QUnit.test 运行 在 for 循环中
QUnit.test running in a for loop
我正在尝试为 QUnit 测试做一些代码重构。我使用 json 数组 TestCaseSource 来存储测试用例的输入和预期输出，并像这样编码，
var data = TestCaseSource.data;
for (var i in data) {
console.log(data[i]);
QUnit.test(data[i].TestCaseName, function () {
DoProcess(data[i].TestCaseName, "", data[i]);
});
}
使用这段代码，我只能得到第一个和最后一个测试用例运行。 QUnit 省略了中间的所有情况。如果我删除了 for 循环，并硬编码 QUnit.test 就像
QUnit.test('TestCaseName1', function () {
DoProcess('TestCaseName1', "", TestCaseSource.data[0]);
});
QUnit.test('TestCaseName2', function () {
DoProcess('TestCaseName2', "", TestCaseSource.data[1]);
});
...
那时一切都很好。为什么 for 循环不起作用？
我在 Asynchronous Process inside a javascript for loop 中找到了答案。正如它在这个非常好的 post:
中所说的
You have to freeze the value of i by passing it into a function
somewhere so it's value exists uniquely for each iteration of the loop
in a function closure. Otherwise, all asynch callbacks will just see
the value of i at the end of the loop which is the value it has when
they execute their callbacks (sometime later when the loop has
finished), not each their own value.
我正在尝试为 QUnit 测试做一些代码重构。我使用 json 数组 TestCaseSource 来存储测试用例的输入和预期输出，并像这样编码，
var data = TestCaseSource.data;
for (var i in data) {
console.log(data[i]);
QUnit.test(data[i].TestCaseName, function () {
DoProcess(data[i].TestCaseName, "", data[i]);
});
}
使用这段代码，我只能得到第一个和最后一个测试用例运行。 QUnit 省略了中间的所有情况。如果我删除了 for 循环，并硬编码 QUnit.test 就像
QUnit.test('TestCaseName1', function () {
DoProcess('TestCaseName1', "", TestCaseSource.data[0]);
});
QUnit.test('TestCaseName2', function () {
DoProcess('TestCaseName2', "", TestCaseSource.data[1]);
});
...
那时一切都很好。为什么 for 循环不起作用？
我在 Asynchronous Process inside a javascript for loop 中找到了答案。正如它在这个非常好的 post:
中所说的You have to freeze the value of i by passing it into a function somewhere so it's value exists uniquely for each iteration of the loop in a function closure. Otherwise, all asynch callbacks will just see the value of i at the end of the loop which is the value it has when they execute their callbacks (sometime later when the loop has finished), not each their own value.