findOverrideMapId = async (overrides: CaseOverrideInfo[]) => { let page = 1; const pageSize = 40; let mapIdTowardIndex = -1; while (page) { const batchMaps = await this.getMaps(page, pageSize); for (const item of overrides) { if (item.missingMap && !item.overrideMapId) { const mapIdTowardId = batchMaps.items.find((ele) => ele?.name === item.mapId)?.id; mapIdTowardIndex = batchMaps.items.findIndex((ele) => ele?.name === item.mapId) + (page - 1) * pageSize; item.overrideMapId = mapIdTowardId || ""; } } page++; if (!batchMaps || batchMaps.items.length < pageSize) { break; } } this.setState({ initialPage: Math.ceil(mapIdTowardIndex / (pageSize + 2)) }); return overrides; };我想找到这个后就停止发请求。
时间: 2024-02-14 21:29:00 浏览: 22
您可以在以下位置添加一个 `break` 语句来停止请求循环:
```javascript
for (const item of overrides) {
if (item.missingMap && !item.overrideMapId) {
const mapIdTowardId = batchMaps.items.find((ele) => ele?.name === item.mapId)?.id;
mapIdTowardIndex =
batchMaps.items.findIndex((ele) => ele?.name === item.mapId) + (page - 1) * pageSize;
item.overrideMapId = mapIdTowardId || "";
}
// 添加以下代码
if (mapIdTowardIndex !== -1) {
break;
}
}
```
这样,当找到匹配的 `mapId` 并更新 `overrideMapId` 后,`mapIdTowardIndex` 的值将不再是 -1,循环会被终止。请注意,这只是一个简单的示例,具体的实现可能需要根据您的代码逻辑进行调整。
相关问题
findOverrideMapId = async (overrides: CaseOverrideInfo[]) => { let page = 1; const pageSize = 40; let mapIdTowardIndex = -1; let foundAllOverrides = overrides.length; while (page && foundAllOverrides) { const batchMaps = await this.getMaps(page, pageSize); for (const item of overrides) { if (item.missingMap && !item.overrideMapId) { const mapIdTowardId = batchMaps.items.find((ele) => ele?.name === item.mapId)?.id; mapIdTowardIndex = batchMaps.items.findIndex((ele) => ele?.name === item.mapId) + (page - 1) * pageSize; item.overrideMapId = mapIdTowardId || ""; item.initialPage = Math.ceil(mapIdTowardIndex / (pageSize + 2)); if (item.overrideMapId) --foundAllOverrides; } } page++; if (!batchMaps || batchMaps.items.length < pageSize) { break; } } return overrides; }; 有点问题,我希望overrides对象找一遍后尽量减少时间复杂度。
要尽量减少时间复杂度,可以考虑使用一个哈希表来存储已经处理过的 `mapId`,避免重复查找。以下是修改后的代码示例:
```javascript
findOverrideMapId = async (overrides: CaseOverrideInfo[]) => {
let page = 1;
const pageSize = 40;
let mapIdTowardIndex = -1;
const processedMapIds = new Set();
let foundAllOverrides = overrides.length;
while (page && foundAllOverrides) {
const batchMaps = await this.getMaps(page, pageSize);
for (const item of overrides) {
if (item.missingMap && !item.overrideMapId && !processedMapIds.has(item.mapId)) {
const mapIdTowardId = batchMaps.items.find((ele) => ele?.name === item.mapId)?.id;
mapIdTowardIndex =
batchMaps.items.findIndex((ele) => ele?.name === item.mapId) + (page - 1) * pageSize;
item.overrideMapId = mapIdTowardId || "";
item.initialPage = Math.ceil(mapIdTowardIndex / (pageSize + 2));
if (item.overrideMapId) {
--foundAllOverrides;
processedMapIds.add(item.mapId);
}
}
}
page++;
if (!batchMaps || batchMaps.items.length < pageSize) {
break;
}
}
return overrides;
};
```
在这个示例中,我们使用 `processedMapIds` 哈希表来存储已经处理过的 `mapId`,通过 `has()` 方法来判断是否已经处理过。如果已经处理过,就跳过当前循环。这样可以避免重复查找和处理相同的 `mapId`,从而减少时间复杂度。
请注意,这只是一种优化的方式,具体的实现可能需要根据您的代码逻辑进行调整。
findOverrideMapId = async (overrides: CaseOverrideInfo[]) => { const allMapsList = []; let page = 1; const pageSize = 40; while (page) { const batchMaps = await this.getMaps(page, pageSize); allMapsList.push(...batchMaps.items); page++; if (!batchMaps || batchMaps.items.length < pageSize) { break; } } let mapIdTowardIndex = -1; for (const item of overrides) { if (item.missingMap) { const mapIdTowardId = allMapsList.find((ele) => ele?.name === item.mapId)?.id; mapIdTowardIndex = allMapsList.findIndex((ele) => ele?.name === item.mapId); item.overrideMapId = mapIdTowardId || ""; } } this.setState({ initialPage: Math.ceil(mapIdTowardIndex / (pageSize + 2)) }); return overrides; }; 我想改一下逻辑,不用获取所有列表allMapsList。接口一找到就返回并且停止,并且还有序号需要正确
您想改变逻辑,不再获取所有的列表allMapsList,并且在找到匹配项后立即返回,并且确保序号正确。以下是您修改后的代码:
```javascript
findOverrideMapId = async (overrides: CaseOverrideInfo[]) => {
let mapIdTowardIndex = -1;
for (const [index, item] of overrides.entries()) {
if (item.missingMap) {
const batchMaps = await this.getMaps(1, index + 1); // 获取对应位置的地图项
const mapIdTowardId = batchMaps.items.find((ele) => ele?.name === item.mapId)?.id;
mapIdTowardIndex = batchMaps.items.findIndex((ele) => ele?.name === item.mapId);
item.overrideMapId = mapIdTowardId || "";
break; // 找到匹配项后立即停止循环
}
}
this.setState({ initialPage: Math.ceil(mapIdTowardIndex / (pageSize + 2)) });
return overrides;
};
```
这样修改后的代码会在找到第一个匹配项后立即停止循环,并且使用正确的序号进行处理。注意,我假设您的`getMaps`方法会根据提供的页码和页面大小返回对应位置的地图项。