"MYSQL_ROW row" Data Type

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The reason this is not working is that MYSQL_ROW row and an integer of 1 or 0 are not of the same data type. How would i make them the same data type?

           mysql_query(conn,"SELECT COUNT(*) FROM tblURL WHERE IP = '192.168.1.399' AND Status = 'Active'");
   my_ulonglong i = 0;
   res_set = mysql_store_result(conn);
   my_ulonglong numrows = mysql_num_rows(res_set);

   {
   row = mysql_fetch_row(res_set);
   if (row[i] = 1) //////THIS IS THE ISSUE??????/////////
       printf("YES: %s\n",row[i]);
   else
       printf("NO: %s\n",row[i]);
   }

   mysql_free_result(res_set);
   mysql_close(conn);
   system("PAUSE");
   /*End Works*/
}

Posted 12-May-11 12:23pm

Member 7766180

Updated 12-May-11 23:26pm

Kim Togo

v2

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Richard MacCutchan · Answer 1 · 2011-05-13T02:32:00

Solution 2

Take a look at the documentation here[^], which explains the data types returned by MySQL functions. mysql_fetch_row() returns a database row not an integral value.

Posted 13-May-11 2:32am

Richard MacCutchan

Comments

Albert Holguin 13-May-11 10:23am

See, being productive with responses is not so bad... :) ...my 5

Richard MacCutchan 13-May-11 10:52am

I am always productive; but sometimes (as in this case) brute force is needed.

Albert Holguin 13-May-11 11:22am

I see frustration is reaching an all-time high... lol

Member 7766180 13-May-11 10:50am

Right, But that return is always a "1" or a "0" in my case. That is all that it will ever be. A "1" or "0".

Richard MacCutchan 13-May-11 10:55am

You still seem incapable or unwilling to make an effort to read properly what you have been told. Go to the documentation and read carefully what it states about the return type of this function. Then go and find out how you can interpret that data and its content. Simply repeating this nonsense about 0 and 1 is getting you nowhere.

Member 7766180 13-May-11 11:19am

my_ulonglong

The type used for the number of rows and for mysql_affected_rows(), mysql_num_rows(), and mysql_insert_id(). This type provides a range of 0 to 1.84e19.

On some systems, attempting to print a value of type my_ulonglong does not work. To print such a value, convert it to unsigned long and use a %lu print format. Example:
printf ("Number of rows: %lu\n",
(unsigned long) mysql_num_rows(result));

ulonglong returns either a "1" or a "0" which is perfect. This is what its doing, however; I get the warning. Also the if statement is not reading it. I need to get the if statemnet to read if its a "1" or a "0".

Richard MacCutchan 13-May-11 13:03pm

What warning? What do you mean by "the if statement is not reading it"?
Comparing an unsigned long against a constant value is a simple expression of the form:
if (variable == constant)
One of the reasons I previously suggested you take a look at your C++ reference documentation.

Albert Holguin · Answer 2 · 2011-05-12T13:52:00

Solution 1

first off...

in an if statement:

if(x = 1)  //<-- this sets x to 1, not compares
if(x == 1) //<-- this compares

second (assuming the rest of your code is kosher, i'm pretending it is)...

you have to look up what MYSQL_ROW is typedef'ed from, if its an int, then a comparator should work, but if you need to cast it, you need to do an explicit cast (usually if you have to do this you should know what you're doing as far as data types, could lead to crashes, bugs, or loss of data):

MYSQL_ROW row;
int x= (int) row; //<-- again this sets it equal to

//In a statement
if(x == row) //<--this compares, but if they're compatible types the compiler should be ok with it
if(x == (int)row) //<--comparator with explicit cast

I'd help you more with the SQL portion of it but its been years since I last did SQL.

*****Update:*****

int value; //<--single integer value
int row[10]; //<-- array of 10 integers

if(row == 1) //<-- comparing pointer of array to one (this will likely never be one)
if(row[2] == 1) //<-- comparing third element to one (array indexing starts at zero)

Posted 12-May-11 13:52pm

Albert Holguin

Updated 12-May-11 14:37pm

v3

Comments

Member 7766180 12-May-11 20:25pm

Thank you so much Albert! Informative and Learnative (made that up) at the same time! I tried several of the suggestions and this one debugs...
if (int(row) == 1)
printf("PASS: %s\n",row[i]);
else
printf("FAIL: %s\n",row[i]);

However; if the answer is either '0' or '1' it always returns "FAIL", I'm going to search some more on the "==" that you gave me. Once again, thank you.
DS

Albert Holguin 12-May-11 20:29pm

int should be in parentheses, not row, now.... if row[i] is actually working (as in giving you a value that makes sense), then that means row must be a pointer to an array, so if you look at just row (without the []), then the value is some memory address that will likely never be 1

Member 7766180 12-May-11 20:29pm

This always returns "PASS" Somehow I'll split the difference!
if (int(row) >= 1)
printf("PASS: %s\n",row[i]);
else
printf("FAIL: %s\n",row[i]);

Albert Holguin 12-May-11 20:30pm

see my reply above, row is probably a pointer, hence always greater than 1

Albert Holguin 12-May-11 20:37pm

expanded on explanation a little in the solution itself...

Member 7766180 12-May-11 21:02pm

If I use the printf("RESULT: %s\n",row[i]); line and change the IP address to 192.168.1.33 which I know is in the database I get a "1" returned. If I use 192.168.1.433 which I know IS NOT in the database I get a "0", so I assume the row is returning a value.

Albert Holguin 12-May-11 23:04pm

row[i] has the value, not row... row itself if the address of the array

Member 7766180 12-May-11 23:38pm

Thats what I thought...Digging furthur I think it goes back to the ulonglong problem of earlier days. See my new post.