PDF-kit 使用 forEach 循环并排显示图像

发布于 2025-01-15 03:34:46 字数 595 浏览 2 评论 0原文

我有 API 可以在向数据库发送请求后将图片添加到 pdf 中。如果我发送一张图片，我就能成功，但如何并排显示图像？有可能吗？因为目前我的图像每秒堆叠一个，依此类推...

这是我的代码：

const photosToPdf = [];
for(let j=0;j<uploadsPath.length;j++){
  photosToPdf.push(uploadsFileNames[j]);
}

photosToPdf.forEach(v => {
  doc.image(`uploads/${v}`, 30, 350, {width: 100});
});

它将采用空数组并用用户上传的图像名称填充它。然后它应该循环遍历该数组并使用文件名作为 img 源。它正在工作，但它看起来像这样，一秒一秒。

我有想法，我将使用另一个循环来生成 X 和 Y 位置并每次都增加它，但也许有更简单的解决方案。

原文

I have API that can add picture to pdf after sending request to database. I managed to make it work if I send one picture, but how can I show images side by side? Is it even possible? Because currently I have images stack one on second and so on...

Here is my code:

const photosToPdf = [];
for(let j=0;j<uploadsPath.length;j++){
  photosToPdf.push(uploadsFileNames[j]);
}

photosToPdf.forEach(v => {
  doc.image(`uploads/${v}`, 30, 350, {width: 100});
});

It will take empty array and fill it with Images name uploaded by user. Then It should loop through that array and use file name as img source. It is working, but It is looking like this, one on second.

I have idea, that I will use another loop to generate X and Y position and increase it every time, but maybe there is more simple solution.

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酷遇一生 2025-01-22 03:34:46

通过每次迭代增加位置来解决这个问题。

const photosToPdf = [];
let positionX = 30;
let positionY = 240;

for(let j=0;j<uploadsPath.length;j++){
   photosToPdf.push(uploadsFileNames[j]);
}

photosToPdf.forEach(img => {
   if(positionX == 570){
      positionX = 30
      positionY = positionY + 160
   }
   doc.image(`uploads/${img}`, positionX, positionY, {width: 80});
   positionX = positionX + 90
   });

Fixed this problem by incrementing position every iteration.

const photosToPdf = [];
let positionX = 30;
let positionY = 240;

for(let j=0;j<uploadsPath.length;j++){
   photosToPdf.push(uploadsFileNames[j]);
}

photosToPdf.forEach(img => {
   if(positionX == 570){
      positionX = 30
      positionY = positionY + 160
   }
   doc.image(`uploads/${img}`, positionX, positionY, {width: 80});
   positionX = positionX + 90
   });

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