英文:
Reading from reader until a string is reached
问题
我正在尝试编写一个函数,从缓冲读取器中持续读取,直到遇到特定的字符串,然后停止读取并返回之前读取的所有内容。
换句话说,我想做的与reader.ReadString()
相同,只是接受一个字符串而不是单个字节。
例如:
mydata, err := reader.ReadString("\r\n.\r\n") //显然无法编译
我该如何做到这一点?
提前感谢,
Twichy
修正1:之前的尝试
这是我之前的尝试;它写得很糟糕,而且不起作用,但希望能够演示我想要做的事情。
func readDotData(reader *bufio.Reader)(string, error){
delims := []byte{ '\r', '\n', '.', '\r', '\n'}
curpos := 0
var buffer []byte
for {
curpos = 0
data, err := reader.ReadSlice(delims[0])
if err!=nil{ return "", err }
buffer = append(buffer, data...)
for {
curpos++
b, err := reader.ReadByte()
if err!=nil{ return "", err }
if b!=delims[curpos]{
for curpos >= 0{
buffer = append(buffer, delims[curpos])
curpos--
}
break
}
if curpos == len(delims){
return string(buffer[len(buffer)-1:]), nil
}
}
}
panic("unreachable")
}
英文:
I am trying to write a function to keep reading from a buffered reader until I hit a certain string, then to stop reading and return everything read prior to that string.
In other words, I want to do the same thing as reader.ReadString()
does, except taking a string instead of a single byte.
For instance:
mydata, err := reader.ReadString("\r\n.\r\n") //obviously will not compile
How can I do this?
Thanks in advance,
Twichy
Amendment 1: Previous attempt
Here is my previous attempt; its badly written and doesnt work but hopefully it demonstrates what I am trying to do.
func readDotData(reader *bufio.Reader)(string, error){
delims := []byte{ '\r', '\n', '.', '\r', '\n'}
curpos := 0
var buffer []byte
for {
curpos = 0
data, err := reader.ReadSlice(delims[0])
if err!=nil{ return "", err }
buffer = append(buffer, data...)
for {
curpos++
b, err := reader.ReadByte()
if err!=nil{ return "", err }
if b!=delims[curpos]{
for curpos >= 0{
buffer = append(buffer, delims[curpos])
curpos--
}
break
}
if curpos == len(delims){
return string(buffer[len(buffer)-1:]), nil
}
}
}
panic("unreachable")
}
答案1
得分: 11
package main
import (
"bytes"
"fmt"
"log"
)
type reader interface {
ReadString(delim byte) (line string, err error)
}
func read(r reader, delim []byte) (line []byte, err error) {
for {
s := ""
s, err = r.ReadString(delim[len(delim)-1])
if err != nil {
return
}
line = append(line, []byte(s)...)
if bytes.HasSuffix(line, delim) {
return line[:len(line)-len(delim)], nil
}
}
}
func main() {
src := bytes.NewBufferString("123deli456elim789delimABCdelimDEF")
for {
b, err := read(src, []byte("delim"))
if err != nil {
log.Fatal(err)
}
fmt.Printf("%q\n", b)
}
}
英文:
package main
import (
"bytes"
"fmt"
"log"
)
type reader interface {
ReadString(delim byte) (line string, err error)
}
func read(r reader, delim []byte) (line []byte, err error) {
for {
s := ""
s, err = r.ReadString(delim[len(delim)-1])
if err != nil {
return
}
line = append(line, []byte(s)...)
if bytes.HasSuffix(line, delim) {
return line[:len(line)-len(delim)], nil
}
}
}
func main() {
src := bytes.NewBufferString("123deli456elim789delimABCdelimDEF")
for {
b, err := read(src, []byte("delim"))
if err != nil {
log.Fatal(err)
}
fmt.Printf("%q\n", b)
}
}
Output:
"123deli456elim789"
"ABC"
2009/11/10 23:00:00 EOF
答案2
得分: 2
例如,
package main
import (
"bufio"
"bytes"
"fmt"
"strings"
)
var delim = []byte{'\r', '\n', '.', '\r', '\n'}
func ScanLines(data []byte, atEOF bool) (advance int, token []byte, err error) {
if atEOF && len(data) == 0 {
return 0, nil, nil
}
for i := 0; i+len(delim) <= len(data); {
j := i + bytes.IndexByte(data[i:], delim[0])
if j < i {
break
}
if bytes.Equal(data[j+1:j+len(delim)], delim[1:]) {
// We have a full delim-terminated line.
