Southmead Hospital
Tel: 0117 9505050
Address: Southmead Road, Westbury-on-Trym, Bristol, Avon, BS10 5NB
Website: http://www.nbt.nhs.uk
Email: complaints@nbt.nhs.uk
I need the output like this
Southmead Hospital
Tel: 0117 9505050
Address: Southmead Road, Westbury-on-Trym, Bristol, Avon, BS10 5NB
Website: http://www.nbt.nhs.uk
Email: complaints@nbt.nhs.uk
Yoichi
(Yoichi)
2
Hi,
There are some ways to achieve it. Hope the following helps you.
System.Text.RegularExpressions.Regex.Replace(yourString,"(\r?\n){2,}",vbcrlf)
or
String.Join(vbcrlf,yourString.Split(vbcrlf.ToCharArray,System.StringSplitoptions.RemoveEmptyEntries))
Regards,
This is working fine. Thanks
System.Text.RegularExpressions.Regex.Replace(yourString,“(\r?\n){2,}”,vbcrlf)
1 Like
Could you please help me with this example
1.Withybush General Hospital
Fishguard Road
Haverfordwest
SA61 2PZ
01437 764545
Open 24 hours Open
Send to mobile
Services offered
More Information
Map/Directions
Output Should be like this
Withybush General Hospital
Fishguard Road
Haverfordwest
SA61 2PZ
01437 764545
Adrian_Star
(Adrian Starukiewicz)
6
Hi,
example for you, as @Yoichi mentioned:
output = String.Join(vbCrLf, your_String.Split(Environment.NewLine.ToArray, StringSplitOptions.RemoveEmptyEntries))
output = String.Join(vbCrLf, your_String.Split(vbCrLf.ToArray, StringSplitOptions.RemoveEmptyEntries))
Remove New Line from whole string:
Replace(vbCr, "")
- removes \r
[CR (Carriage Return)]
Replace(vbLf, "")
- removes \n
[LF (Line Feed, Environment.NewLine)]
Replace(vbCrLf, "")
- removes \r\n
[CR+LF ( End Of Line)]
Remove New Line only from the end of string:
textString.TrimEnd(vbCr.ToCharArray)
textString.TrimEnd(vbCrLf.ToCharArray)
textString.TrimEnd(vbCrLf.ToCharArray)
Check this post:
1 Like
system
(system)
Closed
8
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