NodeJs 异步等待 1 个调用而不是 2 个不同的函数

Question

我可以减少这些仅在一个中协同工作的功能吗？如果我能做到，该怎么做？

编辑：createMedia 函数需要在 wordpress 中上传媒体文件，而 createMediaAsync 用于 await 完成上传过程，尽管这对这个问题并不重要。

const createMedia = (postImgGalleryToUpload) => {
  return wp.media()
    // Specify a path to the file you want to upload, or a Buffer
    .file(postImgGalleryToUpload)
    .create({
      title: 'My awesome image',
      alt_text: 'an image of something awesome',
      caption: 'This is the caption text',
      description: 'More explanatory information'
    })
    .then(function (response) {
      return response.link;
    });
}

const createMediaAsync = async (postImgGalleryToUpload) => {
  try {
    var linkImage = await createMedia(postImgGalleryToUpload);
    console.log("IMMAGINE INSERITA" + linkImage);
  } catch (error) {
    console.log("Errore in createMediaAsync() - " + error);
  }
} 

在我的主文件中，我调用了异步文件：

createMediaAsync(postwp.postImgGallery1);

谢谢大家可以给我小费。

为了让问题更清楚，我想做这样的事情：

const createMedia = async (postImgGalleryToUpload) => {
    return await wp.media()
        // Specify a path to the file you want to upload, or a Buffer
        .file(postImgGalleryToUpload)
        .create({
            title: 'My awesome image',
            alt_text: 'an image of something awesome',
            caption: 'This is the caption text',
            description: 'More explanatory information'
        })
        .then(function (response) {
            return response.link;
        })
        .catch(function (error) {
            return error;
        });
}

我知道这是错误的，实际上它根本不起作用，有没有替代方法。

谢谢

Answer 1

当然，您可以直接内联函数定义来替换调用：

const createMediaAsync = async (postImgGalleryToUpload) => {
  try {
    var linkImage = await wp.media()
      .file(postImgGalleryToUpload)
      .create({
        title: 'My awesome image',
        alt_text: 'an image of something awesome',
        caption: 'This is the caption text',
        description: 'More explanatory information'
      })
      .then(function (response) {
        return response.link;
      });
    console.log("IMMAGINE INSERITA" + linkImage);
  } catch (error) {
    console.log("Errore in createMediaAsync() - " + error);
  }
} 

但是，你should not mix async/await syntax with .then(…) syntax！使用任一

function createMediaAsync(postImgGalleryToUpload) {
  return wp.media()
    // Specify a path to the file you want to upload, or a Buffer
    .file(postImgGalleryToUpload)
    .create({
      title: 'My awesome image',
      alt_text: 'an image of something awesome',
      caption: 'This is the caption text',
      description: 'More explanatory information'
    })
    .then(response => {
      const linkImage = response.link;
      console.log("IMMAGINE INSERITA" + linkImage)
    }, error => {
      console.log("Errore in createMediaAsync() - " + error);
    });
}

或

async function createMediaAsync(postImgGalleryToUpload) {
  try {
    const response = await wp.media()
      .file(postImgGalleryToUpload)
      .create({
        title: 'My awesome image',
        alt_text: 'an image of something awesome',
        caption: 'This is the caption text',
        description: 'More explanatory information'
      });
    const linkImage = response.link;
    console.log("IMMAGINE INSERITA" + linkImage);
  } catch (error) {
    console.log("Errore in createMediaAsync() - " + error);
  }
}