Question

我是json的新手，我试图从json发布我的表单值来更新mysql数据库。当我提交时，我有一个成功警报，但是当我查看我的数据库时，似乎我的值没有通过实际传递，使我的大多数字段都空白。需要帮助使用json和php将表单数据传递到我的数据库。

JAVASCRIPT

$('#save').on('click', function () {
        $.ajax({

            type: "POST",
            url: "http://localhost/v_warehouse_1/inc/updateprofile.php",
            data: {
                detailid: id,
                titleid: $('#selectmenu').val(),
                name: $('#txtname').val(),
                surname: $('#txtsurname').val(),
                contact_no: $('#txtcontact_no').val(),
                email: $('#txtemail').val(),
                category:$('#txtcategory').val(),
                package: $('#txtpackage').val(),
                password: $('#txtpassword').val()
            },
            datatype: "json",
            success: function (status) {
                if (status.success == false) {
                    //alert a failure message
                    alert("Your details we not saved");
                } else {
                    //alert a success message
                    alert("Details Updated");

                    location.href='profiledetails.html?id='+id;

                }

            }
        });
    });

PHP

require_once("database.php");
$mydb = new MySQLDatabase();

//set varables from json data
    $id = json_decode($_POST['detailid']);
    $titleid = json_decode($_POST['titleid']);
    $name = json_decode($_POST['name']);
    $surname = json_decode($_POST['surname']);
    $contact_no = json_decode($_POST['contact_no']);
    $email = json_decode($_POST['email']);
    $category = json_decode($_POST['category']);        
    $package = json_decode($_POST['package']);
    $password = json_decode($_POST['password']);


$mydb->query("UPDATE tblprofile SET title_fk = '$titleid',`name` = '$name',surname = '$surname',contact_no ='$contact_no',email = '$email',category_fk = '$category',package_fk = 'package_fk' ,`password` = 'password' WHERE id = '$id' ;");
$mydb->close_connection();

Answer 1

无需解码数据。它们将作为普通的发布数据发布。只需访问它们 -

$id = $_POST['detailid'];

Answer 2

你不需要json_decode来自$ _POST的值。将代码更改为此

$id = $_POST['detailid'];
$titleid = $_POST['titleid'];
$name = $_POST['name'];
$surname = $_POST['surname'];
$contact_no = $_POST['contact_no'];
$email = $_POST['email'];
$category = $_POST['category'];
$package = $_POST['package'];
$password = $_POST['password'];

虽然你是通过ajax调用发送json但是没有在服务器中编码

Answer 3

除非您从客户端发送JSON格式的数据，否则请勿使用json_decode()

例如：

在ajax调用中，而不是 data:{} 如果你试图以这种方式发送，

var Jdata = JSON.parse("{'detailid':'"+id+"'");
$.ajax({
    type: "POST",
    url: "http://localhost/v_warehouse_1/inc/updateprofile.php",
    data:Jdata,
    datatype: "json",
    success: function (status) {
        //your stuff..
    }
});

然后在php中使用json_decode()

将表单数据从json转换为php mysql

3 个答案: