2015-04-29 08:05
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I download a multiline file from Amazon S3 in format like:

ColumnAv1 ColumnBv1 ColumnCv1 ...
ColumnAv2 ColumnBv2 ColumnCv2 ...

the file is of type byte. Then I want to parse this with regex:

matches := re.FindAllSubmatch(file,-1)

then I want to feed result row by row to function which takes []string as input (string[0] is ColumnAv1, string[1] is ColumnBv2, ...).

How should I convert result of [][][]byte to []string containing first, second, etc row? I suppose I should do it in a loop, but I cannot get this working:

for i:=0;i<len(len(matches);i++{

BTW, Why does function FindAllSubmatch return [][][]byte whereas FindAllStringSubmatch return [][]string?

(Sorry I don't have right now access to my real example, so the syntax may not be proper)

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2条回答 默认 最新

  • dqy006150
    dqy006150 2015-04-29 08:30

    It's all explained extensively in the package's documentation.
    Read the parapgraph which explains :

    There are 16 methods of Regexp that match a regular expression and identify the matched text. Their names are matched by this regular expression:


    In your case, you probably want to use FindAllStringSubmatch.

    In Go, a string is just a read-only []byte.
    You can choose to either keep passing []byte variables around,
    or cast the []byte values to string :

    var byteSlice = []byte{'F','o','o'}
    var str string
    str = string(byteSlice)
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  • dongzhan2461
    dongzhan2461 2015-04-29 09:55

    You can simply iterate through the bytes result as you would do for strings result using two nested loop, and just convert slice of bytes to a string in the second loop:

    package main
    import "fmt"
    func main() {
        f := [][][]byte{{{'a', 'b', 'c'}}}
        for _, line := range f {
            for _, match := range line { // match is a type of []byte


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