10 Replies Latest reply on Aug 29, 2013 6:04 AM by MakeFast

# Find more than 9 digits

HI Forum.

I want to find digits more than 9 characters and retain...

eg. 12345678910

app.findGrepPreferences.findWhat = (\\d)+(\\d)+(\\d)+(\\d)+(\\d)+(\\d)+(\\d)+(\\d)+(\\d)+(\\d)

app.changeGrepPreference.changeTo = \$1\$2\$3\$4\$5\$6\$7\$8\$9 //then how to include group for 10th character...

thanks & regards...

• ###### 1. Re: Find more than 9 digits

guh.. maby you want to be a bit more clear about what tou are trying to do, because right now doesn't make much sense.

to find more than 9 digits:

find: (\\d{9,})

• ###### 2. Re: Find more than 9 digits

HI sir,

actually im find digits more than 9 and retaining the same 9 digits with some space given inbetween.

so uptill 9 digits, i can retain with \$9... so what is the option to retain 10th digits, 11th , 12th digits..

eg. if 123456789123  //here 12 digits are there

and i want to change to

12 34 567 891 23

// for this i m using

app.changeGrepPreference.changeTo = \$1\$2//s\$3\$4//s\$5\$6\$7//s\$8\$9

for nine digits to retain

after that if i give \$10,    it'sagain pasting the first digits...

Many thanks

• ###### 3. Re: Find more than 9 digits

oh now i see. as far as i know you can't use more than 9 groups in grep.

but you are going at it the wrong way. you are not looking for 12 digits, but looking for 2 digits, followed by another 2, followed by 3 more.. you see the pattern here, i guess. so your searcg should be:

(\\d{2})(\\d{2})(\\d{3})(\\d{3})(\\d{2})

and replace with:

\$1 \$2 \$3 \$4 \$5

notes: \\d{2} can also be written as \\d\\d, and \\d{3} like \\d\\d\\d

\s means any kind oh whitespace, so using it in the replace box is.. wierd. use plain space, or ~S if you need a nonbreaking space.

• ###### 4. Re: Find more than 9 digits

Hi,

If your 12 (or even more) digits are mixed with spaces irregularly you should split it in 2 steps.

First: clean it up using find: (\d)\s*(\d)     change: \$1\$2;

Jarek

• ###### 5. Re: Find more than 9 digits

Thanks you sir(s).

First used \s\$ to remove space before paragraph marks after digits.....

then....

Using Jarek idea 1st: find: (\d)\s*(\d)

(replaced * with +)

(\d)\s+(\d) ( to remove multiple spaced between any number of digits..... more than 10 digits 12 or 25 etc.,.

but what happens after the spaces are removed...

all the digits combine as single line, instead they stand on their own line....

eg.. 1 23 232 50-459459904       3434   12                   line 1

 1 23 23     2 50-459             4599      04 3434   12     line 2

now  converted to single line....

 12323250-45945990434341212323250-459459904343412 LIKE THIS

12323250-459459904343412 line one

12323250-459459904343412 line two

should be like above....

Many thanks.....

sorry if my request is confusing...... you...

• ###### 6. Re: Find more than 9 digits

Hi,

find: (\\d)?[\\t+\\x{20}+](\\d)?

change: \$1\$2

\\t      is for tabs

\\x{20}      is for regular spaces (spacebar key)

include more kinds of space into [ ] if needed

Jarek

• ###### 7. Re: Find more than 9 digits

But Jarek G..

Using \$1\$2 for retaining....gives the result...

some of the digits missing.....

also it searches and removes the spaces inside the paragraphs......

So, would it be great to find only the multiple digits and its multiple spaces accompanied....

then keep all the digits without spaces......

Manythanks.... Jarek...

• ###### 8. Re: Find more than 9 digits

Hi,

yeah, "?" means 0 or 1..., so this one:

(\\d)[\\t+\\x{20}+]+(\\d)

Jarek

• ###### 9. Re: Find more than 9 digits

Check This:

example:

1234567891012234567891012

1 23 456 7 89 101 2234567891012

(\d{1})(\d{2})(\d{3})

\$1~S\$2~S\$3~S

• ###### 10. Re: Find more than 9 digits

Thank you Jarek,

for all the reply and support....