Quotations marks and strstr

This is a discussion on Quotations marks and strstr within the C Programming forums, part of the General Programming Boards category; Hi. Quick question. I am trying to debug a program and I can't seem to find anything on google. The ...

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    Registered User Tommo's Avatar
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    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?

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    Code Goddess Prelude's Avatar
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    >Any idea?
    Not a clue. Do you have any sample code that makes answering your question anything but an exercise in futility?
    My best code is written with the delete key.

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    Registered User Tommo's Avatar
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    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.
    Last edited by Tommo; 10-04-2007 at 01:48 PM.

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    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.

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

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

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  7. #7
    Registered User Tommo's Avatar
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    Sorry, I'm not sure I know what you mean.

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    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.

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  9. #9
    Just Lurking Dave_Sinkula's Avatar
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    Quote Originally Posted by Tommo View Post
    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;
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*

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    Registered User Tommo's Avatar
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    @ 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.

  11. #11
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    Quote Originally Posted by Tommo View Post
    @ 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.
    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
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  12. #12
    Registered User Tommo's Avatar
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    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.

  13. #13
    Registered User Tommo's Avatar
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    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;
    	}	
    
    }
    Thanks for the help.

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