_gral= "Problema al acceder a los datos del stock";
            }

        }else{echo "no llego bien el post";}
    }

}
echo $resul_gral;
?>

Answer 1

尝试：

$.post("car.php",
            {
              action: "cargar_stock",
              cajas_dulce_de_leche: CajaDulceDeLeche,
              cajas_fruta: CajaFruta
            },
            function(data){
              alert("Data: "+data.alert);
            }, 'json'
        );

并在car.php中尝试：

$varInfo['alert'] = 'Hello';
exit(json_encode($varInfo));

Answer 2

解决了！在mi¨index.php的背景下，当我包括¨register.php时，我失去了运行程序的环境...... 我有一个index.php，它有一个菜单，我在每个选项中有不同的选项包括不同的文件....然后我的index.php包含register.php当car.php想在同一目录中使用此文件你做了参考register.php。就像我就在那里，问题是我的index.php在一个目录中！

Jquery + php，没有回应

2 个答案: