如前所述,这是一个递归对象比较函数。还有一点。假设这种功能的主要用途是对象检查,我有话要说。当某些差异无关紧要时,完整的深度比较是一个坏主意。例如,TDD 断言中的盲目深度比较会使测试变得不必要的脆弱。出于这个原因,我想介绍一个更有价值的partial diff。它是以前对该线程的贡献的递归模拟。它会忽略 a
中不存在的键
var bdiff = (a, b) =>
_.reduce(a, (res, val, key) =>
res.concat((_.isPlainObject(val) || _.isArray(val)) && b
? bdiff(val, b[key]).map(x => key + '.' + x)
: (!b || val != b[key] ? [key] : [])),
[]);
BDiff 允许在容忍其他属性的同时检查预期值,这正是您想要自动检查的内容。这允许构建各种高级断言。例如:
var diff = bdiff(expected, actual);
// all expected properties match
console.assert(diff.length == 0, "Objects differ", diff, expected, actual);
// controlled inequality
console.assert(diff.length < 3, "Too many differences", diff, expected, actual);
回到完整的解决方案。使用 bdiff 构建完整的传统差异是微不足道的:
function diff(a, b) {
var u = bdiff(a, b), v = bdiff(b, a);
return u.filter(x=>!v.includes(x)).map(x=>' < ' + x)
.concat(u.filter(x=>v.includes(x)).map(x=>' | ' + x))
.concat(v.filter(x=>!u.includes(x)).map(x=>' > ' + x));
};
在两个复杂对象上运行上述函数将输出类似于以下内容:
[
" < components.0.components.1.components.1.isNew",
" < components.0.cryptoKey",
" | components.0.components.2.components.2.components.2.FFT.min",
" | components.0.components.2.components.2.components.2.FFT.max",
" > components.0.components.1.components.1.merkleTree",
" > components.0.components.2.components.2.components.2.merkleTree",
" > components.0.components.3.FFTResult"
]
最后,为了了解这些值的不同之处,我们可能希望直接eval() diff 输出。为此,我们需要一个更丑陋的 bdiff 版本,它可以输出语法正确的路径:
// provides syntactically correct output
var bdiff = (a, b) =>
_.reduce(a, (res, val, key) =>
res.concat((_.isPlainObject(val) || _.isArray(val)) && b
? bdiff(val, b[key]).map(x =>
key + (key.trim ? '':']') + (x.search(/^\d/)? '.':'[') + x)
: (!b || val != b[key] ? [key + (key.trim ? '':']')] : [])),
[]);
// now we can eval output of the diff fuction that we left unchanged
diff(a, b).filter(x=>x[1] == '|').map(x=>[x].concat([a, b].map(y=>((z) =>eval('z.' + x.substr(3))).call(this, y)))));
这将输出类似于以下内容:
[" | components[0].components[2].components[2].components[2].FFT.min", 0, 3]
[" | components[0].components[2].components[2].components[2].FFT.max", 100, 50]
MIT 许可证 ;)