我在parse.com上遇到问题,我在其中获取类型的id并将其传递给另一个上传图像的函数。然后在图像中取出type.id并将其发布到另一个将数据保存到类中的函数。
这是我迄今为止尝试过的但没有成功。
- OnClick代码
$('#submitId').on("click", function(e, f) {
e.preventDefault();
typeSave(typeid1);
//var objnew1 = typeSave();
console.log("inside onclick " + type2);
var fileUploadControl = $("#profilePhotoFileUpload")[0];
var file = fileUploadControl.files[0];
var name = file.name; //This does *NOT* need to be a unique name
var parseFile = new Parse.File(name, file);
parseFile.save().then(
function() {
//typeSave();
type2 = typeid1;
saveJobApp(parseFile, type2);
console.log("inside save onclick " + type2);
},
function(error) {
alert("error");
}
);
});
- 输入代码
var type;
var typeid1;
var type2;
function typeSave() {
var type = new Parse.Object("type");
var user = new Parse.Object("magazia");
//var bID = objbID;
//user.id = bID;
var cafebar = document.getElementById('cafe_bar').checked;
if (cafebar) {
var valueCafebar = true;
} else {
var valueCafebar = false;
}
var club = document.getElementById('club').checked;
if (club) {
var valueClub = true;
} else {
var valueClub = false;
}
var restaurant = document.getElementById('restaurant').checked;
if (restaurant) {
var valueRestaurant = true;
} else {
var valueRestaurant = false;
}
var pistes = document.getElementById('pistes').checked;
if (pistes) {
var valuePistes = true;
} else {
var valuePistes = false;
}
type.set("cafebar", valueCafebar);
type.set("club", valueClub);
type.set("restaurant", valueRestaurant);
type.set("pistes", valuePistes);
type.save(null, {
success: function(type) {
//saveJobApp(type.id);
var typeid1 = type.id;
console.log("inside type save " + typeid1);
//return ;
},
error: function(type, error) {
alert('Failed to create new object, with error code: ' + error.description);
}
});
}
- 将数据发送到parse.com类代码
function saveJobApp(objParseFile, type2) {
var jobApplication = new Parse.Object("magazia");
var email = document.getElementById('email').value;
var name = document.getElementById('name').value;
var description = document.getElementById('description').value;
var website = document.getElementById('website').value;
var phone = document.getElementById('phone').value;
var address = document.getElementById('address').value;
var latlon = document.getElementById('latlon').value;
var area = document.getElementById('area').value;
var value = latlon;
value = value.replace(/[\(\)]/g, '').split(', ');
console.log("inside saveJobApp " + type2);
var x = parseFloat(value[0]);
var y = parseFloat(value[1]);
var point = new Parse.GeoPoint(x, y);
jobApplication.set("image", objParseFile);
jobApplication.set("email", email);
jobApplication.set("phone", phone);
jobApplication.set("address", address);
jobApplication.set("name", name);
jobApplication.set("website", website);
jobApplication.set("description", description);
jobApplication.set("area", area);
jobApplication.set("latlon", point);
jobApplication.set("typeID", type2);
jobApplication.save(null, {
success: function(gameScore) {
// typeSave(jobApplication.id);
},
error: function(gameScore, error) {
alert('Failed to create new object, with error code: ' + error.description);
}
});
}
所以当我点击按钮首先运行typesave()
函数,然后它在解析时在类型类上发布类型,从成功函数获取type.id并发送它时,我正在尝试恢复到parseFile.save().then
然后将objectFile
和type2
(type.id
)发送到saveJobApp
并将它们保存在课程magazia
中
我从console.logs得到的是这个
这意味着我的代码发布到类型类并采用type.id
但它并没有通过parsefile保存将它发送到
magazia
类。
知道我错过了什么吗?
答案 0 :(得分:1)
我注意到你的错误不是关于函数,而是关于尝试将type.id作为字符串传递而不是作为saveJobApp函数中的函数传递。
如果你尝试这样做
function saveJobApp(objParseFile , objtype) {
var jobApplication = new Parse.Object("magazia");
var type = new Parse.Object("type");
type.id = objtype;
jobApplication.set("typeID", type);
我认为它会奏效。
并且还将onclick和ParseFile保存代码更新为此
$('#submitId').on("click", function(e) {
typeSave();
});
function PhotoUpload(objtype){
var fileUploadControl = $("#profilePhotoFileUpload")[0];
var file = fileUploadControl.files[0];
var name = file.name; //This does *NOT* need to be a unique name
var parseFile = new Parse.File(name, file);
parseFile.save().then(
function() {
saveJobApp(parseFile, objtype);
},
function(error) {
alert("error");
}
);
}
typeSave()中的成功函数 应该是这样的
type.save(null, {
success: function(type) {
PhotoUpload(type.id);
},
希望这会有所帮助:)