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in your original code snippet, you're using a CString , you need to use a BSTR
use the AllocSysString() method in CString
that may help
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Ok, so are you saying I need to convert CString to BSTR and AllocSysString() to do this?
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Just to let you know that I followed what you said, but it threw an unhandled exception.
Here is the code as it on screen
OpenOutlookNewMessageWithMsgFile(CString filePath)
//Call CreateItemFromTemplate to create new mailitem using late binding
LPDISPATH mailItem
static BYTE params[] = VTS_BSTR VTS_VARIANT ;
BSTR bFilePath = filePath.AllocSysString();
outlookApp->InvokeHelper(0x10b, DISPATCH_METHOD, VT_DISPATCH, (void*)&mailItem, params, bFilePath, NULL);
outlookMailItem = new COleDispatchDriver(mailItem);
//Set User Properties
SetProperties();
//Set Display of Mail Item
VARIANT_BOOL modal = VARIANT_FALSE;
static BYTE params2[] = VTS_I4;
outlookMailItem->InvokeHelper(oxf0a6,DISPATCH_METHOD,VT_EMPTY, NULL, params2, &modal);
So, this is the whole for creating an email from template, but it is on outlookApp->InvokeHelper that I am having a problem. What am I doing wrong?
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I believe your problem is with the last parameter to "CreateItemFromTemplate". NULL is not the correct way to pass an [in, optional] VARIANT. There is no need to convert filePath to a BSTR, presuming that "outlookApp" is indeed an COleDispatchDriver. Do cast it to an LPCTSTR, InvokeHelper expects an in param defined as a VTS_BSTR to be an LPCTSTR, MFC's implementation of COleDispathDrvier will do the BSTR conversion for you inside the InvokeHelper call.
OpenOutlookNewMessageWithMsgFile(CString filePath)
{
COleVariant varOptional((long)DISP_E_PARAMNOTFOUND, VT_ERROR);
LPDISPATCH mailItem;
static BYTE params [] = VTS_BSTR VTS_VARIANT ;
outlookApp->InvokeHelper(0x10b, DISPATCH_METHOD, VT_DISPATCH, (void*)&mailItem, params, (LPCTSTR)filePath, &varOptional);
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Thank you for your reply.
Because I do not have a folder for the email to be created in and wish to use the default folder inside Outlook, what parameter should I use, or should I only pass in the parameters I am using and Outlook can then interpret this?
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Ignore my last message - I tried your solution and it worked!
Thanks for your help!
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Hello All,
I am new with c!! Can anybody give me the code for following result
Input:aababba
Output:a2bab2a
Its like encrypt given value.
Thanks in advance.
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Well since it would appear to be RLE [Run Length Encoding], I would google for that.
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Thank you.I got code for same on google!!
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here is an algorithm that should do it, in pseudo-code:
char prev=0;
int count=0;
foreach(char c in the input) {
if (c!=prev) {
if (count>1) output(count);
output(c);
prev=c;
count=0;
}
count++;
}
It is a bit tricky, I suggest you study it first (test it by hand!), then implement it in C and test it again.
Luc Pattyn [Forum Guidelines] [My Articles] Nil Volentibus Arduum
Please use <PRE> tags for code snippets, they preserve indentation, improve readability, and make me actually look at the code.
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Well, you're always right: after all, C# code is pseudo-code.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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CPallini wrote: you're always right
Luc Pattyn [Forum Guidelines] [My Articles] Nil Volentibus Arduum
Please use <PRE> tags for code snippets, they preserve indentation, improve readability, and make me actually look at the code.
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hi
i am trying to encrypting the existing file and folder using the zip archive lib.
here is the code
CZipArchive sar;
sar.Open(_T("C:\\1234.zip"), CZipArchive::zipCreate);
sar.SetCompressionMethod(CZipCompressor::methodBzip2);
sar.AddNewFile(_T("C:\\1234.txt"));
sar.AddNewFile(_T("C:\\sar1"));
sar.AddNewFile(_T("C:\\afzal.txt"));
but i want to do 1234.txt itself an zip file or directory ie existing not by adding the file or folder to a zip file .......
i studied encrypt file method i did't understood the code which they mention in the help file how to use it ..
is there any other way to encrypt it using the zip archive lib
plz help me out this
thanking you
sarfaraz
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I have done this by using an open source
Info Zip->[^]
"Every Little Smile can touch Somebody's Heart...
May we find Hundreds of Reasons to Smile Everyday... and
May WE be the Reason for someone else to smile always!" (ICAN)
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Try asking your question of the author of CZipArchive here[^].
The best things in life are not things.
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how to use it this function CZipArchive::EncryptFile()in MFC in ZIP archive lib they have mention that this function is used encrypt the existing file.i am not able to understand the what parameter does it take uindex means please make it clear with example code ......
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Did you do what I suggested?
The best things in life are not things.
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Hi,
I am working Fodero eclipse c++. I have one unsigned char(one byte unsigned integer) variable.
i am passing some integer value. like
unsigned char aa= 10;
but aa is not assign that value? How to pass that value on unsigned char.
same like i am using unsigned long, i add 'u' on the end the passing number.
unsigned long bb= 10u;
here i got bb value is 10.
how to get unsigned char value?. Any ideas or link share to me.
please urgent
Regards,
M.Mathivanan
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mathivanaan wrote: but aa is not assign that value?
how do you know this?
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mathivanaan wrote: please urgent
Sorry for not understanding: Is it the question that is urgent or the answer, or maybe both? Could you elaborate?
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If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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Your assignment looks valid. How did you test that aa does not hold the correct value? If you checked in the debugger, or pass it to cout, please remember that the type is still a character, and the standard representation is as a character. The value 10 is a non-prinitable character, and as such would show no visible output either in the debugger or in a printout, unless you convert aa to int first.
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unsigned char aa = (unsigned char)10;
BTW: when you get build-time messages or run-time errors/exceptions, you should mention them verbatim in your question.
Luc Pattyn [Forum Guidelines] [My Articles] Nil Volentibus Arduum
Please use <PRE> tags for code snippets, they preserve indentation, improve readability, and make me actually look at the code.
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mathivanaan wrote: unsigned char aa= 10;
The statement is indeed correct. Probably the fault is in the way you check it.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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You are getting a 10 as a binary value with that assignment.
If you use cout<< to print it out, however, you won't see a 10. That's because cout<< called on a char or unsigned char value always treats the value as a character, not a number. What you will see is that the next thing you output will begin on a new line. The 10 translates to a '\n' newline character.
To see the binary value of a char as a unsigned decimal value, use a typecast first.
unsigned char bb = 10;
cout << "** bb looks like [" << bb <<] this." << endl;
cout << "** (unsigned)bb looks like: " << (unsigned)bb << endl;
Which should display:
** bb looks like [
] this.
** (unsigned)bb looks like: 10
I put the unsigned char version in [] brackets so you can see what the effect is on the output.
Cheers,
Mike
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