Parse custom data by JavaScript regular expression

Question

I have input like below in a js variable var str

MZ 09/10/2020 Zvwi‡L AbywôZ MCQ Test G DËxY© cÖv_©x‡`i ‡ivj b¤^‡ii ZvwjKv t
(cÖKvwkZ djvd‡j †Kv‡bv Kvi‡Y †Kv‡bv ms‡kva‡bi cÖ‡qvRb n‡j KZ©„cÿ Zv ms‡kva‡bi AwaKvi msiÿY K‡i)
9007 1027 1028 1029
7038 8040 1054 1055
0058 3062 1067 1069
3078 7097 1098 1106 = 16 Rb Kvi‡Y †Kv‡bv ms‡kva‡bi cÖ‡qvRb 3457,00867,1122 jtukh $3308

I wanna parse only all 4 digit numeric values from 9007 to 1106 (before = 16 Rb) in an array.

[9007,1027,1028,1029,7038,8040,1054,1055,0058,3062,1067,1069,3078,7097,1098,1106]

I tried with str.match(/\d{4}/g) but it will give me all 4 digits numeric values and 2020, null and unexpected result also.

First, extract that substring with var block = s.match(/9007[\s\S]*?1106/)[0] then use your regex against block. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Oct 22, 2020 at 13:45
Match parts of 4 digits till 16 Rb and then split on a space ^\d{4}(?:\s+\d{4})+(?= = 16 Rb) regex101.com/r/1mWmQJ/1 — The fourth bird
– The fourth bird, Commented Oct 22, 2020 at 13:48
Values are not static. so that, I can not bind from 9007 to 1106. — Akash khan
– Akash khan, Commented Oct 22, 2020 at 13:49

Mark Reed · Accepted Answer · 2020-10-22 14:47:42Z

2

What's the predictable pattern? If it's as simple as "four digits with space on both sides", then you can use something like /(?<=\s)\d{4}(?=\s)/ (using lookaround assertions stops any number's match from "eating" the space around it and preventing that space from matching for the adjacent number):

const t='MZ 09/10/2020 Zvwi‡L AbywôZ MCQ Test G DËxY© cÖv_©x‡`i ‡ivj b¤^‡ii ZvwjKv t\n(cÖKvwkZ djvd‡j †Kv‡bv Kvi‡Y †Kv‡bv ms‡kva‡bi cÖ‡qvRb n‡j KZ©„cÿ Zv ms‡kva‡bi AwaKvi msiÿY K‡i)\n9007 1027 1028 1029\n7038 8040 1054 1055\n0058 3062 1067 1069\n3078 7097 1098 1106 = 16 Rb Kvi‡Y †Kv‡bv ms‡kva‡bi cÖ‡qvRb 3457,00867,1122 jtukh $3308\n'
console.log(t.match(/(?<=\s)\d{4}(?=\s)/g))
[
  '9007', '1027', '1028',
  '1029', '7038', '8040',
  '1054', '1055', '0058',
  '3062', '1067', '1069',
  '3078', '7097', '1098',
  '1106'
]

If the pattern isn't so simple, then of course the solution won't be either. It depends on what you can assume about the text.

Also, if this is client-side code, be aware that some browsers (e.g. Safari) still don't support lookbehind assertions like (?<=\s), and even those that support it do so only in their more recent versions, since the feature was only added to the language in the ES2018 specification.

