Re: Select and compare problems still ...

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()); ?>