listview characters

[vistar86]

Hi,

i am using Microsoft.XMLHTTP to pull information from a php page on my server, and they look exactly how they should look via firefox webbrowser, but via Microsoft.XMLHTTP outputted to a Listview the special characters aren't formmated correctly.

eg,

& #39; (had to put a space after & as it turns into a apostrophe on the site)

instead of

'

This is a encoding issue? how do i encode it to utf-8 if this is the case?

thanks in advance

[minitech]

& #39; is an HTML character code, not an encoding issue... you'll probably need a regular expression, like so:

Dim re As New System.Text.RegularExpressions.Regex("&#(\d+);", System.Text.RegularExpressions.RegexOptions.SingleLine)

Then you can match it with each string you're outputting and access the code:

For Each m As System.Text.RegularExpressions.Match In re.Match(my string)
     Dim cc As Integer = Integer.Parse(re.Groups(1).Value)
     Dim chr As Char = AscW(cc)
     (my string) = (my string).Remove(m.Index,m.Length).Insert(m.Index,chr)
Next

You've got to take into account things like ' too, though.

[vistar86]

For Each m As System.Text.RegularExpressions.Match In re.Match(my string)
     Dim cc As Integer = Integer.Parse(re.Groups(1).Value)
     Dim chr As Char = AscW(cc)
     (my string) = (my string).Remove(m.Index,m.Length).Insert(m.Index,chr)
Next

ok this is good, but when i put string into "my string" it states "cannot be converted to -1 dimension bla bla"

this is the changed code

Dim download_list() As String
        download_list = Split(listmain, "<##>")
 
        For Each line As String In download_list
            'if line is null, skip it
            If line = vbNullString Then Exit For
            'Dim info As New UserControl1()
            Dim values() As String
            values = Split(line, "<**>")
 
For Each m As System.Text.RegularExpressions.Match In re.Match(values)
     Dim cc As Integer = Integer.Parse(re.Groups(1).Value)
     Dim chr As Char = AscW(cc)
     (values) = (my string).Remove(m.Index,m.Length).Insert(m.Index,chr)
Next
 
            Dim str(3) As String
            Dim itm As ListViewItem
            str(0) = values(2)
            str(1) = values(1)
            str(2) = values(3)
 
            itm = New ListViewItem(str)

whats the correct syntax for multiple values?

[vistar86]

i just also tried it before the splitting so its all one string still like so:

listmain = oHTTP.responseText
 
 
        For Each m As System.Text.RegularExpressions.Match In re.Match(listmain)
            Dim cc As Integer = Integer.Parse(re.Groups(1).Value)
            Dim chr As Char = AscW(cc)
            listmain = (listmain).Remove(m.Index, m.Length).Insert(m.Index, chr)
        Next

still dont work, gives error:

Expression is of type 'System.Text.RegularExpressions.Match', which is not a collection type.

[vistar86]

i have done it now, like this (it replaces the most common characters):

Please tell me if there is an easier (more smaller/simpler way .

listmain = oHTTP.responseText

        listmain = Regex.Replace(listmain, "&", "&")
        listmain = Regex.Replace(listmain, "'", "'")
        listmain = Regex.Replace(listmain, " ", " ")
        listmain = Regex.Replace(listmain, "!", "!")
        listmain = Regex.Replace(listmain, """, "'")
        listmain = Regex.Replace(listmain, "#", "#")
        listmain = Regex.Replace(listmain, "$", "$")
        listmain = Regex.Replace(listmain, "%", "%")
        listmain = Regex.Replace(listmain, "&", "&")
        listmain = Regex.Replace(listmain, "'", "'")
        listmain = Regex.Replace(listmain, "(", "(")
        listmain = Regex.Replace(listmain, ")", ")")
        listmain = Regex.Replace(listmain, "*", "*")
        listmain = Regex.Replace(listmain, "+", "+")
        listmain = Regex.Replace(listmain, ",", ",")
        listmain = Regex.Replace(listmain, "-", "-")
        listmain = Regex.Replace(listmain, ".", ".")
        listmain = Regex.Replace(listmain, "/", "/")
        listmain = Regex.Replace(listmain, ":", ":")
        listmain = Regex.Replace(listmain, ";", ";")
        listmain = Regex.Replace(listmain, "<", "<")
        listmain = Regex.Replace(listmain, "=", "=")
        listmain = Regex.Replace(listmain, ">", ">")
        listmain = Regex.Replace(listmain, "?", "?")
        listmain = Regex.Replace(listmain, "@", "@")
        listmain = Regex.Replace(listmain, "[", "[")
        listmain = Regex.Replace(listmain, "\", "\")
        listmain = Regex.Replace(listmain, "]", "]")
        listmain = Regex.Replace(listmain, "&#94;", "^")
        listmain = Regex.Replace(listmain, "_", "_")
        listmain = Regex.Replace(listmain, "{", "{")
        listmain = Regex.Replace(listmain, "|", "|")
        listmain = Regex.Replace(listmain, "}", "}")
        listmain = Regex.Replace(listmain, "~", "~")
        listmain = Regex.Replace(listmain, "¡", "¡")
        listmain = Regex.Replace(listmain, "¢", "¢")
        listmain = Regex.Replace(listmain, "£", "£")
        listmain = Regex.Replace(listmain, "¤", "¤")
        listmain = Regex.Replace(listmain, "¥", "¥")
        listmain = Regex.Replace(listmain, "¦", "¦")
        listmain = Regex.Replace(listmain, "§", "§")
        listmain = Regex.Replace(listmain, "¨", "¨")
        listmain = Regex.Replace(listmain, "©", "©")
        listmain = Regex.Replace(listmain, "ª", "ª")
        listmain = Regex.Replace(listmain, "«", "«")
        listmain = Regex.Replace(listmain, "¬", "¬")
        listmain = Regex.Replace(listmain, "®", "®")
        listmain = Regex.Replace(listmain, "¯", "¯")
        listmain = Regex.Replace(listmain, "°", "°")
        listmain = Regex.Replace(listmain, "±", "±")
        listmain = Regex.Replace(listmain, "´", "´")
        listmain = Regex.Replace(listmain, "·", "·")
        listmain = Regex.Replace(listmain, "»", "»")
        listmain = Regex.Replace(listmain, "¿", "¿")
        listmain = Regex.Replace(listmain, "&#x22;", "\")
        listmain = Regex.Replace(listmain, "&#x2039;", "<")
        listmain = Regex.Replace(listmain, "&#x203A;", ">")
        listmain = Regex.Replace(listmain, "&gt;", ">=")
        listmain = Regex.Replace(listmain, "&lt;", "<=") 

