不传递给PHP的AJAX数据

Question

我很难将AJAX数据传递到PHP。我对PHP有经验，但对JavaScript还不熟悉。

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>

$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和PHP代码是可以工作的，因为我可以手动创建$commodity_code变量，并且脚本工作得很好。我认为我的问题在于将AJAX数据传递给我的PHP脚本。

Answer 1

忘记在AJAX调用中添加method: 'POST'。你的电话有点问题。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);
});

Answer 2

这一行错误data: "commodity_code: commodity_code", ..。您可以简单地传递commodity_code变量。

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