请找到代码,
json = JSON.parse(e.target.result);
$.ajax({
type: "POST",
enctype: 'multipart/form-data',
url: "jsnservlt",
data: json,
processData: false,
contentType: false,
cache: false,
timeout: 600000,
success: function(dataa) {
var resp =JSON.parse(data);
console.log(resp);
alert(dataa);
},
error: function(error) {
console.log(error);
}
});
如何在servlet中检索json?