PHP没有从$ .ajax

时间:2016-04-18 17:59:55

标签: javascript php jquery html ajax

我有一个运行POST方法的JavaScript,一旦我的datepicker失去焦点(我也在常规提交按钮上尝试了这个)并运行脚本rent-fetch-pick-up-point.php。 PHP运行,但它没有超过if语句,因为我没有获得POST数据。 datepicker绑定到输入字段time-period-from

datepickerTo.blur(function(){
  if (selectedDateFrom.length > 0) {

    datepickerFrom.delay(500).queue(function(){

      $.ajax({
        type: "POST",
        url: "include/rent-fetch-pick-up-point.php",
        data: {action: selectedDateFrom},
        success: function(data) {
          $("#pick-up-point-container").html(data);
        }
      });
    });
  }
});

这是PHP代码:

if (isset($_POST['time-period-from'])) {
  require '../include/connection.php';

  $dateFrom = $_POST['time-period-from'];
  $sql = "SELECT * FROM order WHERE $dateFrom BETWEEN date_from AND date_to";
  $result = mysqli_query($connection, $sql);
  $numRows = mysqli_num_rows($result);

  echo $sql; // For testing purposes
}

以下是HTML:

  <input type="text" name="time-period-from" id="datepicker-from" class="datepicker"></p>

我也尝试使用$ .post()而不是$ .ajax(),但我遇到了同样的问题:

$.post("include/rent-fetch-pick-up-point.php", {name: selectedDateTo}, function(data) {

  $("#pick-up-point-container").text(data)

});

3 个答案:

答案 0 :(得分:2)

gl.glMatrixMode(GL2.GL_PROJECTION); gl.glLoadIdentity(); gl.glOrtho(left, right, bottom, top, zNear, zFar); gl.glMatrixMode(GL2.GL_MODELVIEW); gl.glLoadIdentity(); 的键来自您传递给$_POST选项的对象的键,而不是最初来源的表单字段的名称。自从您使用:

data:

值将在data: { action: selectedDateFrom } ,而不是$_POST['action']。所以你需要使用:

$_POST['time-period-from']

if (isset($_POST['action']))

或者您可以将Javascript更改为:

$dateFrom = $_POST['action'];

答案 1 :(得分:0)

我认为你的selectedDateFrom变量是一个数组,导致你的帖子信息无法正常使用。

data: {action: $('#selectedDateFrom').serializeArray()}

然后您可以正确获取表单数据

答案 2 :(得分:0)

你并没有在PHP方面抓住正确的变量:

if (isset($_POST['action'])) {
  require '../include/connection.php';

  $dateFrom = $_POST['action'];
  $sql = "SELECT * FROM order WHERE $dateFrom BETWEEN date_from AND date_to";
  $result = mysqli_query($connection, $sql);
  $numRows = mysqli_num_rows($result);
  echo $sql; // For testing purposes
}