问题
I came across a strange behaviour when doing some regular expressions in JavaScript today (Firefox 3 on Windows Vista).
var str = "format_%A";
var format = /(?:^|\s)format_(.*?)(?:\s|$)/.exec(str);
console.log(format); // ["format_%A", "%A"]
console.log(format[0]); // "format_undefined"
console.log(format[1]); // Undefined
There's nothing wrong with the regular expression. As you can see, it has matched the correct part in the first console.log call.
Internet Explorer 7 and Chrome both behave as expected: format[1] returns "%A" (well, Internet Explorer 7 doing something right was a bit unexpected...)
Is this a bug in Firefox, or some "feature" I don't know about?
回答1:
This is because console.log() works like printf(). The first argument to console.log() is actually a format string which may be followed with additional arguments. %A is a placeholder. For example:
console.log("My name is %A", "John"); // My name is "John"
See console.log() documentation for details. %A and any other undocumented placeholders seem to do the same as %o.
回答2:
Seems like %A somehow translates into the string undefined.
Try escaping the %A part, I think that will solve the problem.
来源：https://stackoverflow.com/questions/432826/why-is-a-matched-substring-returning-undefined-in-javascript