On 17.04.2013 17:09, jorge ivan poot diaz wrote:
I don't see a definition of aName, but from the error message assume that
it is String.
http://imagebin.org/254354
That is because the String class has no << operator defined. Try to use
aName.getStr() instead of just aName.
http://imagebin.org/254355
As you can see from the picture, I already made the change, but I have the
following error:
http://imagebin.org/254357
I try with aEdtName.GetText () instead of aName.getStr() but yet does not
compile:
http://imagebin.org/254361
Help, ragards.
Sorry, I do not use String anymore and had to guess and apparently
guessed wrong. If you can, you probably should use ::rtl::OUString
instead of String. The conversion to char* looks like this:
::rtl::OUString sText (::rtl::OUString::createFromAscii("some text"));
::std::cout << ::rtl::OUStringToOString(sText,
RTL_TEXTENCODING_ASCII_US).getStr() << ::std::endl;
or something very similar to this.
If you still want to use String then cosv/inc/cosv/string.hxx suggests
to use c_str() to obtain a char*.
-Andre
2013/4/17 Andre Fischer <awf....@gmail.com>
On 17.04.2013 01:46, jorge ivan poot diaz wrote:
Hello,
I am modifying this code in AOO:
http://opengrok.adfinis-**sygroup.org/source/xref/aoo-**
trunk/main/cui/source/**tabpages/tpcolor.cxx#465<http://opengrok.adfinis-sygroup.org/source/xref/aoo-trunk/main/cui/source/tabpages/tpcolor.cxx#465>
I'm trying to concatenate with stringstream but I have errors when I'am
building:
1- I already included:
#include <sstream>
2- I have added the following code:
http://imagebin.org/254288
I don't see a definition of aName, but from the error message assume that
it is String.
3- This is the error:
http://imagebin.org/254289
That is because the String class has no << operator defined. Try to use
aName.getStr() instead of just aName.
-Andre
Help me, regards!!
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