natsu Posted December 8, 2011 Share Posted December 8, 2011 Ok so I need to create a form to accept the users EmailAddress and Password as credentials to your site then use an SQL Query to determine if the person has an account <?php require "connectionInfo.php"; $error = ""; if(!isset($_POST["personId"]) || !isset($_POST["firstName"]) || !isset($_POST["lastName"]) || !isset($_POST["emailAddress"]) || !isset($_POST["telephoneNumber"]) || !isset($_POST["socialInsuranceNumber"]) || !isset($_POST["password"]) ) { $error = "Please fill in the info"; } else { if($_POST["personId"] != "" && $_POST["firstName"] != "" && $_POST["lastName"] != "" && $_POST["emailAddress"] != "" && $_POST["telephoneNumber"] != "" &&$_POST["socialInsuranceNumber"] != "" && $_POST["password"] != "") { $dbConnection = mysql_connect($host, $username, $password); if(!$dbConnection) die("Could not connect to the database. Remember this will only run on the Playdoh server."); mysql_select_db($database); $sqlQuery = "INSERT INTO persons (personId, FirstName, LastName, emailAddress, telephoneNumber, socialInsuranceNumber, password) VALUES('".$_POST["personId"]."', '".$_POST["firstName"]."', '".$_POST["lastName"]."', '".$_POST["emailAddress"]."', '".$_POST["telephoneNumber"]."', '".$_POST["socialInsuranceNumber"]."', '".$_POST["password"]."')"; if(mysql_query($sqlQuery)) $error = "Person Successfully Added"; else $error = "Person Could not be added ".mysql_error(); mysql_close($dbConnection); } else $error = "Please enter all the information"; } ?> <form action="createAccount.php" method="post"> Person ID: <input type="text" name="personId" /> <br /> First Name: <input type="text" name="firstName" /> <br /> Last Name: <input type="text" name="lastName" /> <br /> Email: <input type="text" name="emailAddress" /> <br /> Telephone: <input type="text" name="telephoneNumber" /> <br /> Social Insurance Number: <input type="text" name="socialInsuranceNumber" /> <br /> Password: <input type="text" name="password" /> <br /> <input type="submit" value="Submit to Database" /> </form> -----EDIT----- Ok I was able to create the html code for it, but how do I use an sql query to determine if the person has an account? <form method='post' action='login.php'> <table><tr><td>Email Address:</td><td><input type='text' name='emailAddress'></td></tr> <tr><td>Password:</td><td><input type='password' name='password'></td></tr> <tr><td></td><td><input type='submit' name='submit' value='Log in'></td></tr></table> </form> Quote Link to comment Share on other sites More sharing options...
natsu Posted December 8, 2011 Author Share Posted December 8, 2011 This is what I got so far <?php require "connectionInfo.php"; $dbConnection = mysql_connect($host, $username, $password); if(!$dbConnection) die("Could not connect to the database. Remember this will only run on the Playdoh server."); mysql_select_db($lab9_hadd0076); $sqlQuery = "SELECT * FROM persons"; $result = mysql_query($sqlQuery); //$loginDetail = emailAddress, password; How do I specify the login credentials, I know this is wrong if($loginDetail == 0) echo "*** There is no accounts made ***"; else { echo "Yes you have created an account"; } mysql_close($dbConnection); ?> Quote Link to comment Share on other sites More sharing options...
kney Posted December 8, 2011 Share Posted December 8, 2011 <?php require "connectionInfo.php"; $dbConnection = mysql_connect($host, $username, $password); if(!$dbConnection) die("Could not connect to the database. Remember this will only run on the Playdoh server."); mysql_select_db($lab9_hadd0076); $login = $_POST['whatever name your form login field has']; $pwd = $_POST['whatever name your form login field has']; $sqlQuery = "SELECT * FROM persons WHERE userName= '$login' AND password = '$pwd'"; $result = mysql_query($sqlQuery); if(mysql_num_rows($result) == 0) echo "There is no user with these credentials"; }else{ $user = mysql_fetch_array($result); echo "Welcome, " . $user['userName']; } mysql_close($dbConnection); ?> Quote Link to comment Share on other sites More sharing options...
