Re: Select and compare problems still ...

On Wed, Jul 1, 2009 at 11:23 AM, Andrew Ballard<aballard@gmail.com> wrote: > On Wed, Jul 1, 2009 at 10:56 AM, Miller, > Terion<tmiller@springfi.gannett.com> wrote: >> Why doesn't this work? >> >> >>    $query = "SELECT * FROM restaurants WHERE name >> ='$ucName' AND >> address = '$ucAddress'  " ; >> >> $result = mysql_query($query) or die(mysql_error()); >> >> >>  echo $result; >>     $row = mysql_fetch_array ($result); >> >> >> >>  $sql = "INSERT INTO restaurants (name, address, >> inDate, inType, notes, >> critical, cviolations, noncritical)  VALUES (" ; >>     $sql .= " '$ucName', >> '$ucAddress', '$inDate', '$inType', '$notes', >> '$critical', >> '$cleanViolations', '$noncritical')"; >> >> >> >>        $result = mysql_query($sql) or die(mysql_error()); >> >> The error I keep getting is: >> >> You have an error in your SQL syntax; check the manual that corresponds to >> your MySQL server version for the right syntax to use >> >> And I have gone in the mySQL panel and let it make the query ....so I'm >> really stumped why it hangs ... >> >> >> > > The last example you posted said: > > You have an error in your SQL syntax; check the manual that > corresponds to your MySQL server version for the right syntax to use > near 's Roast Beef Restaurant #9459', ' 1833 W Republic Rd ', > '3/2/09', '' at line 1 > > This indicated that the value for $ucName contained an > apostrophe/single-quote character. (Perhaps it was supposed to be > "Arby's Roast Beef Restaurant #9459"?). > > Try this: > > <?php > > $data = array($ucName, $ucAddress, $inDate, $inType, $notes, > $critical, $cleanViolations, $noncritical); > > $sql = vprintf("INSERT INTO restaurants (name, address, > inDate, > inType, notes, critical, cviolations, noncritical) VALUES ('%s', '%s', > '%s', '%s', '%s', '%s', '%s', > '%s')", > array_map('mysql_real_escape_string', $data)); > > $result = mysql_query($sql) or die(mysql_error()); > > ?> > oops = make that vsprintf(), not vprintf(); Andrew