无法在jQuery post中获取select选项值

HTML:

<select id="s2_basic" class="form-control" id="userid" name="userid">
    <option value="" selected>-- Choose option --</option>
    <option value="1">One</option>
    <option value="2">Two</option>
    <option value="3">Three</option>
</select>

Javascript:

function submitForm() {
    var userid = $('#userid').attr('selected', 'selected');

    $.ajax({
        type: "POST",
        url: "submit.php",
        data: "userid=" + userid,
        success: function(text) {
            if (text == "success") {
                formSuccess();
            } else {
                formError();
                submitMSG(false,text);
            }
        }
    });
}

Submit PHP returns error message:

提交PHP返回错误消息:

Value: ([object Object])

I have also tried

我也试过

var userid = $('#userid').val();
var userid = $('#userid option').val();

Still, the option value never gets passed correctly. What am I doing wrong?

不过，选项值永远不会被正确地传递。我做错了什么?

I want just the value="x" number to be posted as userid=x.

我只想让值="x"号作为userid=x。

2 个解决方案

#1

remove id="s2_basic" from your select.

从您的选择中删除id="s2_basic"。

change :

变化:

<select id="s2_basic" class="form-control" id="userid" name="userid">

来

<select class="form-control" id="userid" name="userid">

#2

$('#userid').on('change',function(){

console.log('value '+$('#userid option:selected').val())
console.log('text '+$('#userid option:selected').text())
console.log('index '+$('#userid option:selected').index())

})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
<select class="form-control" id="userid" name="userid">
    <option value="" selected>-- Choose option --</option>
    <option value="1">One</option>
    <option value="2">Two</option>
    <option value="3">Three</option>
</select>

try this

试试这个

#1