Kendo ui grid - 在我的php上获取READ数据发送

时间:2016-02-05 17:11:29

标签: php kendo-ui kendo-grid

我已经看到this回答在执行READ操作时发送额外的参数,但是如何在我的php上访问该变量?

这是我在客户端的代码:

          dataSource = new kendo.data.DataSource({
          transport: {
              read: 
              {
                url:"basedados.php",
                type: "POST",
                dataType: "json",
                data: { "my_param": 1, access_token : "my_token" },  // send parameter "access_token" with value "my_token" with the `read` request
              }, 

但现在在我的.php文件中,如何访问它?:

  else
  {
    $verb = $_SERVER["REQUEST_METHOD"];

    if ($verb == "POST") 
    {
        header("Content-type: application/json");

        if($_POST['access_token']) // do i say it like this?
        {

            $formData = $_POST["access_token"];
            echo $formData;

        }

编辑:感谢您回答Rick,但让我们向您展示更多代码:

parameterMap: function(options, operation) 
              {
                if (operation == "read" && options.models) 
                {     
                   return {models: kendo.stringify(options.models)};
                 };
                if (operation == "create" && options.models) 
                {
                   options.models[0].idPai = currentId;
                  $('#gridBaseDados').data('kendoGrid').dataSource.read();

                  return {models: kendo.stringify(options.models)};
                };

                  if (operation !== "read" && options.models) 
                  {
                      return {models: kendo.stringify(options.models)};
                  }
              }

在我的服务器端,我(通常)是这样的:

else
{
    $verb = $_SERVER["REQUEST_METHOD"];


    if ($verb == "GET") 
    {
        header("Content-type: application/json");

        $arr = array();
        $statement = $pdo->query("SELECT a.idDocumento, a.idPai, a.tipo,a.nome,a.dataCriacao,
                a.dataModificacao,b.nome as NomeUtilizadorCriador,
                c.nome as NomeUtilizadorUpdate
                FROM Documento a left outer join Utilizador b on a.idUtilizadorCriador=b.idUtilizadores  
                                  left outer join Utilizador c on a.idUtilizadorUpdate=c.idUtilizadores");

        $statement->execute();
        $result = $statement->fetchAll(PDO::FETCH_ASSOC);

        echo json_encode(array("data" => $result, "user" => $_SESSION["user"]));
    } 

这是有效的,但我需要额外的参数,因为我需要将它与我的" idPai"进行比较。

再次感谢。

1 个答案:

答案 0 :(得分:1)

你不需要在这里发帖。使用parameterMap选项发送数据。

$formData = $_GET["access_token"];
echo $formData;

<强> PHP

{{1}}