表单使用Ajax上传图片
HTML代码
<form id="up-form">
<input type="file" name="img" onchange="ajaxUp()" />
</form>
Javascript代码
function ajaxUp() {
var data = new FormData($("#up-form")[0]);
var url = "upload.php";
$.ajax({
url: url,
type: 'POST',
data: data,
async: false,
cache: false,
contentType: false,
processData: false,
dataType: "json",
success: function (result) {
console.log(result.msg,false);
},
error: function (result) {
console.log("网络传输错误",false);
}
});
}