If it's a char, then you can just use atoi([var]) or (int)[var]+1. I'm
wondering what your database field is though. If it's like MySQL, then
every field returned is a string value, in which case you'd use something
like this:
int val =3D atoi(returned_field)+1;
You'd have to have your own updating code, as your description doesn't give
much to go by.
At 10:51 AM 11/05/2003 +0100, Daniel Glenfield wrote:
>I'm interested to know why you would add 1 to a char, but you can convert=
=20
>using atoi()
>
>it's something like
>
>int Buffer[10]; // create a buffer
>atoi(Buffer, StringName, 10); // Where to store the int, your string to=
=20
>convert, and the base type (10 for decimal).
>
>Do a search on google or msdn.
>
>
>>From: "Fernando Correa da Concei;=E1o" <fer...@te...>
>>To: dev...@li...
>>Subject: [Dev-C++] Converting types(How to do it?)
>>Date: Sat, 10 May 2003 17:08:29 -0300
>>
>>Hi!
>>I=B4m trying to use a database, but I need this, I have a field(a year by=
=20
>>example), it=B4s a char.
>>What I need is convert it to an int, add +1 and convert again to char.
>>
>>Thanks!
>>
>>
>>
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>-------------------------------------------------------
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>www.enterpriselinuxforum.com
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