javascript 下拉菜单选择后自动发布
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dropdown menu to auto post when selected
提问by crashwap
How would I make a drop down menu post automatically when an option has been chosen?
选择一个选项后，如何自动制作下拉菜单帖子？
<SELECT NAME="select_page"><?php echo $options1;?></SELECT>
What is the best way to achieve this?
实现这一目标的最佳方法是什么？
回答by Nick Rolando
You can utilize the onchangeevent for the selectelement, and code a form submit:
您可以将onchange事件用于select元素，并编写表单提交代码：
<form method="post">
<select name="myselect" onchange="this.form.submit();">
<option>blue</option>
<option>red</option>
</select>
</form>
That will automatically submit the form when the value is changed.
这将在更改值时自动提交表单。
回答by crashwap
Note that jQuery is referencing the <select>tag by its ID, so put one in the tag. Also, you can use AJAX to post your data if you wish to receive something back.
请注意，jQuery<select>通过其 ID引用标记，因此在标记中放置一个。此外，如果您希望收到回复，您可以使用 AJAX 发布您的数据。
<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('#Sel').change(function() {
var opt = $(this).val();
$.ajax({
type: "POST",
url: "receiving_file.php",
data: 'selected_opt=' + opt,
success:function(data){
alert('This was sent back: ' + data);
}
});
});
});
</script>
</head>
<body>
<select id = "Sel">
<option value ="Song1">default value<br>
<option value ="Song2">Break on through<br>
<option value ="Song3">Time<br>
<option value ="Song4">Money<br>
<option value ="Song5">Saucerful of Secrets
</select>