natsu Posted December 8, 2011 Author Share Posted December 8, 2011 what do you mean when u say "whatever name your form login field has", would that be the 'emailAddress' and 'password' Quote Link to comment Share on other sites More sharing options...
kney Posted December 9, 2011 Share Posted December 9, 2011 Yes then it would be: <?php require "connectionInfo.php"; $dbConnection = mysql_connect($host, $username, $password); if(!$dbConnection) die("Could not connect to the database. Remember this will only run on the Playdoh server."); mysql_select_db($lab9_hadd0076); $email = $_POST['emailAddress']; $pwd = $_POST['password']; // if your columnnames are different, change it $sqlQuery = "SELECT * FROM persons WHERE emailAddress = '$email' AND password = '$pwd'"; $result = mysql_query($sqlQuery); if(mysql_num_rows($result) == 0) echo "There is no user with these credentials"; }else{ $user = mysql_fetch_array($result); // between the brackets is a columnname (so if it's different, change it) echo "Welcome, " . $user['userName']; } mysql_close($dbConnection); ?> Quote Link to comment Share on other sites More sharing options...
natsu Posted December 10, 2011 Author Share Posted December 10, 2011 It's still not working Here is the HTML on the same page (in first post too) <form method='post' action='login.php'> <table><tr><td>Email Address:</td><td><input type='text' name='emailAddress'></td></tr> <tr><td>Password:</td><td><input type='password' name='password'></td></tr> <tr><td></td><td><input type='submit' name='submit' value='Log in'></td></tr></table> </form> The emailAddress and password was made using this on my createAccounte.php <?php require "connectionInfo.php"; $error = ""; if(!isset($_POST["personId"]) || !isset($_POST["firstName"]) || !isset($_POST["lastName"]) || !isset($_POST["emailAddress"]) || !isset($_POST["telephoneNumber"]) || !isset($_POST["socialInsuranceNumber"]) || !isset($_POST["password"]) ) { $error = "Please fill in the info"; } else { if($_POST["personId"] != "" && $_POST["firstName"] != "" && $_POST["lastName"] != "" && $_POST["emailAddress"] != "" && $_POST["telephoneNumber"] != "" && $_POST["socialInsuranceNumber"] != "" && $_POST["password"] != "") { $dbConnection = mysql_connect($host, $username, $password); if(!$dbConnection) die("Could not connect to the database. Remember this will only run on the Playdoh server."); mysql_select_db($database); $sqlQuery = "INSERT INTO persons (FirstName, LastName, EmailAddress, TelephoneNumber, SocialInsuranceNumber, Password) VALUES('".$_POST["firstName"]."', '".$_POST["lastName"]."', '".$_POST["emailAddress"]."', '".$_POST["telephoneNumber"]."', '".$_POST["socialInsuranceNumber"]."', '".$_POST["password"]."');"; if(mysql_query($sqlQuery)) $error = "Person Successfully Added"; else $error = "Person Could not be added : ".mysql_error(); mysql_close($dbConnection); } else $error = "Please enter all the information"; } ?> <form action="createAccount.php" method="post"> Person ID: <input type="text" name="personId" /> <br /> First Name: <input type="text" name="firstName" /> <br /> Last Name: <input type="text" name="lastName" /> <br /> Email: <input type="text" name="emailAddress" /> <br /> Telephone: <input type="text" name="telephoneNumber" /> <br /> Social Insurance Number: <input type="text" name="socialInsuranceNumber" /> <br /> Password: <input type="text" name="password" /> <br /> <input type="submit" value="Submit to Database" /> </form> <br /> <br /> <?php echo $error; ?> Quote Link to comment Share on other sites More sharing options...
natsu Posted December 10, 2011 Author Share Posted December 10, 2011 nvm, I got it working Quote Link to comment Share on other sites More sharing options...
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