Replacing an escaped HTML character using StringRegExpReplace

[DobraGolonka]

I am trying to format some lines of imported HTML code which contain escaped characters, e.g.

Denial & # 38; Deprivation

(spaces introduced to allow the problem to be viewed)

should be

Denial & Deprivation

(A rather apt title, since I have been tearing my hair out on this!)

I have the following snippet of code:

$x = "Denial & Deprivation"
   $y = _FormatTitle($x)
   msgbox(0,"Why oh why doesn't this work!?!?!?!?",$y)
   
   Func _FormatTitle($s_input)
       Local $s_output = StringRegExpReplace($s_input, "&#(\d{2,3});", "\1")
       Return $s_output
   EndFunc

which extracts the decimal ascii of an ampersand quite nicely, but how on Earth do I convert/replace the decimal ascii with its equivalent (&) character????

I have tried StringRegExpReplace ( "test", "pattern", Chr("\1")) and StringRegExpReplace ( "test", "pattern", Chr(Eval("\1"))) and similar variants, but I cannot work this out for the life of me.

Please help!

Regards

DG

[Aceguy]

im not very good at this but try....

$string = "Denial & # 38; Deprivation"

$res=StringRegExpReplace($string,"[!£$%^#;:'&*()]"," ")

$ss=StringSplit($res," ")
$nn=''
for $x=1 to $ss[0]
    if $ss[$x] <> "" Then
    if StringIsDigit($ss[$x])=0 then $nn&= $ss[$x]&" "
EndIf
Next
$str=StringTrimRight($nn,1)
ConsoleWrite(">>>"&$str&"<<<"&@lf)
MsgBox(0,"",$str,0)

EDIT: just impooved it

Edited August 29, 2008 by Aceguy

[Szhlopp]

I am trying to format some lines of imported HTML code which contain escaped characters, e.g.

(spaces introduced to allow the problem to be viewed)

should be

(A rather apt title, since I have been tearing my hair out on this!)

I have the following snippet of code:

$x = "Denial & Deprivation"
   $y = _FormatTitle($x)
   msgbox(0,"Why oh why doesn't this work!?!?!?!?",$y)
   
   Func _FormatTitle($s_input)
       Local $s_output = StringRegExpReplace($s_input, "&#(\d{2,3});", "\1")
       Return $s_output
   EndFunc

which extracts the decimal ascii of an ampersand quite nicely, but how on Earth do I convert/replace the decimal ascii with its equivalent (&) character????

I have tried StringRegExpReplace ( "test", "pattern", Chr("\1")) and StringRegExpReplace ( "test", "pattern", Chr(Eval("\1"))) and similar variants, but I cannot work this out for the life of me.

Please help!

Regards

DG

Not quite sure what you want... but this does what your example says:

$x = "DE & # 38; Dep"
   $y = _FormatTitle($x)
   msgbox(0,"Why oh why doesn't this work!?!?!?!?",$y)
   
   Func _FormatTitle($s_input)
       Local $s_output
       $s_output = StringRegExpReplace($s_input, "\s#\s([0-9])*;", "")
       Return $s_output
   EndFuncoÝ÷ Ù6¤xqá®Ël{¦¦WÙ¨v'òx"µ«­¢+ØÀÌØíàôÅÕ½ÐíµÀìÌàìÀÅÕ½Ðì(ÀÌØíäô}½ÉµÑQ¥Ñ± ÀÌØíà¤(µÍ½à À°ÅÕ½Ðí]¡ä½ Ý¡ä½Í¸ÌäíÐÑ¡¥ÌݽɬÌÌìüÌÌìüÌÌìüÌÌìüÅÕ½Ðì°ÀÌØíä¤((Õ¹}½ÉµÑQ¥Ñ± ÀÌØíÍ}¥¹ÁÕФ(1½°ÀÌØíÍ}½ÕÑÁÕÐ(ÀÌØíÍ}½ÕÑÁÕÐôMÑÉ¥¹IáÁIÁ± ÀÌØíÍ}¥¹ÁÕаÅÕ½ÐìÀäÈíÍlÌÌîÀÌØìxìèÌäìµÀì¨ ¥tÀäÈíÌ¡lÀ´åt¤¨ìÅÕ½Ðì°ÅÕ½ÐìÅÕ½Ðì¤(IÑÕɸÀÌØíÍ}½ÕÑÁÕÐ(¹Õ¹

Does one of these work? If not post here with more situations and we can help you better=)

Edited August 29, 2008 by Szhlopp

[Aceguy]

StringRegExpReplace baffles me too......

getting there ,,,,,, what would be the answer to getting rid of the spaces and just leaving one.?