This should come to some use for someone alse also with same problem.

EDIT: cannot get it to show as characters are being converted when submitted :/ you get the idea anyway.

[minitech]

What in the world is that code supposed to do???! The problem with when you replaced "my string" is that you kept the parenthesis. Delete them.

What your code is currently doing is replacing a few characters with themselves using Regex. That is 100% bad coding and useless. Remove it!

[vistar86]

no the forums here re-encodes the html special characters so does not look like above, an example of how the above should look:

listmain = Regex.Replace(listmain, "& #35;", "#")

and so on...

had to put a space after & because like i say it convertes the characters on this forum.

I know i dont even need regex, only replace would work.

oh and if your solution worked id have used that, but you see here is the code and i dont put parenthesis, and get the error look:

Dim values() As String
values = Split(line, "<**>")

For Each m As System.Text.RegularExpressions.Match In re.Match(values)
 
     Dim cc As Integer = Integer.Parse(re.Groups(1).Value)
 
     Dim chr As Char = AscW(cc)
 
     (values) = (my string).Remove(m.Index,m.Length).Insert(m.Index,chr)
 
Next

[minitech]

Look, you're doing it completely wrong. What I gave you originally is the easiest solution. Just take out the parenthesis.

[vistar86]

Where's the parethesis??

Dim re As New System.Text.RegularExpressions.Regex("&#(\d+);", System.Text.RegularExpressions.RegexOptions.SingleLine)
 
        For Each line As String In download_list
            'if line is null, skip it
            If line = vbNullString Then Exit For
            Dim values() As String
            values = Split(line, "<**>")
 
     For Each m As System.Text.RegularExpressions.Match In re.Match(values)
 
           Dim cc As Integer = Integer.Parse(re.Groups(1).Value)
 
           Dim chr As Char = AscW(cc)
 
           (values) = (my string).Remove(m.Index,m.Length).Insert(m.Index,chr)
 
      Next 
 
          Dim str(3) As String
            Dim itm As ListViewItem
            str(0) = values(2)
            str(1) = values(1)
            str(2) = values(3)
 
            itm = New ListViewItem(str)

[minitech]

Here is your code, fixed:

Dim re As New System.Text.RegularExpressions.Regex("&#(\d+);", System.Text.RegularExpressions.RegexOptions.SingleLine)
         For Each line As String In download_list
            Dim values() As String
            values = Split(line, "<**>")
            For Each m As System.Text.RegularExpressions.Match In re.Match(values)
               Dim cc As Integer = Integer.Parse(re.Groups(1).Value)
               Dim chr As Char = AscW(cc)
               line = line.Remove(m.Index,m.Length).Insert(m.Index,chr)
            Next
            Dim str(3) As String
            Dim itm As ListViewItem
            str(0) = values(2)
            str(1) = values(1)
            str(2) = values(3)
             itm = New ListViewItem(str)

[vistar86]

still says an error on the:

In re.Match(values)

Value of type '1-dimensional array of String' cannot be converted to 'String'.

[minitech]

Ah OK, you weren't supposed to replace it with that. By (my string) I meant the actual string. Use this:

Dim re As New System.Text.RegularExpressions.Regex("&#(\d+);", System.Text.RegularExpressions.RegexOptions.SingleLine)
         For Each line As String In download_list
            For Each m As System.Text.RegularExpressions.Match In re.Match(line)
               Dim cc As Integer = Integer.Parse(re.Groups(1).Value)
               Dim chr As Char = AscW(cc)
               line = line.Remove(m.Index,m.Length).Insert(m.Index,chr)
            Next
            Dim values() As String
            values = Split(line, "<**>")
            Dim str(3) As String
            Dim itm As ListViewItem
            str(0) = values(2)
            str(1) = values(1)
            str(2) = values(3)
             itm = New ListViewItem(str)

I also don't recommend using the VB6 Split() function. But it doesn't make a difference anyways.