php - 通过邮寄方式发送数据和通过邮寄方式接收数据

Question

我已经在我写 ID 的地方输入并点击按钮来检查我是否有这样的 ID(按功能)。数据通过 jquery 和运行功能进入第 2 页进行检查，我什至可以将其打印到屏幕上。我需要将数据从第二页的功能发送到我的主页并使用这些数据。如何在不提交的情况下从第 2 页发送它？

我的第一页:

$(function() {
  $("#checkfin").click(function() {
    var fin = $("#pinofsv").val();

    $.post("checkfin.php", {
      ffin: fin
    }, function(result) {
      $('#result').append(result);
    });
  });
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap-theme.min.css" rel="stylesheet" />
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" rel="stylesheet" />

<form name="formids" id="formids" action="" method="POST">
  <div class="row">
    <div class="col-lg-12">
      <div class="panel panel-default">
        <div class="panel-body">
          <div class="col-sm-12">
            <div class="row">
              <div class="col-sm-3">
                <label>Enter your ID: <span style="color:#FF0000;">*</span></label>
              </div>
              <div class="col-sm-4">
                <div class="input-group">
                  <input type="text" class="form-control input-sm" name="pinofsv" id="pinofsv" required />
                  <div class="input-group-addon"><a href="#" onmouseover="document.getElementById('place-holder-1').src='../svphoto.jpg';" onmouseout="document.getElementById('place-holder-1').src='';"><i class="glyphicon glyphicon-info-sign blue"></i><img src="" id="place-holder-1" style="z-index: 100; position: absolute;right: 1%;" /></a></div>
                </div>
              </div>
              <div class="col-sm-5">
                <div class="col-sm-5">
                  <button class="btn btn-primary" name="checkfin" id="checkfin" onclick="return false">Check </button>
                </div>
                <div class="col-sm-7">

                </div>
              </div>
            </div>
          </div>
        </div>
      </div>
    </div>
  </div>
</form>

<div id="result"></div>

---

我的第二页:

if(isset($_POST["ffin"]) && $_POST["ffin"] != "")
{
  $fintrue = definePINNew($fin);
  if($fintrue != "")
	{
    //do something for example print to screen data from function
  }
  
 }

Answer 1

只要你发送请求，你就会得到响应

if(isset($_POST["ffin"]) && $_POST["ffin"] != "")
{
  $fintrue = definePINNew($fin);
 if($fintrue != "")
   {
   echo 'JUST SEND YOUR DATA IT WILL BE THERE IN JQUERY RESPONSE';

  Eg: echo json_encode($fintrue['anydata']); 
  // it will give you $_POST data in jquery response.
  // which is at your first page.
  // And parse data in JSON can easily get in jquery response


   }
 }

$(function() {
 $("#checkfin").click(function() {
   var fin = $("#pinofsv").val();
  $.post("checkfin.php", {
   ffin: fin
   }, function(result) {
   console.log(result.anydata); // you will get the data from second page here.
  // result.anydata you can place it anywhere.
  });
 });
});