php - 将ajax的变量传递给PHP

Question

<select type="category" id="x1">
  <option value="country">Country</option>
  <option value="city">City</option>
</select>

<script type="text/javascript">
 $(document).ready(function(){
 $('#x1').on('change',function(){
 var x1 = $(this).val();

  if(x1){
   $.ajax({
      type: 'POST',
      url: 'ajaxData.php',
      data: {'x1'},
      success: function(data){
      console.log("success");
      }
    }); 
   }
  });
});

//script.php
<?php
  $var1 = $_POST['x1'];
  print_r ($var1);
?>

我试图从select（国家或城市）获取价值并传递给'script.php'。新手，请原谅任何愚蠢的错误。

标签： phpajax

Answer 1

data参数必须是一个key:value对象，所以：

$.ajax({
  type: 'POST',
  url: 'ajaxData.php',
  data: {x1: x1},  // here first x1 is a key name, second x1 is selected value
  success: function(data){
  console.log("success");
  }
});