Question

有没有办法简化这个或我可以使用的更简洁的形式？包含的逻辑都是正确的。它看起来好像很多return和else if。

mode = (function(mode, current, proposed, origins, destinations) {
            if (mode === 'none') {
                return 'project';
            } else if (proposed.count === 0) {
                return 'unseated';
            } else if (current.count > proposed.count && proposed.count > 0) {
                return 'reducing';
            } else if (proposed.count === destinations.count && destinations.count > 1 && current.count === 0) {
                return 'newplus';
            } else if (proposed.count === destinations.count && current.count === 0) {
                return 'new';
            } else if (proposed.count > destinations.count) {
                return 'increasing';
            } else if (proposed.count === destinations.count && destinations.count > origins.count) {
                return 'moveplus';
            } else {
                return 'move';
            }
        }(moves.register[staff].move, {count: mode.currentDesks}, {count: mode.proposedDesks}, {count: mode.origins}, {count: mode.destinations})));

我之前使用过嵌套的三元缩写，但是我认为更容易出错并且难以阅读（由于代码演变而导致的结果略有不同）： p>

mode = 
(moves.register[staff].move === 'none') ? 'project' :
    (mode.proposedDesks === 0) ? 'unseated' :
        (mode.currentDesks > mode.proposedDesks && mode.proposedDesks > 0) ? 'reducing' :
            (mode.proposedDesks === mode.destinations && mode.destinations > 1 && mode.currentDesks === 0) ? 'newplus' :
                (mode.proposedDesks === mode.destinations && mode.currentDesks === 0) ? 'new' :
                    (mode.proposedDesks > mode.destinations) ? 'additional' :
                        (mode.proposedDesks === mode.destinations && mode.destinations === mode.origins) ? 'move' :
                            (mode.proposedDesks === mode.destinations && mode.destinations > mode.origins) ? 'moveplus' :
                                'other';

因此，虽然我喜欢if…then…elseif堆栈的易读性，但感觉它比它更加冗长。由于比较变量的数量，我不认为我正在寻找switch…case版本并没有完全削减它，并且在if内嵌套switch…case语句感觉不对或if…then…else内的三元运算符。

我本能地想，我想要一种矩阵形式，其中返回值在网格中，并且以某种方式对各种按位条件的矩阵计算返回正确的结果。我怀疑这将是一个紧凑的代码胜过易读性的胜利。

有什么建议吗？

NB。为了易读性，选择变量名称，包括向每个变量添加count属性，而不是指定变量，或者不指示它是计数。

Answer 1

您可以使用var存储您的选择并在最后返回

    mode = (function(mode, current, proposed, origins, destinations) {
                var varName='move';
                if (mode === 'none') {
                    varName='project';
                } else if (proposed.count === 0) {
                    varName='unseated';
                } else if (current.count > proposed.count && proposed.count > 0) {
                    varName= 'reducing';
                } else if (proposed.count === destinations.count && destinations.count > 1 && current.count === 0) {
                    varName= 'newplus';
                } else if (proposed.count === destinations.count && current.count === 0) {
                    varName='new';
                } else if (proposed.count > destinations.count) {
                    varName= 'increasing';
                } else if (proposed.count === destinations.count && destinations.count > origins.count) {
                    varName= 'moveplus';
                }             
}(moves.register[staff].move, {count: mode.currentDesks}, {count: mode.proposedDesks}, {count: mode.origins}, {count: mode.destinations})));

什么是在javascript中实现if then else if else堆栈的简单方法

1 个答案: