[英]Find an integer followed by a string with Golang regexp
I want to find an integer which is followed by the term "Price: ", whether in the output, I only need to print the integer which must be excluded the term "Price: ". 我想找到一个整数,其后紧跟术语“ Price:”,无论是否在输出中,我只需要打印必须排除术语“ Price:”的整数。 Right now, my code is like this and the output is [Price: 100], but I only need 100 in the output.
现在,我的代码是这样的,输出是[Price:100],但是我只需要100。
package main
import (
"regexp"
"fmt"
)
const str = "Some strings. Price: 100$. Some strings123"
func main() {
re := regexp.MustCompile("Price:[[:space:]][0-9]+")
fmt.Println(re.FindAllString(str, -1))
}
You may use a capturing group around the number pattern and call re.FindStringSubmatch
: 您可以在数字模式周围使用捕获组,然后调用
re.FindStringSubmatch
:
package main
import (
"regexp"
"fmt"
)
const str = "Some strings. Price: 100$. Some strings123"
func main() {
re := regexp.MustCompile(`Price:\s*(\d+)`)
match := re.FindStringSubmatch(str)
if match != nil {
fmt.Println(match[1])
} else {
fmt.Println("No match!")
}
}
Note that `Price:\\s*(\\d+)`
is a raw string literal where you do not have to extra-escape backslashes that form regex escapes, so \\s*
matches zero or more whitespaces and (\\d+)
matches and captures 1+ digits into Group 1 in this pattern string literal. 请注意,
`Price:\\s*(\\d+)`
是原始字符串文字,您无需额外转义形成正则表达式转义的反斜杠,因此\\s*
匹配零个或多个空格,而(\\d+)
匹配并捕获在此模式字符串文字中,第1组中的1+位数字。
Try to use next regexp: 尝试使用下一个正则表达式:
re := regexp.MustCompile(`Price:[[:space:]]([0-9]+)`)
matches := re.FindStringSubmatch(str)
The only difference - is parentheses around [0-9]
, now you can access 100 by: matches[1]
. 唯一的区别-是
[0-9]
左右的括号,现在您可以通过以下方式访问100: matches[1]
。
Also you can replace: 您也可以替换:
[[:space:]]
with \\s
[[:space:]]
与\\s
[0-9]
with \\d
[0-9]
与\\d
so your regex will look simpler, like: Price:\\s(\\d+)
因此您的正则表达式看起来会更简单,例如:
Price:\\s(\\d+)
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