String

[magix]

Hi,

If I have string like below:
InputString= "@123456^BI LL GATES^APR-2011 ?"

where @ - Start Character
^ - Separator
? - End Character

and it contains ID, NAME, and DATE

Pseudocode:

char * ID;
char * NAME;
char * DATE;

if ( (start char is @) AND (End Char is ?) AND (There are two ^ in string))
Then
{ //This is valid string

strncpy(ID, &InputString[1], 6); // ID has fix length
ID[6] = '\0'; // terminate the string

strncpy(NAME, &InputString[8], (Length of variable NAME Length) );
// NAME has variable length, but it is between two ^
NAME[strlen(NAME)] = '\0'; // terminate the string

strncpy(DATE, &InputString[position after the 2nd ^], (Length of
date between 2nd ^ and ?)); // Date is between 2nd ^ and ?
DATE[strlen(DATE)] = '\0'; // terminate the string
}

can help to transform into code (some string functions..i'm not too sure to
achieve and finding the length /position) ? Many thanks.

Regards.

[Peter Nilsson]

Re: String

"magix" <ma...@asia.com wrote:

Hi,
>
If I have string like below:
InputString= "@123456^BI LL GATES^APR-2011        ?"
>
where @ - Start Character

[Note that @ need not be a character in the basic
execution character set.]

          ^  - Separator
           ? - End Character
>
and it contains ID, NAME, and DATE
>
Pseudocode:
>
 char * ID;
 char * NAME;
 char * DATE;
>
if ( (start char is @) AND (End Char is ?) AND (There are two ^ in string))
Then
{    //This is valid string
>
        strncpy(ID, &InputString[1], 6);    // ID has fix length
        ID[6] = '\0';    // terminate the string
>
        strncpy(NAME, &InputString[8], (Length of variable NAME Length) );
// NAME has variable length, but it is between two ^
        NAME[strlen(NAME)] = '\0'; // terminate the string
>
        strncpy(DATE, &InputString[position after the 2nd ^], (Length of
date between 2nd ^ and ?));  // Date is between 2nd ^ and ?
        DATE[strlen(DATE)] = '\0'; // terminate the string
>
}
>
can help to transform into code (some string functions..i'm not too sure to
achieve and finding the length /position) ? Many thanks.

Untested...

const char *pc1, *pc2, *pcm;

if ( InputString[0] == '@'
&& (pc1 = strchr(InputStr ing + 1, '^')) != 0
&& pc1 - &InputString[1] == 6
&& (pc2 = strchr(++pc1, '^')) != 0
&& (pcm = strchr(++pc2, '?')) != 0
&& pcm[1] == 0)
{
strncpy(ID, InputString + 1, 6);
ID[6] = 0;

strncpy(NAME, pc1, pc2 - pc1 - 1);
NAME[pc2 - pc1 - 1] = 0;

strncpy(DATE, pc2, pcm - pc2);
DATE[pcm - pc2] = 0;
}

--
Peter

[Eric Sosman]

Re: String

magix wrote:

Hi,
>
If I have string like below:
InputString= "@123456^BI LL GATES^APR-2011 ?"
>
where @ - Start Character
^ - Separator
? - End Character
>
and it contains ID, NAME, and DATE
>
Pseudocode:
>
char * ID;
char * NAME;
char * DATE;
>
if ( (start char is @) AND (End Char is ?) AND (There are two ^ in string))
Then
{ //This is valid string
>
strncpy(ID, &InputString[1], 6); // ID has fix length
ID[6] = '\0'; // terminate the string
>
strncpy(NAME, &InputString[8], (Length of variable NAME Length) );
// NAME has variable length, but it is between two ^
NAME[strlen(NAME)] = '\0'; // terminate the string
>
strncpy(DATE, &InputString[position after the 2nd ^], (Length of
date between 2nd ^ and ?)); // Date is between 2nd ^ and ?
DATE[strlen(DATE)] = '\0'; // terminate the string
}
>
can help to transform into code (some string functions..i'm not too sure to
achieve and finding the length /position) ? Many thanks.

There's more than one way to do this, but here's one
approach with the advantage of simplicity. I'll assume
the indicated fixed maximum sizes for NAME and DATE; other
approaches could cope with variable sizes.

char ID[6+1];
char NAME[40+1];
char DATE[8+1];
int len;

/* Grab the three fields and get the total length */
if (sscanf(InputSt ring,
"@%6[^^]^%40[^^]^%8[^^?]?%n",
ID, NAME, DATE, &len) != 3)
return BADSTRING;

/* Expected length for ID? */
if (strlen(ID) != 6)
return BADSTRING;

/* Optional: Any garbage after the end? */
if (InputString[len] != '\0')
return BADSTRING;

/* All's well */
return GOODSTRING;

--
Eric Sosman
esosman@ieee-dot-org.invalid