寻找子字符串替代javascript

Question

Basically my problem is I am using the substring method on the variable version for the result then to be used inside a URL using ng-href:

substring(0, 3)
version 9.1.0 = 9.1 (good)
version 9.2.0 = 9.2 (good)
version 9.3.0 = 9.3 (good)
..
version 9.10.0 = 9.1 (breaks here)
version 10.1.0 = 10. (breaks here)

As you can see eventually the substring method stops working, how can I fix this??

Answer 1

等效的是substring()检查 MDN 文档

Answer 2

Use split and join on the dot, and during the time you manipulate an array, use slice to remove the last item:

 const inputs = ['9.1.0', '9.2.0', '9.3.0', '9.10.0', '10.1.0', '22.121.130']; inputs.forEach(input => { const result = input.split('.').slice(0, -1).join('.'); console.log(input, '=>', result); }) 

Simple enough and it will work whatever your version number :)

Hoping that will help you!

Answer 3

/^\\d+\\.\\d+/ will match the first 2 digits with dot between.

The regex will not have to process the entire input like the split approaches do.

It will also catch consecutive . like 30..40 . And spaces.

It will even catch alphabetic parts like 10.B

This also will extend if you want to start allowing segments such as -alpha , -beta , etc.

 const rx = /^\\d+\\.\\d+/ const inputs = ['9.1.0', '9.2.0', '9.3.0', '9.10.0', '10.1.0', , '22.121.130', '10.A', '10..20', '10. 11', '10 .11']; inputs.forEach(input => { const m = rx.exec(input) console.log(input, m ? m[0] : 'not found') }) 

Answer 4

You can get the substring the other way by removing the last two character which will be a dot and a numeric character:

 function version(val){ console.log(val.substring(0,val.length-2)) } version('9.1.0'); version('9.2.0'); version('9.3.0'); version('9.10.0'); version('10.1.0'); 

But what if there are two numeric character and a dot at the end? Here is that solution:

 function version(val){ var tempVersion = val.split('.'); tempVersion.pop(); console.log(tempVersion.join('.')) } version('9.1.0'); version('9.2.0'); version('9.3.1020'); version('9.10.0'); version('10.1.0123');