mannrazz Posted December 14, 2015 Share Posted December 14, 2015 (edited) predict.php <body> <!--contact form start here--> <h1>Smart Medical Application</h1> <div class="contact"> <div class="contact-top"> <h2>PREDICTION - Enter Your Symptom</h2> </div> <div class="fiel"> <form> <form action="what.php" method = "post" > <input placeholder="First Symptom" name = "first" type="text" > <input placeholder="Second Symptom" name = "second" type="text" > <input placeholder="Third Symptom" name = "third" type="text" > </form> </div> <div class="send"> <form> <input type="submit" name="submit" value="PREDICT!"/> </form> <br></br> <a href="../main/main.php"><button>BACK</button></a> </div> </div> <div class="copyright"> <p>© 2015 All rights reserved | MannRazz Dev Team</p> </div> <!--contact form end here--> </body> what.php <?php $first = $_POST['first']; $second = $_POST['second']; $first = stripcslashes($first); $second = stripcslashes($second); $first = mysql_real_escape_string($first); $second = mysql_real_escape_string($second); mysql_connect("localhost", "root", ""); mysql_select_db("predict"); $result = mysql_query("SELECT * FROM predict WHERE first = '$first' AND second = '$second'"); or die ("Failed to query database".mysql_error()); $row = mysql_fetch_array($result); if ($row['first'] == $first && $row['second'] == $second) { echo "LOGIN SUCCES WELCOME ".$row['first']; } else { "Failed to login"; } ?> **** - i want to conect with the database when both of the query is correct.. and when both is correct... i wanted to call and display the third data from the database (etc : welcome, SAM - SAM is the data that i wanted to call) but it seem doesnt working at the moments... any help will be much appreciated....thanz Edited December 14, 2015 by Ch0cu3r Quote Link to comment https://forums.phpfreaks.com/topic/299738-cant-connect-with-database/ Share on other sites More sharing options...
cyberRobot Posted December 14, 2015 Share Posted December 14, 2015 The database connection needs to be made before you can use mysql_real_escape_string(). http://php.net/manual/en/function.mysql-real-escape-string.php Also, there is a syntax error here: $result = mysql_query("SELECT * FROM predict WHERE first = '$first' AND second = '$second'"); or die ("Failed to query database".mysql_error()); There shouldn't be a semi-colon before "or". $result = mysql_query("SELECT * FROM predict WHERE first = '$first' AND second = '$second'") or die ("Failed to query database".mysql_error()); And in case you're not aware, mysql_* functions have been deprecated and will be removed in PHP 7. More information can be found here: http://php.net/manual/en/mysqlinfo.api.choosing.php Quote Link to comment https://forums.phpfreaks.com/topic/299738-cant-connect-with-database/#findComment-1527905 Share on other sites More sharing options...
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