natsu

nvm, I got it working

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; ?>

what do you mean when u say "whatever name your form login field has", would that be the 'emailAddress' and 'password'

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); ?>

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>

I solved this. Thanks a lot

I solved this. Thanks a lot

Trying to get 2 per line and into a html table what I have so far <?php for ($i = 0; $i <= 9; $i++) { echo "$i &nbsp"; } ?>

Sorry, but I dont understand what u mean by concentrating the function to the string.

Another problem I am having same trouble with scope for displaying max within a function, I tried global and return, and my output is ----> max(Array), it should show 99 <?php $a = array(9, 3, 1, 0, 99, 2, 5, 6, 32, 1, 55); function highestValue () { global $a; echo "The highest value is at index: (insert index), The value at index (insert index) is: max($a)"; } highestValue(); ?> OUTPUT: The highest value is at index: (insert index), The value at index (insert index) is: max(Array) I still don't know the code to display the index also

Thank you, I was able to do it <?php echo "The average is " . array_sum($a)/count($a) . "\n"; ?> As for the max and min's I know how to find the actual highest and lowest, but how would u indicate the index of the highest and lowest? <?php $a = array(9, 3, 1, 0, 99, 2, 5, 6, 32, 1, 55); echo max($a); echo "<br>"; echo min($a); ?>

Yes I understand scope's now but I am trying to do http://www.phpfreaks.com/forums/index.php?topic=348076.msg1642492#msg1642492

Wow that's really interesting! <?php $a = array(9, 3, 1, 0, 99, 2, 5, 6, 32, 1, 55); function getAverage () { global $a; echo "The average is " . array_sum($a) . "\n"; } getAverage(); ?> I tried do divide that number by the amount of value's in the array like the following <?php echo "The average is " . array_sum($a)/11 . "\n"; // try 1 echo "The average is " . array_sum($a) . "\n"; // try 2 echo "array_sum($a)/11" ?> or is there some sort of code that knows how many elements are in the array? And take the array_sum and divide it by that.

Ok I am going to tackle this piece by piece... <?php $a = array(9, 3, 1, 0, 99, 2, 5, 6, 32, 1, 55); function getAverage () { echo "The average is " . array_sum($a) . "\n"; } getAverage(); ?> I know that array_sum doesn't return the average, it returns the sum of the array. But even that is not displaying the total why? if I display the following line of code outside the function <?php echo "The average is " . array_sum($a) . "\n"; ?> It works, but if I write it inside the function, it does not work why? And to get the average, what is the term for diving by the number of value's in the array?

I'm not lying. My freind told me he needs to do this, I am not sure if it is his homework or not lol, probebly is