这是一个创建于 805 天前的主题,其中的信息可能已经有所发展或是发生改变。
功能描述
当 const mergeTable=[2]时,
函数返回
{ rowSpan: 2 }
{ rowSpan: 0 }
当 const mergeTable=[2,3]时,函数返回
{ rowSpan: 2 }
{ rowSpan: 0 }
{ rowSpan: 3 }
{ rowSpan: 0 }
{ rowSpan: 0 }
当 const mergeTable=[2,3,3,2]函数返回
{ rowSpan: 2 }
{ rowSpan: 0 }
{ rowSpan: 3 }
{ rowSpan: 0 }
{ rowSpan: 0 }
{ rowSpan: 3 }
{ rowSpan: 0 }
{ rowSpan: 0 }
{ rowSpan: 2 }
{ rowSpan: 0 }
需要优化的函数
功能已经实现,代码冗余,请问如何优化
```
// const mergeTable=[2] // 测试 ok
const mergeTable = [2, 3]; // 测试 ok
// const mergeTable=[2,3,3,2]; // 测试 ok
// 如何简化这个函数
function renderRowSpan(index) {
for (let i in mergeTable) {
i=Number(i);
if (i === 0) {
if (index === 0) {
return { rowSpan: mergeTable[0] };
} else if (index < mergeTable[0]) {
return { rowSpan: 0 };
}
}
if (i === 1) {
if (index === mergeTable[0]) {
return { rowSpan: mergeTable[1] };
}
if (index < mergeTable[0] + mergeTable[1]) {
return { rowSpan: 0 };
}
}
if (i === 2) {
if (index === mergeTable[0] + mergeTable[1]) {
return { rowSpan: mergeTable[2] };
}
if (index < mergeTable[0] + mergeTable[1] + mergeTable[2]) {
return { rowSpan: 0 };
}
}
if (i === 3) {
if (index === mergeTable[0] + mergeTable[1]+mergeTable[2]) {
return { rowSpan: mergeTable[3] };
}
if (index < mergeTable[0] + mergeTable[1] + mergeTable[2]+ mergeTable[3]) {
return { rowSpan: 0 };
}
}
}
}
// 外部方法会渲染 index
for (let index = 0; index < 10; index++) {
console.log(renderRowSpan(index));
}
```
7 条回复 • 2022-02-26 23:24:55 +08:00
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1
MegrezZhu 2022-02-26 17:34:52 +08:00 1
啥花里胡哨的
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2
binux 2022-02-26 17:55:43 +08:00
function renderRowSpan(index) { for (let i of mergeTable) { if (index == 0) return { rowSpan: i }; if (index < 0) return { rowSpan: 0 }; index -= i; } if (index < 0) return { rowSpan: 0 }; }
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5
sweetcola 2022-02-26 20:59:17 +08:00
const renderRowSpan = data => data.reduce((t, c) => ([...t, { rowSpan: c }, ...new Array(c - 1).fill({ rowSpan: 0 })]), []);
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7
des 2022-02-26 23:24:55 +08:00
用 flat 更简洁易懂一点,顺便给你做了错误处理 const makeRowSpan = i => ({ rowSpan: i }) const makeRowBySpanNum = num => [makeRowSpan(num), ...new Array(--num).fill(makeRowSpan(0))] const makeSpan = rows => rows?.filter(Number.isFinite).map(makeRowBySpanNum).flat() ?? [] makeSpan([2,3,3,null,2])
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