jQuery Validator,以编程方式显示错误

Question

我可以这样做:

validator.showErrors({ "nameOfField" : "ErrorMessage" });

这工作正常,但如果我尝试做这样的事情:

var propertyName = "nameOfField";
var errorMessage = "ErrorMessage";
validator.showErrors({ propertyName : errorMessage });

它会抛出'element is undefined'错误.

Answer 1

关于什么:

var propertyName = "nameOfField";
var errorMessage = "ErrorMessage";

var obj = new Object();
obj[propertyName] = errorMessage;

validator.showErrors(obj);

值得注意的是,以下三种语法是等效的:

var a = {'a':0, 'b':1, 'c':2};

var b = new Object();
b['a'] = 0;
b['b'] = 1;
b['c'] = 2;

var c = new Object();
c.a = 0;
c.b = 1;
c.c = 2;