Quotations marks and strstr

Hi. Quick question. I am trying to debug a program and I can't seem to find anything on google.

The question relates to strstr(). I am connecting to a dictionary site using sockets and seeing if the given word is valid, by evaluating the html. Now, I have a pattern variable which contains the pattern I am looking for in the html. The only problem is, the pattern string has quotation marks in it so I have to escape them with '\'. Now I have tried to do strstr( buffer, pattern ) but it always seems to come out NULL, and I think it may be something to do with the quotation marks. Any idea?

>Any idea?
Not a clue. Do you have any sample code that makes answering your question anything but an exercise in futility?

Here you go. See attached.

Code:

---

int WordisinDictionary( char *buffer, char *pattern ) {

        if( strstr( buffer, pattern ) == NULL ) {
                return 1;
        }
        else {
                return 0;
        }       

}

---

Let me explain. The pattern variable is "class=\"results\">No" which appears in the html when the word ISN'T valid. So the if/else statement above isn't very intuitive. Another possible explanation for it not working is that the GET request isn't producing the correct html.

I'm 99.9% sure that strstr doesn't care the least bit about whether your text is quotes or not - why should it treat those special. The only places where quotes make a difference is in computer languages (such as C, PHP, HTML, etc),where they have a meaning (of separating text strings from keywoards, variables and other "programming" bits). All string functions will quite happily take strings that contain any character sequence you can think of.

--
Mats

So it should work? Hmm, I'll have to look a bit further at my code then.

Why don't you do a partial strstr for "class" or some such, and print the string you get from that.

--
Mats

Sorry, I'm not sure I know what you mean.

I mean, search for a part of your string "class=\"results\">No" , such as "class". THen check that you are getting the right thing and not missing spaces etc.

--
Mats

Quote:

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Originally Posted by Tommo

So the if/else statement above isn't very intuitive.

---

I'll say. (Are you sure you haven't confused yourself?) WordisinDictionary returns 1 if the word is NOT in the dictionary, and returns 0 if it is.

Code:

---

#include <stdio.h>
#include <string.h>

int WordisinDictionary( char *buffer, char *pattern )
{
  if ( strstr( buffer, pattern ) == NULL )
  {
      return 1;
  }
  else
  {
      return 0;
  } 
}

int main( int argc, char *argv[] )
{
  char *text[] =
  {
      "<A class=\"results\">No</A>", /* MATCH */
      "<A class=\"results\">Yes</A>" /* NO MATCH*/
  };
  char pattern[25] = "class=\"results\">No";
  size_t i;
  for ( i = 0; i < sizeof text / sizeof *text; ++i )
  {
      int result = WordisinDictionary( text[i], pattern );
      printf("&#37;s is %s the dictionary\n", text[i], result ? "in" : "NOT in" );
  }
  return 0;
}

/* my output
<A class="results">No</A> is NOT in the dictionary
<A class="results">Yes</A> is in the dictionary
*/

---

I might just code it similar to this:

Code:

---

      int result = strstr( text[i], pattern ) != 0;

---

@ Dave_Sinkula: Sorry I seem to have confused you. If the pattern matches then the word was not found, so if strstr( buffer, pattern ) is NULL, then the pattern was not found and so the word does exist. Basically I am saying, if I don't get "not found" then the word exists.

@matsp: Ah ok. The thing is, I have gone to the site, searched for an invalid word, then looked at the html for the result, so I shouldn't really need to do this.

Quote:

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Originally Posted by Tommo

@ Dave_Sinkula: Sorry I seem to have confused you. If the pattern matches then the word was not found, so if strstr( buffer, pattern ) is NULL, then the pattern was not found and so the word does exist. Basically I am saying, if I don't get "not found" then the word exists.

@matsp: Ah ok. The thing is, I have gone to the site, searched for an invalid word, then looked at the html for the result, so I shouldn't really need to do this.

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Yes, but it's sometimes easy to make mistakes here, and if you print what you get back from a partial match that is almost certain to match, then you can CAREFULLY check that it's actually as you think it should be. In your applicaiton, not in a web-browser that MAY do something else.

--
Mats

Thanks matsp.

I realise why it doesn't work now. The pattern may not appear when I fill my buffer for the first time, say, so wordisindictionary() returns 1 and breaks from the loop, therefore giving me a message saying invalid words are valid.

OK, got it working by changing 'pattern' to "class=\"results\">" and changing the function to:

Code:

---

int WordisinDictionary( char *buffer, char *pattern ) {

        char *ptr;
        int len;
        if( ( ptr = strstr( buffer, pattern ) ) != NULL ) {
                len = strlen( pattern );
                if( ptr[len] == 'N' && ptr[len + 1] == 'o' ) {
                        return 0;
                }
                else {
                        return 1;
                }
        }
        else {
                return 0;
        }       

}

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Thanks for the help.