You could get away with using only lookahead and allowing the preceding space to be consumed by the regex, but then you'll want to use a capture group so that the space isn't part of the returned match, which means you have to use matchAll instead of match and grab the capture group from each result. So this slightly more complicated expression works in more browsers:

Array.from(t.matchAll(/\s(\d{4})(?=\s)/g)).map(m => m[1])

Either way, what you wind up with is an array of Strings. Your example output has unquoted numbers with leading zeroes, so it's not clear if that's what you want. If you want an array of Numbers, you can get that by either adding a map call to the first answer or modifying the existing map call in the second:

t.match(/(?<=\s)\d{4}(?=\s)/g).map(Number)
//or
Array.from(t.matchAll(/\s(\d{4})(?=\s)/g)).map(m => Number(m[1]))

Which in your example gets you this array of numeric values:

[
  9007, 1027, 1028, 1029,
  7038, 8040, 1054, 1055,
    58, 3062, 1067, 1069,
  3078, 7097, 1098, 1106
]

edited Oct 22, 2020 at 14:47

answered Oct 22, 2020 at 13:51

Mark Reed

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2 Comments

The fourth bird Over a year ago

But that matches all 4 digits in the text surrounded by whitespace chars, and it would not match only 4 digits at the start of the string.

Akash khan Over a year ago

In 16 Rb where 16 is the count of all 4 digit number. Your solution works fine until 999 Rb up to 3 digit count. When the count reaches 4 digits then it'll include 4 digit Rb number also. @mark-reed

The fourth bird · Accepted Answer · 2020-10-22 13:57:30Z

1

You could match parts of 4 digits from the start of the string and then split on whitespace chars.

^\d{4}(?:\s+\d{4})+(?= = \d+ Rb\b)

^ Start of string
\d{4} Match 4 digits
(?:\s+\d{4})+ Repeat matching 1+ whitespace chars and 4 digitd
(?= = \d+ Rb\b) Positive lookahead, assert what is directly at the right is a space, = 1+ digits and Rb

Regex demo

    const regex = /^\d{4}(?:\s+\d{4})+(?= = \d+ Rb\b)/gm;
    const str = `MZ 09/10/2020 Zvwi‡L AbywôZ MCQ Test G DËxY© cÖv_©x‡\`i ‡ivj b¤^‡ii ZvwjKv t
(cÖKvwkZ djvd‡j †Kv‡bv Kvi‡Y †Kv‡bv ms‡kva‡bi cÖ‡qvRb n‡j KZ©„cÿ Zv ms‡kva‡bi AwaKvi msiÿY K‡i)
9007 1027 1028 1029
7038 8040 1054 1055
0058 3062 1067 1069
3078 7097 1098 1106 = 16 Rb Kvi‡Y †Kv‡bv ms‡kva‡bi cÖ‡qvRb 3457,00867,1122 jtukh \$3308`;
    console.log(str.match(regex)[0].split(/\s+/));

edited Oct 22, 2020 at 13:57

answered Oct 22, 2020 at 13:52

The fourth bird

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2 Comments

Akash khan Over a year ago

16 Rb not static value. 16 is the count of all 4 digit number.

The fourth bird Over a year ago

@Akashkhan Then use ^\d{4}(?:\s+\d{4})+(?= = \d+ Rb\b) See regex101.com/r/7HGM9d/1

Nooruddin Lakhani · Accepted Answer · 2020-10-22 13:55:49Z

0

Try this

const str = 'MZ 09/10/2020 Zvwi‡L AbywôZ MCQ Test G DËxY© cÖv_©x‡`i ‡ivj b¤^‡ii ZvwjKv t (cÖKvwkZ djvd‡j †Kv‡bv Kvi‡Y †Kv‡bv ms‡kva‡bi cÖ‡qvRb n‡j KZ©„cÿ Zv ms‡kva‡bi AwaKvi msiÿY K‡i) 9007 1027 1028 1029 7038 8040 1054 1055 0058 3062 1067 1069 3078 7097 1098 1106 = 16 Rb Kvi‡Y †Kv‡bv ms‡kva‡bi cÖ‡qvRb 3457,00867,1122 jtukh $3308';

var n = str.indexOf("9007");
var s = str.indexOf("1106");

var res = str.substring(n, s).trim();
var data = res.split(" ");

answered Oct 22, 2020 at 13:55

Nooruddin Lakhani

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