我试图通过ajax获取数据,然后通过表单发送。但它似乎不起作用。你知道我做错了什么吗?
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<form method="get" action="test.php">
<input id="myvar" type="hidden" name="albumid" />
<button type="submit" id="btnsubmit">Submit</button>
</form>
<script type="text/javascript">
$('form').submit(function() {
$.ajax({
url: "newAlbum.php",
success: function(data){
var album = data;
$('#myvar').val(album);
}
});
});
</script>
newAlbum.php
<?PHP echo '11'; ?>
test.php
<?php echo $_GET["albumid"]; ?>
好的,尝试添加如下:
$('form').submit(function() {
$.ajax({
url: "newAlbum.php",
async:false,
success: function(data){
var album = data;
$('#myvar').val(album);
}
});
});
data
参数。但我也会使用$(this).serialize()
,因为这将获取所有表单输入元素。
<script type="text/javascript">
$('form').submit(function() {
$.ajax({
url: "newAlbum.php",
data: $(this).serialize(),
success: function(data){
var album = data;
$('#myvar').val(album);
},
error : function(data) {
console.log(data);
}
});
});
</script>
尝试以JSON 形式接收数据
$.getJSON()