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AJAX数据没有传递给PHP

[英]AJAX data not passing to PHP

I am having trouble passing AJAX data to PHP. 我无法将AJAX数据传递给PHP。 I am experienced with PHP but new to JavaScript. 我对PHP很有经验,但对JavaScript很新。

HTML / JavaScript 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 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);

I know the general AJAX fetch and PHP code is working as I can manually create the $commodity_code variable and the script works fine. 我知道一般的AJAX提取和PHP代码正在工作,因为我可以手动创建$ commodity_code变量,脚本工作正常。 I think my issue lies somewhere in passing the AJAX data to my PHP script. 我认为我的问题在于将AJAX数据传递给我的PHP脚本。

You forgot to add the method: 'POST' in your AJAX Call. 你忘了在你的AJAX Call中添加method: 'POST' And you have some issues with your call. 你的电话有问题。 Check below: 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);
  }  
});

Or to make it simple, use the shorthand function: 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);
});

error in this line data: "commodity_code: commodity_code", .. you can simple pass the commodity_code variable.. 这行data: "commodity_code: commodity_code",错误data: "commodity_code: commodity_code", ..你可以简单地传递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);
  }  
});

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