return j + len(delim), data[0:j], nil
}
i = j + 1
}
// If we're at EOF, we have a final, non-terminated line. Return it.
if atEOF {
return len(data), data, nil
}
// Request more data.
return 0, nil, nil
}
func main() {
delims := string(delim)
input := "1234" + delims + "5678" + delims + "1234567901234567890" + delims
scanner := bufio.NewScanner(strings.NewReader(input))
scanner.Split(ScanLines)
for scanner.Scan() {
fmt.Printf("%s\n", scanner.Text())
}
if err := scanner.Err(); err != nil {
fmt.Printf("Invalid input: %s", err)
}
}
输出:
1234
5678
1234567901234567890
英文:
For example,
package main
import (
"bufio"
"bytes"
"fmt"
"strings"
)
var delim = []byte{'\r', '\n', '.', '\r', '\n'}
func ScanLines(data []byte, atEOF bool) (advance int, token []byte, err error) {
if atEOF && len(data) == 0 {
return 0, nil, nil
}
for i := 0; i+len(delim) <= len(data); {
j := i + bytes.IndexByte(data[i:], delim[0])
if j < i {
break
}
if bytes.Equal(data[j+1:j+len(delim)], delim[1:]) {
// We have a full delim-terminated line.
return j + len(delim), data[0:j], nil
}
i = j + 1
}
// If we're at EOF, we have a final, non-terminated line. Return it.
if atEOF {
return len(data), data, nil
}
// Request more data.
return 0, nil, nil
}
func main() {
delims := string(delim)
input := "1234" + delims + "5678" + delims + "1234567901234567890" + delims
scanner := bufio.NewScanner(strings.NewReader(input))
scanner.Split(ScanLines)
for scanner.Scan() {
fmt.Printf("%s\n", scanner.Text())
}
if err := scanner.Err(); err != nil {
fmt.Printf("Invalid input: %s", err)
}
}
Output:
1234
5678
1234567901234567890
答案3
得分: 2
package main
import (
"bytes"
"fmt"
)
func main() {
b := bytes.NewBuffer([]byte("Hello, playground!\r\n.\r\nIrrelevant trailer."))
c := make([]byte, 0, b.Len())
for {
p := b.Bytes()
if bytes.Equal(p[:5], []byte("\r\n.\r\n")) {
fmt.Println(string(c))
return
}
c = append(c, b.Next(1)...)
}
}
英文:
http://play.golang.org/p/BpA5pOc-Rn
package main
import (
"bytes"
"fmt"
)
func main() {
b := bytes.NewBuffer([]byte("Hello, playground!\r\n.\r\nIrrelevant trailer."))
c := make([]byte, 0, b.Len())
for {
p := b.Bytes()
if bytes.Equal(p[:5], []byte("\r\n.\r\n")) {
fmt.Println(string(c))
return
}
c = append(c, b.Next(1)...)
}
}
答案4
得分: 0
因为字符串中有相同的字节,所以可以按照以下方式进行操作:
func readWithEnd(reader *bufio.Reader) ([]byte, error) {
message, err := reader.ReadBytes('#')
if err != nil {
return nil, err
}
a1, err := reader.ReadByte()
if err != nil {
return nil, err
}
message = append(message, a1)
if a1 != '\t' {
message2, err := readWithEnd(reader)
if err != nil {
return nil, err
}
ret := append(message, message2...)
return ret, nil
}
a2, err := reader.ReadByte()
if err != nil {
return nil, err
}
message = append(message, a2)
if a2 != '#' {
message2, err := readWithEnd(reader)
if err != nil {
return nil, err
}
ret := append(message, message2...)
return ret, nil
}
return message, nil
}
这是一个可以在TCP连接中识别"#\t#"
的示例。
英文:
Because you have the same byte in the string, you can do it as below:
func readWithEnd(reader *bufio.Reader) ([]byte, error) {
message, err := reader.ReadBytes('#')
if err != nil {
return nil, err
}
a1, err := reader.ReadByte()
if err != nil {
return nil, err
}
message = append(message, a1)
if a1 != '\t' {
message2, err := readWithEnd(reader)
if err != nil {
return nil, err
}
ret := append(message, message2...)
return ret, nil
}
a2, err := reader.ReadByte()
if err != nil {
return nil, err
}
message = append(message, a2)
if a2 != '#' {
message2, err := readWithEnd(reader)
if err != nil {
return nil, err
}
ret := append(message, message2...)
return ret, nil
}
return message, nil
}
This is the sample that can recognize the "#\t#"
in TCP connection
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