unsigned int mysql_num_fields(MYSQL_RES
*result)
Or:
unsigned int mysql_num_fields(MYSQL *mysql)
The second form doesn't work on MySQL 3.22.24 or newer. To pass
a MYSQL*
argument, you must use
unsigned int mysql_field_count(MYSQL *mysql)
instead.
Description
Returns the number of columns in a result set.
Note that you can get the number of columns either from a
pointer to a result set or to a connection handle. You would use
the connection handle if
mysql_store_result()
or
mysql_use_result()
returned
NULL
(and thus you have no result set
pointer). In this case, you can call
mysql_field_count()
to determine
whether mysql_store_result()
should have produced a nonempty result. This allows the client
program to take proper action without knowing whether the query
was a SELECT
(or
SELECT
-like) statement. The
example shown here illustrates how this may be done.
Return Values
An unsigned integer representing the number of columns in a result set.
Errors
None.
Example
MYSQL_RES *result; unsigned int num_fields; unsigned int num_rows; if (mysql_query(&mysql,query_string)) { // error } else // query succeeded, process any data returned by it { result = mysql_store_result(&mysql); if (result) // there are rows { num_fields = mysql_num_fields(result); // retrieve rows, then call mysql_free_result(result) } else // mysql_store_result() returned nothing; should it have? { if (mysql_errno(&mysql)) { fprintf(stderr, "Error: %s\n", mysql_error(&mysql)); } else if (mysql_field_count(&mysql) == 0) { // query does not return data // (it was not a SELECT) num_rows = mysql_affected_rows(&mysql); } } }
An alternative (if you know that your query should have returned
a result set) is to replace the
mysql_errno(&mysql)
call
with a check whether
mysql_field_count(&mysql)
returns 0. This happens only if something went wrong.
User Comments
In the code in this example you have to watch out for the meaning of mysql_query() results. In fact it seems like from the example an INSERT that failed would return from mysql_query() successfully. You would also seem to guess that the num_rows value would be 0 in that case. Nothing could be further from the trust.
In fact, in an INSERT that fails because of duplicate keys, you will see mysql_query() fail and mysql_errno will return ER_DUP_ENTRY. This is ia little "gotcha" in this code.
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