I need to replace \ with / in a path string, but following code failed.
package main
import (
"fmt"
"strings"
)
func main() {
string := "P:\Project\project-name/content/topic/"
fmt.Println(strings.Replace(string, "\\", "/", -1))
}
Any helpful suggestions?
Use the function filepath.ToSlash to replace the operating system path separator with '/' in a path.
On Windows, the function returns strings.Replace(path, string(filepath.Separator), "/", -1). On other operating systems, the function returns the path argument as is.
You didn't escape backslashes in string. The following code works:
package main
import (
"fmt"
"strings"
)
func main() {
string := "P:\\Project\\project-name/content/topic/"
fmt.Println(strings.Replace(string, "\\", "/", -1))
}
Play this on playground: https://play.golang.org/p/T3XE5uiIkk
You can also use back-quotes (`) to create a raw string:
func main() {
string := `P:\Project\project-name/content/topic/`
fmt.Println(strings.Replace(string, "\\", "/", -1))
}
Note that the raw string above would still have its internal representation as
"P:\\Project\\project-name/content/topic/"
Hence the need to use "\\" in strings.Replace function.
Related
How to get the correct number of fields when using NewReader ?
package main
import (
"encoding/csv"
"fmt"
"log"
"strings"
)
func main() {
parser := csv.NewReader(strings.NewReader(`||""FOO""||`))
parser.Comma = '|'
parser.LazyQuotes = true
record, err := parser.Read()
if err != nil {
log.Fatal(err)
}
fmt.Printf("record length: %v\n", len(record))
}
https://go.dev/play/p/gg-KYRciWFH
It should return 5, but instead I'm getting 3:
record length: 3
Program exited.
EDIT
I'm actually working with a big CSV file containing many double quotes.
After examining your code, I decided to modify it slightly and then print the results:
package main
import (
"encoding/csv"
"fmt"
"log"
"strings"
)
func main() {
parser := csv.NewReader(strings.NewReader(`x||""FOO""|x|x\n`))
parser.Comma = '|'
parser.LazyQuotes = true
record, err := parser.Read()
if err != nil {
log.Fatal(err)
}
fmt.Printf("record length: %v, Data: %v\n", len(record), strings.Join(record, ", "))
}
When you run this, the data is printed as x, , "FOO"||x|x\n". My thought is that when you end your entry with two double-quotes, the parser is assuming the string is still being quoted and therefore lumps the rest of the line into the third entry. This appears to be a bug with how lazy-quoting works in the csv package, however, when examining the documentation for LazyQuotes, you'll see this:
If LazyQuotes is true, a quote may appear in an unquoted field and a non-doubled quote may appear in a quoted field.
This doesn't mention anything about finding double quotes within double quotes. To fix this, you should either remove the quotes altogether or replace the double double-quotes ("") with double quotes (").
One other thing you might consider would be using the gocsv package. I've worked with this package in the past and it's reasonably stable. I'm not sure how it would respond to this specific issue, but it might be worth your time checking it out.
Note:
The encoding/csv package implements the RFC 4180 standard. If you have such input, that's not an RFC 4180 compliant CSV file and encoding/csv will not parse it properly.
You're misusing the quotes. Quoting a single field FOO is like this:
parser := csv.NewReader(strings.NewReader(`||"FOO"||`))
If you want the field to have the "FOO" value, you have to use 2 double quotes in a quoted field, so it should be:
parser := csv.NewReader(strings.NewReader(`||"""FOO"""||`))
This will output 5. Try it on the Go Playground.
What you have is this:
parser := csv.NewReader(strings.NewReader(`||""FOO""||`))
Since the second " character is not followed by a separator character, the field is not interrupted and the rest is processed as the content of the quoted field (which will terminate at the end of the line).
If you print the record:
fmt.Println(record)
fmt.Printf("%#v", record)
Output will be (try it on the Go Playground):
[ "FOO"||]
[]string{"", "", "\"FOO\"||"}
Quotes are a part of csv format.
There is a problem with go/csv shielding, you can try something like this:
package main
import (
"encoding/csv"
"fmt"
"log"
"strings"
)
func main() {
parser := csv.NewReader(strings.NewReader(`||FOO||`))
parser.Comma = '|'
parser.LazyQuotes = true
record, err := parser.Read()
if err != nil {
log.Fatal(err)
}
fmt.Printf("record length: %v\n", len(record))
fmt.Println(strings.Join(record, " /SEP/ "))
}
or like this:
package main
import (
"encoding/csv"
"fmt"
"log"
"strings"
)
func main() {
parser := csv.NewReader(strings.NewReader(`||"""FOO"""||`))
parser.Comma = '|'
parser.LazyQuotes = true
record, err := parser.Read()
if err != nil {
log.Fatal(err)
}
fmt.Printf("record length: %v\n", len(record))
fmt.Println(strings.Join(record, " SEP "))
}
I am trying to obtain the first directory in an URL-like string like this: "/blog/:year/:daynum/:postname". I thought splitting it, then retrieving the first directory, would be this simple. But it returns square brackets surrounding the string even though it's not a slice. How can I get that first directory? (I am guaranteed that the string starts with a "/" followed by a valid directory designation and that contains both a leading directory and a string using those permalink properties).
