javascript – AJAX数据没有传递给PHP

我无法将AJAX数据传递给PHP.我对PHP很有经验,但对JavaScript很新.

HTML / JavaScript

<input type="text" id="commodity_code"><button id="button"> = </button>

<script id="source" language="javascript" type="text/javascript">

$('#button').click(function()
{
  var commodity_code = $('#commodity_code').val();

  $.ajax({                                      
  url: 'get_code.PHP',  
  data: "commodity_code: commodity_code",
  dataType: 'json',
  success:function(data) {
        var commodity_desc = data[0];
        alert(commodity_desc);
    }  
  });
}); 

</script>

PHP

$commodity_code = $_POST['commodity_code'];

$result = MysqL_query("SELECT description FROM oc_commodity_codes WHERE code = '$commodity_code'");
$array = MysqL_fetch_row($result);
echo json_encode($array);

我知道一般的AJAX提取和PHP代码正在工作,因为我可以手动创建$commodity_code变量,脚本工作正常.我认为我的问题在于将AJAX数据传递给我的PHP脚本.

解决方法:

你忘了在你的AJAX Call中添加方法：’POST’.你的电话有问题.检查如下：

$.ajax({                                      
  url: 'get_code.PHP',
  method: "POST",                         // Change here.
  data: {commodity_code: commodity_code}, // Change here.
  dataType: 'json',                       
  success:function(data) {
        var commodity_desc = data[0];
        alert(commodity_desc);
  }  
});

或者为了简单起见,使用速记功能：

$.post('get_code.PHP', {commodity_code: commodity_code}, function(data) {
  var commodity_desc = data[0];
  alert(commodity_desc);
});