Question

我想執行提交與AJAX和選擇列表。我做了一個簡單的警報，但沒有。我錯了什麼？ 這是我的html：

  <form id='bill_action' action='post'> 
      <input type='hidden' name='bill_id' value='".$v1['id']."'> 
      <select name='action' onchange='this.form.submit()'> 
       <option value='test1'>test1</option> 
       <option value='test2'>test2</option> 
      </select> 

      </form> 

我的JS：

$("#bill_action").submit(function() { 
    alert('hello world'); //for test 
    $.ajax({ //make ajax request to bill_action.php 
     url: "bill_action.php", 
     type: "POST", 
     dataType:"json", //expect json value from server 
     data: bill_id 
    }).done(function(data){ 
//on Ajax success 
    }) 
}) 

Answer 1

試試這個：

<form id='bill_action' action='post'> 
<input type='hidden' name='bill_id' value='".$v1['id']."'> 
<select name='action' onchange='myfunction'> 
    <option value='test1'>test1</option> 
    <option value='test2'>test2</option> 
</select> 

</form> 


function myfunction() { 
    alert('hello world'); //for test 
    $.ajax({ //make ajax request to bill_action.php 
     url: "bill_action.php", 
     type: "POST", 
     dataType:"json", //expect json value from server 
     data: bill_id 
    }).done(function(data){ 
     //on Ajax success 
    }) 
}