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Fix typos and grammar for 04.5.md

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2014-09-14 19:53:18 -07:00
committed by James Miranda
parent 6473cbad66
commit 0fad90094e
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@@ -1,14 +1,14 @@
# 4.5 File upload
Suppose you have a website like Instagram, and you want users to upload their beautiful photos, how you are going to do?
Suppose you have a website like Instagram and you want users to upload their beautiful photos. How would you implement that functionality?
You have to add property `enctype` to the form that you want to use for uploading photos, and there are three possibilities for its value:
You have to add property `enctype` to the form that you want to use for uploading photos. There are three possible values for this property:
application/x-www-form-urlencoded Trans-coding all characters before upload(default).
multipart/form-data No trans-coding, you have to use this value when your form have file upload controls.
text/plain Convert spaces to "+", but no trans-coding for special characters.
application/x-www-form-urlencoded Transcode all characters before uploading (default).
multipart/form-data No transcoding. You must use this value when your form has file upload controls.
text/plain Convert spaces to "+", but no transcoding for special characters.
Therefore, HTML content of a file upload form should look like this:
Therefore, the HTML content of a file upload form should look like this:
<html>
<head>
@@ -23,7 +23,7 @@ Therefore, HTML content of a file upload form should look like this:
</body>
</html>
We need to add a function in server side to handle this affair.
We need to add a function on the server side to handle this form.
http.HandleFunc("/upload", upload)
@@ -57,17 +57,17 @@ We need to add a function in server side to handle this affair.
}
}
As you can see, we need to call `r.ParseMultipartForm` for uploading files, the argument means the `maxMemory`. After you called `ParseMultipartForm`, file will be saved in your server memory with `maxMemory` size, if the file size is larger than `maxMemory`, rest of data will be saved in system temporary file. You can use `r.FormFile` to get file handle and use `io.Copy` to save to your file system.
As you can see, we need to call `r.ParseMultipartForm` for uploading files. The function `maxMemory` argument. After you call `ParseMultipartForm`, the file will be saved in the server memory with `maxMemory` size. If the file size is larger than `maxMemory`, the rest of the data will be saved in a system temporary file. You can use `r.FormFile` to get the file handle and use `io.Copy` to save to your file system.
You don't need to call `r.ParseForm` when you access other non-file fields in the form because Go will call it when it's necessary. Also call `ParseMultipartForm` once is enough, and no differences for multiple calls.
You don't need to call `r.ParseForm` when you access other non-file fields in the form because Go will call it when it's necessary. Also, calling `ParseMultipartForm` once is enough -multiple calls make no difference.
We use three steps for uploading files as follows:
1. Add `enctype="multipart/form-data"` to your form.
2. Call `r.ParseMultipartForm` in server side to save file in memory or temporary file.
3. Call `r.FormFile` to get file handle and save to file system.
2. Call `r.ParseMultipartForm` on the server side to save the file either to memory or to a temporary file.
3. Call `r.FormFile` to get the file handle and save to the file system.
The file handler is the `multipart.FileHeader`, it uses following struct:
The file handler is the `multipart.FileHeader`. It uses the following struct:
type FileHeader struct {
Filename string
@@ -77,11 +77,11 @@ The file handler is the `multipart.FileHeader`, it uses following struct:
![](images/4.5.upload2.png?raw=true)
Figure 4.5 Print information in server after received file.
Figure 4.5 Print information on server after receiving file.
## Clients upload files
I showed the example of using form to upload file, and we can impersonate a client form to upload file in Go as well.
I showed an example of using a form to a upload a file. We can impersonate a client form to upload files in Go as well.
package main
@@ -143,12 +143,12 @@ I showed the example of using form to upload file, and we can impersonate a clie
postFile(filename, target_url)
}
The above example shows you how to use client to upload file, it uses `multipart.Write` to write file in cache and sends to server through POST method.
The above example shows you how to use a client to upload files. It uses `multipart.Write` to write files into cache and sends them to the server through the POST method.
If you have other field need to write into data like user name, call `multipart.WriteField` as needed.
If you have other fields that need to write into data, like username, call `multipart.WriteField` as needed.
## Links
- [Directory](preface.md)
- Previous section: [Duplicate submissions](04.4.md)
- Next section: [Summary](04.6.md)
- Next section: [Summary](04.6.md)