<script type="text/javascript">
$(function() {
$('#uploadSubmit').click(function() {
//将这行替换为document.getElementById 就解决了---------
//var data = new FormData($('#uploadForm')[0]);
var data = new FormData(document.getElementById('uploadForm')[0]);
console.log(data.toString());
$.ajax({
url: 'http://127.0.0.1:8080/face/uploadImg',
type: 'POST',
data: data,
async: false,
cache: false,
contentType: false,
processData: false,
success: function(data) {
console.log(data);
if(data.status) {
console.log('upload success');
} else {
console.log(data.message);
}
},
error: function(data) {
console.log(data.status);
}
});
});
})
</script>