export let i18n: ReturnType<typeof createI18n>; 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)); }; 报错代码报错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.

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：

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：

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：

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[]];

C:\Qt\Qt5.12.3\5.12.3\mingw73_64\include\QtCore\qobject.h:250: error: no matching function for call to 'QObject::connectImpl(const Object*&, void**, const Object*&, void**, QtPrivate::QSlotObject<void (student::*)(), QtPrivate::List<>, void>*, Qt::ConnectionType&, const int*&, const QMetaObject*)' return connectImpl(sender, reinterpret_cast<void **>(&signal), ~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ receiver, reinterpret_cast<void **>(&slot), ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ new QtPrivate::QSlotObject<Func2, typename QtPrivate::List_Left<typename SignalType::Arguments, SlotType::ArgumentCount>::Value, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ typename SignalType::ReturnType>(slot), ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ type, types, &SignalType::Object::staticMetaObject); ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

这个错误提示是因为连接信号和槽时参数不匹配导致的。

在 Qt 的信号和槽机制中，要确保连接的信号和槽的参数类型和数量完全匹配。根据错误信息，问题可能出现在 `QObject::connectImpl` 这个函数的调用上。

请检查连接的信号和槽的参数类型和数量是否匹配，确保它们一致。如果有需要，可以使用 `QOverload` 或 `QOverload` 宏来解决重载函数的问题。

另外，还要确保信号和槽所属的对象正确，以及 Qt 版本和编译器匹配。

如果问题仍然存在，请提供更多的代码片段或详细信息，以便更好地帮助你解决问题。