Question

使用如下所示的缩小代码，

f&&!f.error?k.button.b==k.button.c.G?k.button.Q(b,e,f,c,d):k.button.b==k.button.c.o&&k.button.P(b,e,f,c,d):(console.error(f),f=f.error.message||chrome.i18n.getMessage("error_tooltip"),k.button.v(b.id,f),d({action:"error"}))

是否有一个自动化工具可以将一行条件运算符转换为一系列if语句？

示例1：

这

(i < 0 ? function1() : function2())

到

if (i < 0) {
    function1();
} else {
    function2();
}

示例2：

这

(i < 0 ? function1() : (i === 0 ? function2() : function3()))

到

if (i < 0) {
    function1();
} else {
    if (i === 0) {
        function2();
    } else {
        function3();
    }
}

Answer 1

这

f&&!f.error?k.button.b==k.button.c.G?k.button.Q(b,e,f,c,d):k.button.b==k.button.c.o&&k.button.P(b,e,f,c,d):(console.error(f),f=f.error.message||chrome.i18n.getMessage("error_tooltip"),k.button.v(b.id,f),d({action:"error"}))

到

if (f && !f.error)
{
    if (k.button.b == k.button.c.G)
    {
        k.button.Q(b, e, f, c, d)
    }
    else
    {
        k.button.b == k.button.c.o && k.button.P(b, e, f, c, d)
    }
}
else
{
    (console.error(f), f = f.error.message || chrome.i18n.getMessage("error_tooltip"), k.button.v(b.id, f), d(
    {
        action: "error"
    }))
}

使用此（然后使用JSBeautifier）：

/*jslint browser: true, vars: true, white: true, maxerr: 50, indent: 4 */
(function (console)
{
    "use strict";

    function transform(string)
    {
        var questionMark = string.indexOf("?");
        var colon = string.indexOf(":", questionMark);

        if (questionMark === -1 || colon === -1)
        {
            return string;
        }

        var condition = string.substring(0, questionMark);
        var expressions = string.substring(questionMark + 1, string.length);
        var trueExpression = null;
        var falseExpression = null;

        console.log("expressions: " + expressions);

        // While looking in pairs, find the location where the colon occurs before the question mark.
        questionMark = expressions.indexOf("?");
        colon = expressions.indexOf(":");
        while ((questionMark !== -1 && colon !== -1) && (questionMark < colon))
        {
            questionMark = expressions.indexOf("?", questionMark + 1);
            colon = expressions.indexOf(":", colon + 1);
        }

        console.log("\t" + "questionMark: " + questionMark);
        console.log("\t" + "colon: " + colon);

        trueExpression = expressions.substring(0, colon);
        falseExpression = expressions.substring(colon + 1, expressions.length);

        console.log("condition: " + condition);
        console.log("trueExpression: " + trueExpression);
        console.log("falseExpression: " + falseExpression);

        console.log("-");

        return ("if (" + condition + ") {\n" + transform(trueExpression) + "\n} else {\n" + transform(falseExpression) + "\n}");
    }

    function unittest()
    {
        console.log(transform("(i < 0 ? function1() : function2())"));
        console.log("---");
        console.log(transform("i < 0 ? function1() : function2()"));
        console.log("---");
        console.log(transform("i < 0 ? function1() : i === 0 ? function2() : function3()"));
        console.log("---");
        console.log(transform("i > 0 ? i === 1 ? function1() : function2() : function3()"));
        console.log("---");
        console.log(transform("i > 0 ? i === 1 ? function1() : i === 2 ? function2() : function3() : function4()"));
        console.log("---");
        console.log(transform("i > 0 ? i === 1 ? function1() : i === 2 ? function2() : function3() : i === 0 ? function4() : function5()"));
        console.log("---");
        console.log(transform("f&&!f.error?k.button.b==k.button.c.G?k.button.Q(b,e,f,c,d):k.button.b==k.button.c.o&&k.button.P(b,e,f,c,d):(console.error(f),f=f.error.message||chrome.i18n.getMessage(\"error_tooltip\"),k.button.v(b.id,f),d({action:\"error\"}))"));
    }

    unittest();
}(window.console));

将三元条件运算符转换为if语句？

1 个答案: