Regexp Expression

[loulou2522]

I want to extract in thos text

AD 10 000 AF 20;000 30 000

the fowlowing text in five variable

AD / 10000/ AF / 20000 / 30000

with regexp but i don't arrive.
Can someone help Me ?
Thanks in davance

[jacdelad]

Assuming you always use this exact pattern:

Code: Select all

myInput$="10 000 20;000 30 000"

CreateRegularExpression(0,"(\d+ \d+) (\d+;\d+) (\d+ \d+)")
If ExamineRegularExpression(0,myInput$) And NextRegularExpressionMatch(0)
  var1.l=Val(ReplaceString(RegularExpressionGroup(0,1)," ",""))
  var2.l=Val(ReplaceString(RegularExpressionGroup(0,2),";",""))
  var3.l=Val(ReplaceString(RegularExpressionGroup(0,3)," ",""))
EndIf

Debug var1
Debug var2
Debug var3

[jacdelad]

Oh come on, you changed your post while I was solving your problem...

Code: Select all

myInput$="AD 10 000 AF 20;000 30 000"

CreateRegularExpression(0,"(.+) (\d+ \d+) (.+) (\d+;\d+) (\d+ \d+)")
If ExamineRegularExpression(0,myInput$) And NextRegularExpressionMatch(0)
  var1.s=RegularExpressionGroup(0,1)
  var2.l=Val(ReplaceString(RegularExpressionGroup(0,2)," ",""))
  var3.s=RegularExpressionGroup(0,3)
  var4.l=Val(ReplaceString(RegularExpressionGroup(0,4),";",""))
  var5.l=Val(ReplaceString(RegularExpressionGroup(0,5)," ",""))
EndIf

Debug var1
Debug var2
Debug var3
Debug var4
Debug var5

But only, if it your data follows your exact definition.

[loulou2522]

Thanks Jacdelad
I try on my computer and apparently variable are empty and regexp is not good. Have you an idea ?

[NicTheQuick]

It's working on my side.
If it does not work on your side, you changed something and you should tell us about that.

[jacdelad]

Thanks Jacdelad
I try on my computer and apparently variable are empty and regexp is not good. Have you an idea ?

Like Nic wrote, it should work. RegExp can be an excellent option, if your input always follows the input you gave us. If not...there might be other options shich suit you better. We can help better with more information.

[loulou2522]

I maje an error on the text
the right text is

myInput$="AS 18 000 AE 25 000 35 000"

[AZJIO]

Code: Select all

Define myInput$="AD 10 000 AF 20;000 30 000"
Define Dim Result$(0)
Define i, NbFound

If CreateRegularExpression(1," \d+\K[\h;](?=\d{3})", #PB_RegularExpression_NoCase)
	myInput$ = ReplaceRegularExpression(1, myInput$, "")
Else
    Debug RegularExpressionError()
EndIf

; Debug myInput$

If CreateRegularExpression(0,"[a-z\d]+", #PB_RegularExpression_NoCase)
    Dim Result$(0)
    NbFound = ExtractRegularExpression(0, myInput$, Result$())
    For i = 0 To NbFound-1
        Debug Result$(i)
    Next
Else
    Debug RegularExpressionError()
EndIf

[Marc56us]

I maje an error on the text
the right text is

myInput$="AS 18 000 AE 25 000 35 000"

For those who don't like RegEx

Code: Select all

myInput$="AS 18 000 AE 25 000 35 000"

A$ = StringField(myInput$, 1, " ")
B$ = StringField(myInput$, 2, " ")
C$ = StringField(myInput$, 3, " ")
D$ = StringField(myInput$, 4, " ")
E$ = StringField(myInput$, 5, " ")
F$ = StringField(myInput$, 6, " ")
G$ = StringField(myInput$, 7, " ")
H$ = StringField(myInput$, 8, " ")

Debug A$
Debug B$ + C$
Debug D$
Debug E$ + F$
Debug G$ + H$

; AS
; 18000
; AE
; 25000
; 35000

[loulou2522]

Thanks jackdelad that's work goof know

[loulou2522]

Now i have another problem if in the variable there is more thant 1 space betwwen 5 000 like 5 000 (two or more space) these expression doen't match

Code: Select all

var11= "AF 35 000 AD 45   000   56 000"

f CreateRegularExpression(1," \d+\K[\h;.](?=\d{3})", #PB_RegularExpression_NoCase)
  var11 = ReplaceRegularExpression(1, var11, "")
Else
  Debug RegularExpressionError()
EndIf
If CreateRegularExpression(0,"[a-f\d]+", #PB_RegularExpression_NoCase)
  Dim Result$(0)
  NbFound = ExtractRegularExpression(0, var11, Result$())
  For i = 0 To NbFound-1
    Debug Result$(i)
  Next
  FreeRegularExpression(0)
EndIf 

Any help would be appreciated?
Thanks

[jacdelad]

Add a + after each space in the regexp or use

Code: Select all

While FindString(MyString$," ")
  MyString$=ReplaceString(MyString$," ","")
Wend

before evaluating.

[loulou2522]

Jackdelad,
Sorry but i don't arrive to make what you indicate me. Can you write the right sentence for regexp. I don't know the meaning of \K and \h

[AZJIO]

if in the variable there is more thant 1 space

Old
" \d+\K[\h;.](?=\d{3})"
New
" \d+\K[\h;.]+(?=\d{3})"

Can you write the right sentence for regexp. I don't know the meaning of \K and \h

\h = [ \t] - Any horizontal space, tabulation - Chr (9), Chr (32), Chr (160)
\K - to the left of \K the previous coincidence. Is it similar (?<=...), but at the same time does not have a restriction of the exact length
My instructions in Russian, use Google translator

[loulou2522]

this sentence is
if CreateRegularExpression(1,"\d+\K[\h]+(?=\d{3})" , #PB_RegularExpression_NoCase)