PHP Forum

**Izzy123** · 05-20-2010, 11:17 AM

Hi I'm new to php and I'm currently stuck on this question:

Create a PHP page that contains a form that contains an xhtml selection list containing all the customer names (last name is sufficient) as the option text and the customer ID as the value for the option. The selection list must be created by
running an SQL query to obtain the data from the database. The form should have a submit and a rest button. The form should use task8.php as the
action and the method should be get.

what I have so far is as below:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> 

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>Prac 3 Task 1</title>
</head>

<?PHP
$conn2 = mysql_connect("localhost", "twastudent", "prac3");
mysql_select_db("warehouse", $conn2)
or die ('Database not found ' . mysql_error() );
$sql = mysql_query("SELECT lastName, customerID FROM customer, $conn2)
or die ('Problem with query' . mysql_error());
?>
       
<body>
  <h1>Prac 3 Task 10 </h1>
  <form id="task10" action="task8.php" method="get">

  </form>
</body>
</html>

where all the information about the name and ID is already stored in a database.

Anyone have any idea where to start or what kind of functions should I use?

**NullPointer** · 11:46 AM

Use mysql_fetch_array in a while loop to generate the list:

PHP Code:


			
<select>

<?php while($row = mysql_fetch_array($sql, MYSQL_NUM)) : ?>

     <option value="<?php echo $row[1]; ?>"><?php echo $row[0]; ?></option>

<?php endwhile; ?>

</select>

**Phunk Rabbit** · 11:43 AM

You would use a while loop to create the select rows, example:

PHP Code:


			
<select>
<?php
while($result = mysql_fetch_array($sql)){
     echo '<option value="' . $result['lastName'] . '">' . $result['lastName'] . '</option>';
}
?>
</select>

**Phunk Rabbit** · 05-20-2010, 11:44 AM

Seems i am too slow, NullPointer beat me to it ^^

**Izzy123** · 05-20-2010, 09:09 PM

I got this so far

Code:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> 

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>...</title>
</head>      
 
<body>
<?PHP
$conn2 = mysql_connect("localhost", "twastudent", "prac3");
mysql_select_db("warehouse", $conn2)
or die ('Database not found ' . mysql_error() );
$sql = mysql_query("SELECT* FROM customer, $conn2)
or die ('Problem with query' . mysql_error());
?>

<select>
<?php while($row = mysql_fetch_array($sql)) { ?>
<option value="<?php echo $row['customerID'] ?>"><?php echo $row['lastName'] ?></option>
<?php } endwhile; ?>
</select>

  <form id="task10" action="task8.php" method="get">
<p><input type="submit" value="submit"> <input type="reset"  value="reset"/></p>
</body>
</html>

but when I try to run it, it state Parse error: syntax error, unexpected '?'on line 21

Anyone know what I did wrong?

**NullPointer** · 09:21 PM

2 things wrong:

1. You forgot 2 semicolons on this line (I added them):

PHP Code:


			
<option value="<?php echo $row['customerID']; ?>"><?php echo $row['lastName']; ?></option>

Either use the while-endwhile syntax or the brackets, not both:

PHP Code:


			
<?php } endwhile; ?>

In this case just remove the endwhile. Ex:

PHP Code:


			
while($foo) :

     //code here

endwhile;



while($foo)

{

     //code here

}

**Izzy123** · 05-20-2010, 10:14 PM

Thanks for the reply. I've corrected those errors, but the output still showing the same thing : Parse error: syntax error, unexpected '?' on line 21

Code:

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> 

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>...</title>
</head>      
 
<body>
<?PHP
$conn2 = mysql_connect("localhost", "twastudent", "prac3");
mysql_select_db("warehouse", $conn2)
or die ('Database not found ' . mysql_error() );
$sql = mysql_query("SELECT* FROM customer, $conn2)
or die ('Problem with query' . mysql_error());
?>

<select>
<?php while($row = mysql_fetch_array($sql)) { ?>
<option value="<?php echo $row['customerID']; ?>"><?php echo $row['lastName']; ?></option>
<?php } ?>
</select>

  <form id="task10" action="task8.php" method="get">
</form>
</body>
</html>

the error is still located at

Code:

<option value="<?php echo $row['customerID']; ?>"><?php echo $row['lastName']; ?></option>

**NullPointer** · 05-20-2010, 11:30 PM

You forgot a quote (") on this line:

PHP Code:


			
$sql = mysql_query("SELECT* FROM customer, $conn2)

Using an editor with syntax highlighting should reveal these types of mistakes.

**Izzy123** · 05-20-2010, 11:49 PM

Thanks so much for your help! it works now!

**juliasida** · 05-21-2010, 03:02 AM

In the get method the data made available to the action page ( where data is received ) by the URL so data can be seen in the address bar. Not advisable if you are sending login info like password etc. In the post method the data will be available as data blocks and not as query string in case of get method.