我收到了来自服务器的响应,如何只打印消息在$('#message'(div响应是jsonencoded php
{
"status":false,
"message":"<div id='errmsg' class='alert alert-danger'><button type='button' class='btn btn-info pull-right' id='remove'>Remove<'/button><p> Username Is Required<'/p>'n<p> Password Is Required<'/p>'n<p> Name Is Required<'/p>'n<p> Designation Is Required<'/p>'n<'/div>"
}
这是我使用iv 的axax代码
<script type="text/javascript">
$("#userform").submit(function(e) {
e.preventDefault();
var url = $(this).attr('action');
var formData = new FormData($(this)[0]);
$.ajax({
type: "POST",
url: url,
data: formData,
processData: false,
contentType: false,
success: function(data)
{
console.log(data);
$('#message').html(data.message);
}
});
});
</script>
如果我尝试data['message']
您必须在ajax请求选项中传递dataType:'json'
,所以jQuery会自动将JSON字符串转换为JSON对象。
<script type="text/javascript">
$("#userform").submit(function(e) {
e.preventDefault();
var url = $(this).attr('action');
var formData = new FormData($(this)[0]);
$.ajax({
type: "POST",
url: url,
data: formData,
processData: false,
dataType: 'json',
success: function(data)
{
console.log(data);
$('#message').html(data.message);
}
});
});
</script>