const tFn: I18nGlobalTranslation = (key: string, ...arg: any[]) => { if (!key) return ''; if (!key.includes('.') && !namespace) return key; return t(getKey(namespace, key), ...(arg as I18nTranslationRestParameters)); }; 代码报错his expression is not callable. Each member of the union type 'ComposerTranslation<Record<string, unknown>, unknown, RemoveIndexSignature<{ [x: string]: LocaleMessageValue<VueMessageType>; }>, never, never, never> | VueI18nTranslation<...>' has signatures, but none of those signatures are compatible with each other.
时间: 2024-03-11 20:49:14 浏览: 153
这个错误通常是由于类型不匹配导致的。可能是因为`t`函数的类型定义不正确,或是在调用`t`函数时传递的参数不符合预期。建议检查以下几点:
1. 确保`t`函数的类型定义正确,应该类似于以下代码:
```typescript
import { t, Composer } from 'vue-i18n';
type I18nGlobalTranslation = Composer['t'];
const tFn: I18nGlobalTranslation = (key: string, ...args: any[]) => {
// ...
};
```
2. 确保在调用`t`函数时传递的参数类型和数量与`t`函数期望的类型和数量相匹配。在这个错误中,可能是`getKey`函数返回了一个错误的键名,导致`t`函数无法找到对应的翻译内容。你可以在调用`t`函数之前,先打印一下`getKey(namespace, key)`的返回值,检查是否符合预期。
如果以上方法都无法解决问题,建议提供更多的代码和错误信息,以便更好地定位问题。
相关问题
import type { createI18n } from 'vue-i18n'; export let i18n: ReturnType<typeof createI18n>; type I18nGlobalTranslation = { (key: string): string; (key: string, locale: string): string; (key: string, locale: string, list: unknown[]): string; (key: string, locale: string, named: Record<string, unknown>): string; (key: string, list: unknown[]): string; (key: string, named: Record<string, unknown>): string; }; type I18nTranslationRestParameters = [string, any]; function getKey(namespace: string | undefined, key: string) { if (!namespace) { return key; } if (key.startsWith(namespace)) { return key; } return `${namespace}.${key}`; } export function useI18n(namespace?: string): { t: I18nGlobalTranslation; } { const normalFn = { t: (key: string) => { return getKey(namespace, key); }, }; if (!i18n) { return normalFn; } const { t, ...methods } = i18n.global; const tFn: I18nGlobalTranslation = (key: string, ...arg: any[]) => { if (!key) return ''; if (!key.includes('.') && !namespace) return key; return t(getKey(namespace, key), ...(arg as I18nTranslationRestParameters)); }; return { ...methods, t: tFn, }; } // Why write this function? // Mainly to configure the vscode i18nn ally plugin. This function is only used for routing and menus. Please use useI18n for other places // 为什么要编写此函数? // 主要用于配合vscode i18nn ally插件。此功能仅用于路由和菜单。请在其他地方使用useI18n export const t = (key: string) => key;这段代码报错,帮忙重写
这段代码报错的原因可能是在使用 useI18n 函数时,参数类型不匹配。可以尝试按照下面的方式重写代码:
```typescript
import { createI18n, LocaleMessages } from 'vue-i18n';
const messages: LocaleMessages = {
en: {
// English translations
},
zh: {
// Chinese translations
},
// Add more languages and translations as needed
};
const i18n = createI18n({
locale: 'en',
messages,
});
export function useI18n(namespace?: string) {
const normalFn = {
t: (key: string) => {
return namespace ? `${namespace}.${key}` : key;
},
};
if (!i18n) {
return normalFn;
}
const { t, ...methods } = i18n.global;
const tFn: I18nGlobalTranslation = (
key: string,
...arg: I18nTranslationRestParameters
) => {
if (!key) return '';
if (!key.includes('.') && !namespace) return key;
return t(namespace ? `${namespace}.${key}` : key, ...arg);
};
return {
...methods,
t: tFn,
};
}
// This function is only used for routing and menus. Please use useI18n for other places
export const t = (key: string) => key;
```
注意,这里还需要定义 `I18nGlobalTranslation` 和 `I18nTranslationRestParameters` 类型,具体代码如下:
```typescript
type I18nGlobalTranslation = {
(key: string): string;
(key: string, locale: string): string;
(key: string, locale: string, list: unknown[]): string;
(key: string, locale: string, named: Record<string, unknown>): string;
(key: string, list: unknown[]): string;
(key: string, named: Record<string, unknown>): string;
};
type I18nTranslationRestParameters = [string, ...unknown[]];
```
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