Strapi：将图像上传到Strapi，图像不会渲染

发布于 2025-02-02 12:08:20 字数 1194 浏览 5 评论 0原文

问题

我想知道您是否可以帮助确定与将图像上传到thumb属于content-type 目录的问题。

图像图像在Strapi面板中呈现后不会呈现，它仅显示名称。

const getFileSize = (filePath) => {
  return statSync(filePath)["size"]; // statSync from `fs-extra` library
};


 strapi.service("api::directory.directory").create({
      data: {},
      files: {
        thumb: {
          path: `${imagesDir}/Crab-1-1-scaled.jpg`,
          name: "Crab-1-1-scaled.jpeg",
          type: mime.lookup(`${imagesDir}/Crab-1-1-scaled.jpg`), // mime from `mime-types` library
          size: getFileSize(`${imagesDir}/Crab-1-1-scaled.jpg`),
        },
      },
    });

console.log

size在上传数据时始终为零

{
  files: {
    thumb: {
      path: 'src/scripts/Directory/images/Crab-1-1-scaled.jpg',
      name: 'Crab-1-1-scaled.jpeg',
      type: 'image/jpeg',
      size: 0
    }
  }
}

输出

，图像未渲染，仅显示图像的名称。

原文

Problem

I would like to know if you can help identify the issue related to uploading an image to the thumb field belonging to the content-type Directory.

The image image does not render once it is in the strapi panel, it only shows the name.

const getFileSize = (filePath) => {
  return statSync(filePath)["size"]; // statSync from `fs-extra` library
};


 strapi.service("api::directory.directory").create({
      data: {},
      files: {
        thumb: {
          path: `${imagesDir}/Crab-1-1-scaled.jpg`,
          name: "Crab-1-1-scaled.jpeg",
          type: mime.lookup(`${imagesDir}/Crab-1-1-scaled.jpg`), // mime from `mime-types` library
          size: getFileSize(`${imagesDir}/Crab-1-1-scaled.jpg`),
        },
      },
    });

console.log

size is always zero

{
  files: {
    thumb: {
      path: 'src/scripts/Directory/images/Crab-1-1-scaled.jpg',
      name: 'Crab-1-1-scaled.jpeg',
      type: 'image/jpeg',
      size: 0
    }
  }
}

output

When uploading the data the image is not rendered, it only shows the name of the image.

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你げ笑在眉眼 2025-02-09 12:08:20

我在Strapi 4中所做的是：我首先使用内置上传插件，该插件返回上传文件的ID。然后，我使用ID创建/更新我的实体（在您的情况下 Directory ）。

这是您可以使用的工作代码段：

    const filename = originalFile.split('.')[0]
    const ext = originalFile.split('.').pop()
    const rootDir = process.cwd();
    const filePath = `${rootDir}/public/uploads/${originalFile}` // the file is moved here already - the upload plugin will not upload anything, only reference the existing file.

    const entity = {
      path: filePath,
      name: originalFile,
      hash: `${filename}${ext}`,
      mime: mime.lookup(filePath),
      provider: 'local',
      ext: ext,
      type: mime.lookup(filePath),
      size: fs.statSync(filePath)["size"],
    };

    const localUpload = await strapi.plugin('upload').service('upload').upload({
      files: entity,
      data: {}
    });
    console.log("LOCAL UPLOAD done", localUpload[0].id)

What I do in Strapi 4 is this: I first use the built-in upload plugin, which returns an id of the uploaded file. I then use the id to create/update my entity (Directory in your case).

Here is a working code snippet you can use:

    const filename = originalFile.split('.')[0]
    const ext = originalFile.split('.').pop()
    const rootDir = process.cwd();
    const filePath = `${rootDir}/public/uploads/${originalFile}` // the file is moved here already - the upload plugin will not upload anything, only reference the existing file.

    const entity = {
      path: filePath,
      name: originalFile,
      hash: `${filename}${ext}`,
      mime: mime.lookup(filePath),
      provider: 'local',
      ext: ext,
      type: mime.lookup(filePath),
      size: fs.statSync(filePath)["size"],
    };

    const localUpload = await strapi.plugin('upload').service('upload').upload({
      files: entity,
      data: {}
    });
    console.log("LOCAL UPLOAD done", localUpload[0].id)

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