upload image processing using gin about gin HOT 13 CLOSED

no, but the http package include an powerful API to handle multipart requests.
http://sanatgersappa.blogspot.com.es/2013/03/handling-multiple-file-uploads-in-go.html

Do you think we should add a higher API in gin.Content?
like: c.Multipart(....)

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@manucorporat Add this feature is cool. but is better to be a middleware.

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@johndeng I agree

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@manucorporat Thanks.

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How are you supposed to handle file uploads in Gin, then? I can't seem to access FormFile from Context:
c.Request.FormFile("file")

[GIN] 2014/09/01 - 03:48:13 | 400 | 248.788us | 172.4.231.9 POST /post
Error #1: unknown content-type: multipart/form-data
Meta: Operation aborted

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Oh nevermind it was an issue with binding. If it helps the OP this is how you handle multipart forms:

c.Request.FormValue("value")
file, fileheader, _ := c.Request.FormFile("file")

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Hmm, i realize that there are many ways to do this.
in my use case, i will need to resize the uploaded image into multiple sizes.
so i cant use the more optimized Buffering method.

// the not so optimized way.
file, _, err := r.FormFile("file")
img, _, err := image.Decode(file)
conf := resize.Resize(55, 0, img, resize.NearestNeighbor)
out, err := os.Create("public" + imgPath)
err = jpeg.Encode(out, conf, nil)

another option is to perform the basic upload image using the optimized buffer method
and then read the uploaded image from the server disk to perform further resizing work if needed..
i.e. can provide image resizing as optional additional function..

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I don't think Middleware is the best option for this kind of stuff, sometimes you want to process the data or save the file, etc... but files sometimes are optional, its too broad of a use case.

What would be pretty useful is a simple function within Gin that takes the current context and saves the file to disk given a route (and handles all temp file and copy for the user). You can perform validations and such before, and just save it if needed.

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dont know what's the plan for the image upload .i am trying to do as above and use c.Request.Formfile("image") ,which may not so convinence.

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I use the following stuff, just like @badoet suggested and it works super fine.

file, _, err := c.Request.FormFile("file")
x, _, err := image.Decode(file)
// Do something with the image + error handling for above 2 
jpeg.Encode(f, file, &settings)

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@nazwa and your settings detail?

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Anything you want really. In my case it's

settings := jpeg.Options{Quality: 80}

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@nazwa thankyou

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