excracting char from string

[erezlevi]

if I have the following string: "lskjfdlksjdflkjsdferezlskjflskjdflkdsjferezlksdjflskdjflkdsjerezlksdjflskdjflkjdserez"

and I want to get the "erez" out of there an put a space instead, what is the way to do it? I though that stringreplace will help here but " " is less then "erez" in the number of char. so it replaces "erez" with " rez"

[Richard Robertson]

That sounds like incorrect behavior. Replacing something with something else should remove the entire entry and replace it. Did you actually get that result?

[erezlevi]

That sounds like incorrect behavior. Replacing something with something else should remove the entire entry and replace it. Did you actually get that result?

yes, and it is because the stringreplace is the count value only gives you this:

count [optional] The number of times to replace the searchstring.

0 = all searchstrings will be replaced (default)

but not to replace from e to z with " ".

[erezlevi]

yes, and it is because the stringreplace is the count value only gives you this:

count [optional] The number of times to replace the searchstring.

0 = all searchstrings will be replaced (default)

but not to replace from e to z with " ".

sorry I was wrong that is not the problem I am facing. the problem is more illustrated in the previous post I made in the first page. can you look there? here is a short version:

this is the string:

"list bcms skill 107 [1;1H[24;0H[K7[1;1H[0;7mlist bcms skill 107 [0m8[23;0H[0;7m["

$m=string($s)
$p1=StringInStr($m,Chr(27))
$p2=StringInStr($m,"H")
$p3=($p2-$p1)+1
$p4=StringMid ($m,$p1,$p3)
MsgBox (0,"$m before",$m)
$m=StringReplace ($m,$p1," ",$p3)
StringReplace (
MsgBox (0,"",$p1&$p2 & @CRLF & $p3 & @CRLF & $p4)
MsgBox (0,"$m after",$m)
FileOpen ("c:\erez.txt",2)
FileWrite ("c:\erez.txt",$m)

I need to extract everything in this line that start with Chr(27) and end with "H" like this: [1;1H or this: [23;0H.

the problem is that with VB script I have the command:

Replace(s, Mid(s, InStr(1, s, Chr(27)), InStr(InStr(1, s, Chr(27)), s, "H") - InStr(1, s, Chr(27)) + 1), " ", 1)

and with AutoIT the StringReplace does not have border! it only have starting point and replacement string but it does not replace the whole string if the replacment is not equal to the total substring in the first place.

how can I use StringRegExpReplace here?

Edited January 29, 2008 by erezlevi

[GEOSoft]

$String = "lskjfdlksjdflkjsdferezlskjflskjdflkdsjferezlksdjflskdjflkdsjerezlksdjflskdjflkjdserez"

$String = StringReplace($String, "erez", " ")

Edited January 29, 2008 by GEOSoft

[erezlevi]

$String = "lskjfdlksjdflkjsdferezlskjflskjdflkdsjferezlksdjflskdjflkdsjerezlksdjflskdjflkjdserez"

$String = StringReplace($String, "erez", " ")

I know I made a mistake when trying to made it easier to understand, look above it is the right scenario.

[Siao]

_StringBetween()

[erezlevi]

_StringBetween()

ok, no escape from using Array this time. I will have to make an Array and put all the strings i found that start with {esc} and ends with "H"

they can be:

" [1;1H "

"[ VB is like AutoIT but have much more options H"

and many more.

putting them inside an Array and make a replace string for each and every one of them..... ohhhhr

When VB is much better the AutoIT you must say so... sorry. (JON...)

[GEOSoft]

ok, no escape from using Array this time. I will have to make an Array and put all the strings i found that start with {esc} and ends with "H"

they can be:

" [1;1H "

"[ VB is like AutoIT but have much more options H"

and many more.

putting them inside an Array and make a replace string for each and every one of them..... ohhhhr

When VB is much better the AutoIT you must say so... sorry. (JON...)

Look at _StringBetween() in the UDFs

[Squirrely1]

I don't know what is wrong with this code, StringReplace works everytime it's tried:

$z = "lskjfdlksjdflkjsdferezlskjflskjdflkdsjferezlksdjflskdjflkdsjerezlksdjflskdjflkjdserez"
ConsoleWrite(@CR & 'Original string length: ' & StringLen($z) & @CR)
$x = StringReplace ($z, "erez", " ", 0 ,0)
ConsoleWrite("Number of Replacements made: " & @extended & @CR)
ConsoleWrite('Final string length: ' & StringLen($x) & @CR)
ConsoleWrite('Final string: ' & $x & @CR & @CR)

Original string length = 85

Four replacements times four characters = 16 + four space characters = 12

Final length = 73

Edited January 29, 2008 by Squirrely1

[GEOSoft]

I don't know what is wrong with this code, StringReplace works everytime it's tried:

$z = "lskjfdlksjdflkjsdferezlskjflskjdflkdsjferezlksdjflskdjflkdsjerezlksdjflskdjflkjdserez"
ConsoleWrite(@CR & 'Original string length: ' & StringLen($z) & @CR)
$x = StringReplace ($z, "erez", " ", 0 ,0)
ConsoleWrite("Number of Replacements made: " & @extended & @CR)
ConsoleWrite('Final string length: ' & StringLen($x) & @CR)
ConsoleWrite('Final string: ' & $x & @CR & @CR)

Original string length = 85

Four replacements times four characters = 16 + four space characters = 12

Final length = 73

If you go back and read post 4 you will see whats wrong with it.

[erezlevi]

I'v solved the problem, using Array's. Thanks for your help.