C++ program to print the path without the file name!
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And what is the question? :)
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] -
Use
PathRemoveFileSpec()or_splitpath()."Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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Use
PathRemoveFileSpec()or_splitpath()."Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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See here.
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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See here.
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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The code you posted does 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] -
The code you posted does 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] -
my boss I don't want to use Microsoft C++ , I am using the lagacy C/C++ actually , my code is working but i wanna print the path without the file name as your suggested function. Thanks
You might want to consider
strrchr()to find the last backslash. Put a'\0'character there."Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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this logic worked for me perfectly.
const char \* path = "/work1/data/xxxx/yyy/file\_name.txt"; char path1\[256\]; strcpy(path1,path);//copying the char array cout << path <<"\\n";//this prints the whole filename with path. int l=strlen(path);//finding the length of the path for(int i=l;i>0;i--)//reverse traversing the array { if(path1\[i\]=='/')//if '/' is found, then replacing it with '\\0'. { path1\[i\]='\\0'; break; } } cout << path1<<"\\n";//this prints only the path.the output is as you desired.
-------------------------------------------- Suggestion to the members: Please prefix your main thread subject with [SOLVED] if it is solved. thanks. chandu.
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the code i posted prints the base file name not the path I want to print the path without file name
see here http://www.codeproject.com/Messages/3118524/Re-Cplusplus-program-to-print-the-path-without-the-file-name.aspx[^] and also try with the David crow's recent post that would be easier than mine.
-------------------------------------------- Suggestion to the members: Please prefix your main thread subject with [SOLVED] if it is solved. thanks. chandu.
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this logic worked for me perfectly.
const char \* path = "/work1/data/xxxx/yyy/file\_name.txt"; char path1\[256\]; strcpy(path1,path);//copying the char array cout << path <<"\\n";//this prints the whole filename with path. int l=strlen(path);//finding the length of the path for(int i=l;i>0;i--)//reverse traversing the array { if(path1\[i\]=='/')//if '/' is found, then replacing it with '\\0'. { path1\[i\]='\\0'; break; } } cout << path1<<"\\n";//this prints only the path.the output is as you desired.
-------------------------------------------- Suggestion to the members: Please prefix your main thread subject with [SOLVED] if it is solved. thanks. chandu.
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the code i posted prints the base file name not the path I want to print the path without file name
you're right:
const char * path ="/work1/data/xxxx/yyy/file_name.txt";
char * basename;
char * filename;basename = _strdup(path);
if ( !basename){/*handle error*/}filename = basename+strlen(basename);
while (filename != basename && *(filename) != '/')
filename--;if ( filename != basename)
{
*(filename+1)='\0';
}
else
*basename='\0';printf("%s\n", basename);
free(basename); /* REMEMBER TO FREE */:)
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] -
this logic worked for me perfectly.
const char \* path = "/work1/data/xxxx/yyy/file\_name.txt"; char path1\[256\]; strcpy(path1,path);//copying the char array cout << path <<"\\n";//this prints the whole filename with path. int l=strlen(path);//finding the length of the path for(int i=l;i>0;i--)//reverse traversing the array { if(path1\[i\]=='/')//if '/' is found, then replacing it with '\\0'. { path1\[i\]='\\0'; break; } } cout << path1<<"\\n";//this prints only the path.the output is as you desired.
-------------------------------------------- Suggestion to the members: Please prefix your main thread subject with [SOLVED] if it is solved. thanks. chandu.