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Upload Files

getUploadFile accepts the name of file in order to get the infos. The function returns the name and contents of the file. For this example, the name is "file". 

<form action="upload" method="post" enctype="multipart/form-data">
  <input type="file" name="file" value="eva">
  <input type="submit" value="Submit" name="submit">
</form>

getUploadFile only works when using form parameters and HttpPost method. Context provides a helper function to save the uploadFile to disks. If you don't specify the name of the file, it will use the original name from the client.

proc upload(ctx: Context) {.async.} =
  if ctx.request.reqMethod == HttpGet:
    await ctx.staticFileResponse("tests/local/uploadFile/upload.html", "")
  elif ctx.request.reqMethod == HttpPost:
    let file = ctx.getUploadFile("file")
    file.save("tests/assets/temp")
    file.save("tests/assets/temp", "set.txt")
    resp fmt"<html><h1>{file.filename}</h1><p>{file.body}</p></html>"

The full example