JasonBruce88 Posted March 13, 2010 Share Posted March 13, 2010 I have a form that connects to a MySQL data base but for some reason when I submit the form, only pressing the button once, the data is added to the form twice? Anybody have any ideas on this? Here is the code HTML <?php include("inc/headerlogin.php"); ?> <form action="submited.php" method="post"> <fieldset> <legend>Customer details</legend> <p> <label for="FirstName">First name</label> <input type="text" name="FirstName" value="" id="FirstName"> </p> <p> <label for="LastName">Last name</label> <input type="text" name="LastName" value="" id="LastName"> </p> <p> <label for="Title">Title</label> <select name="Title" id="Title"> <option>Mr</option> <option>Mrs</option> <option>Miss</option> <option>Ms</option> <option>Dr</option> <option>Rev</option> <option>Sir</option> </select> </p> <p> <label for="PN1">Phone</label> <input type="text" name="PN1" value="" id="PN1"> <span>(primary)</span> </p> <p> <label for="PN2">Phone 2</label> <input type="text" name="PN2" value="" id="PN2"> <span>(secondary)</span> </p> <p> <label for="UserTypeCode">I am a:</label> <select name="UserTypeCode"> <option name="Landlord" value="1"> Landlord </option> <option name="Tenant" value="2"> Tenant </option> <option name="Engineer" value="3"> Engineer </option> </select> <p> <label for="Email">Email</label> <input type="text" name="Email" value="" id="Email"> </p> <p> <label for="Password">Password</label> <input type="Password" name="Password" value="" id="Password"> </p> <p> <label for="Password2">Password</label> <input type="Password" name="Password2" value="" id="Password2"> </p> <p class="alignright"> <input type="submit" name="submit" value="Confirm details" class="button"> </p> </fieldset> </form> <?php include("inc/footer.php"); ?> PHP: <?php include("inc/headerlogin.php"); $con = mysql_connect("*", "*", "*") or die(mysql_error("A MySQL error has occurred")); mysql_select_db("*") or die(mysql_error("A MySQL error has occurred")); if (!$con) { die('Could not connect: ' . mysql_error()); } if (isset($_POST['submit'])) { //ensures fields are not blank if (!$_POST['Email'] | !$_POST['Password'] | !$_POST['Password2']) { die('You did not complete all of the required fields'); } //Checks if email address is allready in use if (!get_magic_quotes_gpc()) { $_POST['Email'] = addslashes($_POST['Email']); } $emailcheck = $_POST['Email']; $chek = mysql_query("SELECT * FROM client WHERE Email = '$emailcheck'") or die(mysql_error()); $check2 = mysql_num_rows($chek); //notifys if email address is already registered if ($check2 != 0) { die('Sorry, the Email Address '.$_POST['Email'].' is already in use. Please try again'); } //makes sure passwords match if ($_POST['Password'] !=$_POST['Password2']) { die('The passwords you have entered do not match.Please try again'); } //encrypt the users password $_POST['Password'] = md5($_POST['Password']); if (!get_magic_quotes_gpc()) { $_POST['Password'] = addslashes($_POST['Password']); $_POST['Email'] = addslashes($_POST['Email']); } //insert into database $insert = "INSERT INTO client (FirstName, LastName, Title, PN1, PN2, Email, Password, UserTypeCode) VALUES ('".$_POST['FirstName']."', '".$_POST['LastName']."', '".$_POST['Title']."', '".$_POST['PN1']."', '".$_POST['PN2']."', '".$_POST['Email']."', '".$_POST['Password']."', '".$_POST['UserTypeCode']."')"; $add_member = mysql_query($insert); if (!mysql_query($insert,$con)) { die('Error: ' . mysql_error()); } echo "Thank you, you have registered - you may now login"; } mysql_close($con) ?> Quote Link to comment https://forums.phpfreaks.com/topic/195103-saturday-morning-php/ Share on other sites More sharing options...
PFMaBiSmAd Posted March 13, 2010 Share Posted March 13, 2010 You are executing the query twice in your code, therefore you get two rows inserted - $add_member = mysql_query($insert); if (!mysql_query($insert,$con)) Quote Link to comment https://forums.phpfreaks.com/topic/195103-saturday-morning-php/#findComment-1025570 Share on other sites More sharing options...
harristweed Posted March 13, 2010 Share Posted March 13, 2010 looks like you are adding it twice $add_member = mysql_query($insert); if (!mysql_query($insert,$con)) you have two insert actions Quote Link to comment https://forums.phpfreaks.com/topic/195103-saturday-morning-php/#findComment-1025571 Share on other sites More sharing options...
JasonBruce88 Posted March 13, 2010 Author Share Posted March 13, 2010 FAIL!!!!!!!! lol Quote Link to comment https://forums.phpfreaks.com/topic/195103-saturday-morning-php/#findComment-1025574 Share on other sites More sharing options...
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