从字符串中删除多余的零

Question

I know the ways to remove leading zeros from string:

'000100'.replace(/^0+/, '')

But if a float like string (eg '00100.2300' or '00100'), how to remove all extra zeros from it?

'00100.2300' => '100.23'  
'00100' => '100'  
'00100.0023' => '100.0023'
'100.00' => '100.' or '100' will better
'-100.002300' => '-100.0023'
'0.50' => '0.5'

Assume the string:

• Only contains numbers or dot or negative symbol '-', no characters and others.
• The decimal point doesn't appear first in the string. And up to one.
• The negative symbol '-' will appear in normal position. And no extra zeros before and after the '-'.
• The float represented by the string may be larger than Number.MAX_SAFE_INTEGER
• Both positive and negative float are possible.

A function used to filter extra zeros may be releatively easy and more steps also.

And a regexp will be simpler.

Answer 1

Regex:

^0+(?!\.)|(?:\.|(\..*?))0+$

Live demo

Breakdown:

• ^0+(?!\\.) Match leading zeros that don't meet a decimal point
• | Or
• (?: Start of non-capturing group
  • \\. Match a decimal point
  • | Or
  • (\\..*?) Capture a decimal point preceding digits
• ) End of NCG
• 0*$ Match trailing zeros

JS code:

 var str = `00100.2300 00100 00100.0023 100.00 100. 00.50 0.5`; console.log( str.replace(/^0+(?!\\.)|(?:\\.|(\\..*?))0+$/gm, '$1') ) 

Answer 2

Something you can try is a native method:

parseFloat("000100.2300")

Then you can convert it back to the string with whatever method you want.

Answer 3

You can use ^0+ to capture leading zeros and (\\.\\d*[1-9])(0+)$ to capture trailing zeros. So your regex should be: /^0+|(\\.\\d*[1-9])(0+)$/g

 var res = '00100.2300'.replace(/^0+|(\\.\\d*[1-9])(0+)$/g, '$1'); console.log(res); var res2 = '100'.replace(/^0+|(\\.\\d*[1-9])(0+)$/g, '$1'); console.log(res2); 

Answer 4

It occurs to me you can do a single find and replace to accomplish
everything.

With this, you match the entire valid number at a time, enabling you to
fix multiple numbers in a single string, if done globally.
Also works the same if your string contains a single number.

Find (?:(-)(?![0.]+(?![\\d.]))|-)?\\d*?([1-9]\\d*|0)(?:(?:(\\.\\d*[1-9])|\\.)\\d*)?(?![\\d.])
Replace $1$2$3

JS demo: https://regex101.com/r/H44t6z/1

Readable / Info version

 # Add behind boundary check here
 # -----------------
 (?:
      ( - )                         # (1), Preserve sign -
      (?!                           # Only if not a zero value ahead
           [0.]+ 
           (?! [\d.] )
      )
   |                              # or
      -                             # Match sign, but dump it
 )?
 \d*?                          # Dump leading 0's
 (                             # (2 start), Preserve whole number 
      [1-9]                         # First non-0 number
      \d*                           # Any number
   |                              # or
      0                             # Just last 0 before decimal
 )                             # (2 end)
 (?:                           # Optional fraction part
      (?:                           # -------------
           (                             # (3 start), Preserve decimal and fraction
                \.                            # Decimal
                \d*                           # Any number
                [1-9]                         # Last non-0 number
           )                             # (3 end)
        |                              # or
           \.                            # Match decimal, but dump it
      )                             # -------------
      \d*                           # Dump trailing 0's
 )?
 (?! [\d.] )                   # No digits or dot ahead

Answer 5

This answer solve task even if string 100 is passed:

'00100.002300'.replace(/^0+|(\.0*\d+[^0]+)0+$/g, '$1')

100.0023

'00100'.replace(/^0+|(\.0*\d+[^0]+)0+$/g, '$1')

100

Answer 6

It might be an idea to use JS type coercion here.

 var trim = s => +s+'', strs = ['00100.2300','00100','00100.0023','100.0023'], res = strs.map(trim); console.log(res);