Why am I seeing inconsistent JavaScript logic behavior looping with an alert() vs. without it?
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Ok, i see it now. The key to your problem is the use of the g (global match) flag: when this is specified for a regex, it will be set up such that it can be executed multiple times, beginning each time at the place where it left off last time. It keeps a "bookmark" of sorts in its lastIndex property:
var testRegex = /blah/ig;
// logs: true 4
console.log(testRegex.test("blah blah"), testRegex.lastIndex);
// logs: true 9
console.log(testRegex.test("blah blah"), testRegex.lastIndex);
// logs: false 0
console.log(testRegex.test("blah blah"), testRegex.lastIndex);
The above example creates an instance of a very simple regex: it matches "blah", upper or lower case, anywhere in the string, and it can be matched multiple times (the g flag). On the first run, it matches the first "blah", and leaves lastIndex set to 4 (the index of the space after the first "blah"). The second run starts matching at the lastIndex , matches the second blah, and leaves lastIndex set to 9 - one past the end of the array. The third run doesn't match - lastIndex is bogus - and leaves lastIndex set to 0. A fourth run would therefore have the same results as the first.
Now, your expression is quite a bit more greedy than mine: it will match any number of any characters before or after "blah". Therefore, no matter what string you test on, if it contains "blah" it will always match the entire string and leave lastIndex set to the length of the string just tested. Meaning, if you were to call test() twice, the second test would always fail:
var filterRegex = /.*blah.*/ig;
// logs: true, 9
console.log(filterRegex.test("blah blah"), filterRegex.lastIndex);
// logs: false, 0
console.log(filterRegex.test("blah blah"), filterRegex.lastIndex);
Fortunately, since you create your regex immediately prior to calling test() , and never call test() more than once, you'll never run into unexpected behavior... Unless you're using a debugger that lets you add in another call to test() on the side. Yup. With Firebug running, a watch expression containing your call to test() will result in intermittent false results showing up, either in your code or in the watch results, depending on which one gets to it first. Driving you slowly insane...
Of course, without the g flag, livin' is easy:
var filterRegex = /.*blah.*/i;
// logs: true, 0
console.log(filterRegex.test("blah blah"), filterRegex.lastIndex);
// logs: true, 0
console.log(filterRegex.test("blah blah"), filterRegex.lastIndex);
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I just can't imagine there is any situation where two JavaScript expressions evaluate to true individually, but not when combined.
Are you sure both expressions actually produce a boolean value every time? (Okay, to make regex.test() not produce a boolean value is difficult, but how about event.show . Might that be undefined at times?
Do you refer to the correct index when saying event[0].show , wouldn't you mean event[i].show ?
看起来你正面临事件数组的一些竞争条件，这就是为什么当你使用alert（）时一切正常。
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