$string = "Denial & # 38; Deprivation"

$res=StringRegExpReplace($string,"\W", " ",0)
$res=StringRegExpReplace($res,"\d"," ",0)

ConsoleWrite($res&@lf)
;lots of spaces up till this point, ************

$ss=StringSplit($res," ")
$nn=''
for $x=1 to $ss[0]
    if $ss[$x] <> "" Then
   $nn&= $ss[$x]&" "
EndIf
Next
$str=StringTrimRight($nn,1)
ConsoleWrite(">>>"&$str&"<<<"&@lf)
MsgBox(0,"",$str,0)

[DaleHohm]

This takes all the fun out of it, but check this out: #301771

Dale

[DobraGolonka]

Hi

I don't think I explained myself very well (as usual ), and it's difficult to describe because I'm using escaped characters, which don't appear in the posts!

Anyway, the example string is "Denial & #38; Deprivation" (I have to introduce the space between the '&' and the '#', otherwise you only see '&').

When I use

StringRegExpReplace($s_input, "&#(\d{2,3});", "\1")

this nicely returns "Denial 38 Deprivation", but the problem I'm having is the syntax on how to convert the 38 to Chr(38) to return '&'.

I have tried different formats i.e. ..., Chr("\1") or ..., Eval(Chr("\1")) but nothing seems to work.

What am I doing wrong???

Regards

DG

[Aceguy]

$string = "Denial & # 38; Deprivation"

$res=StringRegExpReplace($string,"\W", " ",0)
$res=StringReplace($res,"38","&")

ConsoleWrite($res&@lf)
;lots of spaces up till this point, ************

$ss=StringSplit($res," ")
$nn=''
for $x=1 to $ss[0]
    if $ss[$x] <> "" Then
   $nn&= $ss[$x]&" "
EndIf
Next
$str=StringTrimRight($nn,1)
ConsoleWrite(">>>"&$str&"<<<"&@lf)
MsgBox(0,"",$str,0)

Edited August 30, 2008 by Aceguy

[DobraGolonka]

I've found a very similar thread describing exactly the same problem, and you can't pass functions to the StringRegExpReplace function itself!

Similar Thread

So, I need to go back to the drawing board.

Many thanks for your help on this.

DG

[Szhlopp]

I've found a very similar thread describing exactly the same problem, and you can't pass functions to the StringRegExpReplace function itself!

Similar Thread

So, I need to go back to the drawing board.

Many thanks for your help on this.

DG

Obviously this would need some modifying, but here you go

$String = "Escaped & #38; Chars"
$regexp = StringRegExp($String, "&(.*);", 1)
$regexp = Chr(StringTrimLeft($regexp[0], 1))
$regexpRep = StringRegExpReplace($String, "&#(\d{2,3});", $regexp)
MsgBox(0, "", $regexpRep)

I've forgotten how all the escaped characters are formatted, so you would have to fix the regexp/regexrep commands to make it work with all the escaped character sets. But this is the basic idea of how to find one and replace it easily=)

Edited August 30, 2008 by Szhlopp

[DobraGolonka]

Well, I've now moved away from the StringRegExpReplace route, and come up with this, which does the trick:

$x = "Witness: Truth & Lies"
$y = _FormatTitle($x)
MsgBox(0, "Oh Joy! It Does Work!", $y)

Func _FormatTitle($s_input)
    Local $s_new_input = $s_input, $s_output

    For $i = 32 To 255
        If StringInStr($s_new_input, "&#" & $i & ";") Then
            $s_output = StringReplace($s_new_input, "&#" & $i & ";", Chr($i))
            $s_new_input = $s_output
        EndIf
    Next
    Return $s_output
EndFunc

It's just a pity that StringRegExpReplace doesn't accept functions - this would have been by far the more elegant route to take.

Regards

DG

Edited August 30, 2008 by DobraGolonka