Question

我想用AJAX执行提交并选择列表。我做了一个简单的警报，但没有。我错了什么吗？这是我的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

试试这个：

<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
    })
}

jquery提交表单，选择不工作

1 个答案: