数据没有保存到PHP中的两个表
问题描述：
I am attempting to save user details to a database, I give sign up success but when I check my database only the User table is receiving the data. I do know my code is kind of messy but I am a newbie after all, here is the PHP, I cannot see any errors in the INSERT it seems to match my table name and fields. I am trying to make both the User table and the Member table be populated but at the moment only the User table is getting the data. Sorry I posted this earlier but included the wrong code so I deleted the post and here I am again.
if (isset($_POST["u"])) {
// CONNECT TO THE DATABASE
include_once("db_conx.php");
// GATHER THE POSTED DATA INTO LOCAL VARIABLES
$u = preg_replace('#[^a-z0-9]#i', '', $_POST['u']);
$e = mysqli_real_escape_string($db_conx, $_POST['e']);
$p = $_POST['p'];
$ln = preg_replace('#[^a-z]#i', '', $_POST['lastName']);
$fn = preg_replace('#[^a-z]#i', '', $_POST['firstName']);
$g = preg_replace('#[^a-z]#i', '', $_POST['g']);
$c = preg_replace('#[^a-z ]#i', '', $_POST['c']);
$m = preg_replace('#[^0-9]#', '', $_POST['m']);
$ci = preg_replace('#[^a-z]#i', '', $_POST['ci']);
$pc = preg_replace('#[^a-z0-9]#i', '', $_POST['pc']);
$rs = $_POST['relationshipStatus'];
$d = $_POST['d'];
$_POST['accountType'] = 0;
$accountType = mysqli_real_escape_string($db_conx, $_POST['accountType']);
// $ip = preg_replace('#[^0-9.]#', '', getenv('REMOTE_ADDR'));
// DUPLICATE DATA CHECKS FOR USERNAME AND EMAIL
$sql = "SELECT Userid FROM User WHERE username='$u' LIMIT 1";
$query = mysqli_query($db_conx, $sql);
$u_check = mysqli_num_rows($query);
// -------------------------------------------
$sql = "SELECT emailAddress from Member WHERE emailAddress = '$e' LIMIT 1";
$query = mysqli_query($db_conx, $sql);
$e_check = mysqli_num_rows($query);
// FORM DATA ERROR HANDLING
if (u == "" || e == "" || p1 == "" || p2 == "" || c == "" || g == "" || m == "" || ci == "" || pc == "" || d == "") {
echo "The form submission is missing values.";
exit();
} else if ($u_check > 0) {
echo "The username you entered is alreay taken";
exit();
} else if ($e_check > 0) {
echo "That email address is already in use in the system";
exit();
} else if (strlen($u) < 3 || strlen($u) > 16) {
echo "Username must be between 3 and 16 characters";
exit();
} else if (is_numeric($u[0])) {
echo 'Username cannot begin with a number';
exit();
} else {
// END FORM DATA ERROR HANDLING
// Begin Insertion of data into the database
// Hash the password and apply your own mysterious unique salt
$cryptpass = crypt($p);
include_once ("randStrGen.php");
$p_hash = randStrGen(20) . "$cryptpass" . randStrGen(20);
// Add user info into the database table for the main site table
$sql = "INSERT INTO Member (`firstName`, `lastName`, `gender`, `emailAddress`,
`city`, `country`, `postCode`, `relationshipStatus`, `mobileNumber`, `dateOfBirth`)
VALUES('$fn', '$ln', '$g', '$e', '$ci', '$c', '$pc', '$rs', '$m', '$d')";
$query = mysqli_query($db_conx, $sql);
$uid = mysqli_insert_id($db_conx);
$sql = "INSERT INTO User (`Userid`, `username`, `password`, `accountType`) VALUES ('$uid','$u','$p','$accountType')";
$query = mysqli_query($db_conx, $sql);
When I echo out the INSERT into Member statement I get the following even though I populate every field on my form
INSERT INTO Member (`firstName`, `lastName`, `gender`, `emailAddress`, `city`, `country`, `postCode`, `relationshipStatus`, `mobileNumber`, `dateOfBirth`) VALUES('', '', 'Male', 'test11@hotmail.com', '', 'Lebanon', '', '', '', '')signup_success
答
Here's what I'd do - forgive the coding standard, I've been coding all day and I'm rather tired
Also, try to kill the page as you hit each error instead of saving them for later - once the page is killed, no further code is processed (it's read by the server, but not processed)
Just makes more sense to me - personal preference!
if (isset($_POST['u'])) {
// CONNECT TO THE DATABASE
include_once("db_conx.php");
$lazy = array(
'u' => 'username',
'ln' => 'surname',
'fn' => 'forename',
'g' => 'gender',
'c' => 'country',
'ci' => 'city',
'pc' => 'post code',
'rs' => 'relationship status'
);
foreach($lazy as $what => $show) {
$_POST[$what] = isset($_POST[$what]) && is_string($_POST[$what]) ? mysqli_real_escape_string(trim($db_conx,$_POST[$what])) : null;
if(empty($_POST[$what]))
exit("You didn't enter a valid ".$show);
${$what} = $what;
}
if (strlen($u) < 3 || strlen($u) > 16)
exit("Username must be between 3 and 16 characters");
if (ctype_digit($u[0]))
exit('Username cannot begin with a number');
$e = isset($_POST['e']) && filter_var($_POST['e'], FILTER_VALIDATE_EMAIL) ? trim($_POST['e']) : null;
if(empty($e))
exit("You didn't enter a valid email address");
if(filter_var($_POST['d'], FILTER_VALIDATE_INT) !== false)
exit("You didn't enter a valid mobile number");
$d = $_POST['d']; // Seeing as you're intent on creating variables for variabled data..
/*
Need to see how users enter their date of birth before I can properly secure that..
If using <input type='date' ..>
if(!preg_match('/[0-9]+\-[0-9]+\-[0-9]+/', $_POST['d']))
exit("Invalid date of birth");
*/
// DUPLICATE DATA CHECKS FOR USERNAME AND EMAIL
$sql1 = "SELECT `Userid` FROM `User` WHERE `username` = '".$u."' LIMIT 1";
$query = mysqli_query($db_conx, $sql1) or exit(mysqli_error($db_conx));
if(mysqli_num_rows($query))
exit("That username is already in use");
// -------------------------------------------
$sql2 = "SELECT `emailAddress` FROM `Member` WHERE `emailAddress` = '".$e."' LIMIT 1";
$query = mysqli_query($db_conx, $sql2) or exit(mysqli_error($db_conx));
if(mysqli_num_rows($query))
exit("That email address is already in use");
// Begin Insertion of data into the database
// Hash the password and apply your own mysterious unique salt
$cryptpass = crypt($p);
include_once("randStrGen.php");
$p_hash = randStrGen(20) . $cryptpass . randStrGen(20);
// Add user info into the database table for the main site table
$sql3 = "INSERT INTO `Member` (`firstName`, `lastName`, `gender`, `emailAddress`, `city`, `country`, `postCode`, `relationshipStatus`, `mobileNumber`, `dateOfBirth`) VALUES('".$fn."', '".$ln."', '".$g."', '".$e."', '".$ci."', '".$c."', '".$pc."', '".$rs."', '".$m."', '".$d."')";
mysqli_query($db_conx, $sql3) or exit(mysqli_query($db_conx));
$uid = mysqli_insert_id($db_conx);
$sql4 = "INSERT INTO `User` (`Userid`, `username`, `password`, `accountType`) VALUES (".$uid.",'".$u."','".$p."','".$accountType."')";
mysqli_query($db_conx, $sql4) or exit(mysqli_query($db_conx));
echo "Account created"; // Or whatever success message you have
}