silverglade Posted September 2, 2011 Share Posted September 2, 2011 Hi, I am trying to make a search based on information about a user in the database. Here is my code so far, notice any words like "and" or "the" in a multiword entry would mess it up. Any advice greatly appreciated. Ideally I would like a list of the matches by user name, and put the word "match" next to the matching data, because I used the OR statement it is going to bring back all of the user's info if one of the criteria matches. Any tutorials on the subject or keywords for google would be greatly appreciated. thanks. here is the search code. <?php ini_set ("display_errors", "1"); error_reporting(E_ALL); $host = " ; $database = " "; $username = " "; $password = " "; $tbl_name = "users"; $conn = mysql_connect($host, $username, $password) or die("Could not connect: " . mysql_error()); if($conn) { mysql_select_db($database); echo "connected to database!!"; } else { echo "failed to select database"; } //include('bouncer.php'); //$currentUser = $_SESSION['myusername']; if(isset($_POST['submit'])) { $dob = mysql_real_escape_string( $_POST['dob']); $gender = mysql_real_escape_string( $_POST['gender']); $ethnic = mysql_real_escape_string( $_POST['ethnic']); //THE SEARCH FUNCTION $sql = mysql_query ( "SELECT * FROM users WHERE dob LIKE '%$dob%' OR gender LIKE '%$gender%' OR race LIKE '%$ethnic%' OR dob LIKE '%$dob%' ") or die(mysql_error()); $result = mysql_query($sql) or die( "<br>Query: $sql<br>Failed with error: " . mysql_error() ); if (!$result) { echo "Could not successfully run query ($sql) from DB: " . mysql_error(); exit; } if (mysql_num_rows($result) == 0) { echo "No rows found, nothing to print so am exiting"; exit; } // While a row of data exists, put that row in $row as an associative array // Note: If you're expecting just one row, no need to use a loop // Note: If you put extract($row); inside the following loop, you'll // then create $userid, $fullname, and $userstatus while ($row = mysql_fetch_assoc($result)) { echo $row["dob"]; echo $row["gender"]; echo $row["race"]; } }//if(isset($_POST['submit'])) ?> <html> <body> <p>Login Successful</p> <p> </p> <form action="login_success7.php" method="post"> <p> <input name="dob" type="text" size="20" id="dob" /> Date of Birth<br /> <input type="text" name="gender" size="20" id="gender" /> Gender <br /> <input type="text" name="ethnic" size="20" id="ethnic" /> Ethnicity <br /> <input type="submit" name="submit" value=" search" /> <input type="reset" value="Reset fields" /> </p> </form> </body> </html> Quote Link to comment Share on other sites More sharing options...
silverglade Posted September 2, 2011 Author Share Posted September 2, 2011 also when I run that script I get an error, Query: Resource id #2 Failed with error: 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 'Resource id #2' at line 1 Quote Link to comment Share on other sites More sharing options...
silverglade Posted September 2, 2011 Author Share Posted September 2, 2011 nevermind, I got it to work, I was using the double query again. thanks Quote Link to comment Share on other sites More sharing options...
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