(VB) replace quotes in a string

theString = Replace(theString, """", "")

Note: it is not 3 but 4 quotes that are needed.

See you.

The chr() function requires a long integer as a parameter. If your variable A0 is not defined, it will take 0 as its value. If it is a question of a hexadecimal value, you must "mention" it in VB, with the prefix &H...

Str = Replace(machaine, Chr(&HA0), "");

Oh, little mistake :
exact code :

laChaine = Replace(laChaine, """", "")

See you.
Slyer26

Hello,

Thank you for the response.
I had solved my problem but you are right, I had the wrong base.
Ultra Edit gave me "A0" which I transcribed as is hence my mistake.
The correct syntax was therefore Str = replace (machaine, chr(160),"")
So A0 in decimal.
Sorry for not catching on earlier..

Bonjour,
I have a problem converting quotes to ASCII code:
I want to display a string exactly like this:
desired string: (compute commande= define TpItem() "NomDuProduit" !enddefine.)

the product name is stored in a variable V_Product(i). The problem is that I can't display the quotes as string around my variable V_Product(i). Here is what I did:

TpCom = "compute command=" & Chr$(34) & define !TpItem ()"& Chr$(34)& V_Product(i) & Chr(34) & "!enddefine. & Chr(34)

the two Chr$(34) in the middle are read as characters and not as VB instructions and vice versa for the quotes.
Do you have any idea how to do it?
Thank you
Ahmed

I also think you're a bad super programmer.

Hello,

I have a problem with replacing a character in a string.
My character looks like a space, but it's actually the BOM (Byte Order Mark) code Chr(A0).
But when I try
Str = replace(machaine, chr(A0), "")
nothing happens!!!