What's the best way to parse out that first directory?
package main
import (
"fmt"
"strings"
)
// Retrieve the first directory in the URL-like
// string passed in
func firstDir(permalink string) string {
split := strings.Split(permalink, "/")
return string(fmt.Sprint((split[0:2])))
}
func main() {
permalink := "/blog/:year/:daynum/:postname"
dir := firstDir(permalink)
fmt.Printf("leading dir is: %s.", dir)
// Prints NOT "blog" but "[ blog]".
}
Since you said:"(I am guaranteed that the string starts with a "/" followed by a valid directory designation and that contains both a leading directory and a string using those permalink properties)"
Then simply use split[1] to get the root directory.
package main
import (
"fmt"
"os"
"strings"
)
func firstDir(permalink string) string {
split := strings.Split(permalink, string(os.PathSeparator))
return split[1]
}
func main() {
permalink := "/blog/:year/:daynum/:postname"
dir := firstDir(permalink)
fmt.Printf("leading dir is: %s.", dir)
// Prints "blog".
}
https://go.dev/play/p/hCHnrDIsWYE
I have a problem with If Else Condition in Regex. I have a file which contains the below format. I was looking for return value to be either 0.0.985 or 3.3.5-3811.
I was trying to use if else condition in regex but unable to do so, can anyone explain me while solving the problem please.
random-app-0.0.985.tgz
busy-app-7.3.1.2-3.3.5-3811-a19874elkc-123254376584.zip
Below is the Go code I am trying to use
package main
import (
"fmt"
"io/ioutil"
"regexp"
)
func main(){
content, err:= ioutil.ReadFile("version.txt")
if err != nil{
fmt.Println(err)
}
version:= string(content)
re:= regexp.MustCompile(`(\d+).(\d+).(\d+)|(\d+).(\d+).(\d+).(\d+)`)
result:= re.FindAllStringSubmatch(version,-1)
for i:= range(result){
fmt.Println(result[i][0])
}
}
Output is coming like
0.0.985
7.3.1
2-3.3
5-3811
19874
123254376584
The following regexp can be used: [\d\.]+[\.-][\d]{2,}
package main
import (
"regexp"
"fmt"
)
func main() {
var re = regexp.MustCompile(`(?m)[\d\.]+[\.-][\d]{2,}`)
var str = `random-app-0.0.985.tgz
busy-app-7.3.1.2-3.3.5-3811-a19874elkc-123254376584.zip`
for i, match := range re.FindAllString(str, -1) {
fmt.Println(match, "found at index", i)
}
}
The output
0.0.985 found at index 0
3.3.5-3811 found at index 1
playground
?m multi line modifier. Causes ^ and $ to match the begin/end of each line (not only begin/end of string). In this case it does not make to much difference. It will work without it.
[\d\.]+ matches at least once (quantifier +) a sequence of a digit or a dot
[\.-] matches a dot or a hypen
[\d]{2,} matches at least two digits (quantifier {2,})
One problem with your code is that in a regular expression . matches any character but you're intending it to match a literal dot. Use \. or [.] instead.
I'm trying to process the strings inputted by users, and wrote following code.
import (
"bufio"
"fmt"
"os"
"strings"
)
func main() {
var input string
fileScanner := bufio.NewScanner(os.Stdin)
fileScanner.Scan()
input = fileScanner.Text()
replaced := strings.Replace(input, "\n", "", -1)
fmt.Println(replace)
}
But I found "\n" was not replaced to "". "\n" does not seems as a string.
I tried it "." or "," instead of "\n" and it works.
I just started learning Go and this question might be too fundamental, but I appreciate for any advice.
"\" characters are processed as escape characters within a formated string and hence wont be replaced by strings.Replace
If you are really trying to replace the \n character itself, not the new line putting it inside a raw string literal should help. Refer the code below:
package main
import (
"fmt"
"strings"
)
func main() {
var input string
input = `\naaa`
replaced := strings.Replace(input, `\n`, "", -1)
fmt.Println(replaced)
}
I'm trying to parse an URL like:
http://example.com/id/123
I've read through the net/url docs but it seems like it only parses strings like
http://example.com/blah?id=123
How can I parse the ID so I end up with the value of the id in the first example?
This is not one of my own routes but a http string returned from an openid request.
In your example /id/123 is a path and you can get the "123" part by using Base from the path module.
package main
import (
"fmt"
"path"
)
func main() {
fmt.Println(path.Base("/id/123"))
}
For easy reference, here's the docs on the path module. http://golang.org/pkg/path/#example_Base
You can try using regular expression as follow:
import "regexp"
re, _ := regexp.Compile("/id/(.*)")
values := re.FindStringSubmatch(path)
if len(values) > 0 {
fmt.Println("ID : ", values[1])
}
Here is a simple solution that works for URLs with the same structure as yours (you can improve to suit those with other structures)
package main
import (
"fmt"
"net/url"
)
var path = "http://localhost:8080/id/123"
func getFirstParam(path string) (ps string) {
// ignore first '/' and when it hits the second '/'
// get whatever is after it as a parameter
for i := 1; i < len(path); i++ {
if path[i] == '/' {
ps = path[i+1:]
}
}
return
}
func main() {
u, _ := url.Parse(path)
fmt.Println(u.Path) // -> "/id/123"
fmt.Println(getFirstParam(u.Path)) // -> "123"
}
Or, as #gollipher suggested, use the path package
import "path"
func main() {
u, _ := url.Parse(path)
ps := path.Base(u.Path)
}
With this method it's faster than regex, provided you know before hand the structure of the URL you are getting.