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PHP无法从$ .ajax获取POST数据

[英]PHP not getting POST data from $.ajax

I have a JavaScript that runs a POST method once my datepicker has been out of focus (I also tried this on a regular submit button) and runs the script rent-fetch-pick-up-point.php . 我有一个JavaScript,一旦我的datepicker失去焦点(我也在常规rent-fetch-pick-up-point.php按钮上尝试过此操作),它就会运行POST方法并运行脚本rent-fetch-pick-up-point.php The PHP runs, however it doesn't get past the if-statement because my it's not getting the POST data. PHP可以运行,但是它不能通过if语句,因为它没有获取POST数据。 The datepicker is tied to a input field time-period-from 日期选择器绑定到输入字段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);
        }
      });
    });
  }
});

Here is the PHP code: 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
}

And here's the HTML: 这是HTML:

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

I also tried using $.post() instead of $.ajax(), but I ran into the same issue: 我也尝试使用$ .post()而不是$ .ajax(),但是遇到了同样的问题:

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

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

});

The keys of $_POST come from the keys of the object you pass to the data: option, not the names of the form fields where the values originally came from. $_POST的键来自传递给data:选项的对象的键,而不是值最初来自的表单字段的名称。 Since you used: 由于您使用过:

data: { action: selectedDateFrom }

the value will be in $_POST['action'] , not $_POST['time-period-from'] . 该值将在$_POST['action'] ,而不是$_POST['time-period-from'] So you need to use: 因此,您需要使用:

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

and: 和:

$dateFrom = $_POST['action'];

or you could change the Javascript to: 或者您可以将Javascript更改为:

data: { "time-period-from": selectedDateFrom }

I think your selectedDateFrom variable is array that cause your post info can't you get properly . 我认为您的selectedDateFrom变量是数组,导致您的帖子信息无法正确获取。

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

then you get your form data properly 然后您可以正确获取表单数据

You aren't grabbing the right variable on the PHP side: 您没有在PHP方面获取正确的变量:

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
}

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