Passing variables from a form to a php script and STUMPED, help this noob. LOL

kwiqness

Okay, first off, I am pretty much a newbie and coding. I am self taught so obviously there are things that I haven't learned about yet. I am writing a simple html/php application that interacts with a MySQL database. Here is the function of it.

Visit a webpage and be presented with records from that db table, edit, delete, and add records from that html form, as well as filter out your results to slim the list down. This is a license tracking application for software licenses. I have done all of the above except applied the filtering function in its entirety. I have added a two drop down menus that populate based on values currently in the database table. I have one of them working, which is a department filter. It will successfully pass the value from the drop down menu to the next page and as a result only show you records that contain that value.

Here is my code for my filtering drop down options the file name is view.php

<?php       

       $sql2 = "SELECT DISTINCT dept FROM sw_lic WHERE dept != ''";
       $sql3 = "SELECT DISTINCT sw_name FROM sw_lic WHERE sw_name != ''";
       $result2 = mysql_query($sql2);
       $result3 = mysql_query($sql3);
       ?>
       <form action="filter_view.php" method="POST">
       <select name="dept">
<?php
       while($nt=mysql_fetch_array($result2))
       {
       echo "<option name='dept' value=$nt[dept]>$nt[dept]</option>";
       }
?>
       </select>
       <select name="sw_name">
<?php
       while($st=mysql_fetch_array($result3))
       {
       echo "<option name='sw_name' value=$st[sw_name]>$st[sw_name]</option>";
       }
?>
<input type="submit" name="submit" value="Submit">

And here is the script that the variables should be passed to, the file name is filtered_view.php

 
<?php
        // connect to the database
        include('connect-db.php');

    // get results from database
    $nt = $_POST["dept"];
    $st = $_POST["sw_name"];
    $result = mysql_query("SELECT * FROM sw_lic WHERE dept='$nt' AND sw_name='$st'")
            or die(mysql_error());  

    // display data in table
    echo "<table border='1' cellpadding='10' background-color='#000066' color='#FFFFFF'>";
    echo "<tr> <th>Purchased for</th> <th>Installed On</th> <th>Dept.</th> <th>Software Name</th> <th>Software Ver.</th> <th>Software Origin</th> <th>LIC Key</th> <th></th> <th></th></tr>";

    // loop through results of database query, displaying them in the table
    while($row = mysql_fetch_array( $result )) {

            // echo out the contents of each row into a table
            echo "<tr>";

            echo '<td>' . $row['username'] . '</td>';
            echo '<td>' . $row['sysname'] . '</td>';
	echo '<td>' . $row['dept'] . '</td>';
	echo '<td>' . $row['sw_name'] . '</td>';
	echo '<td>' . $row['sw_ver'] . '</td>';
            echo '<td>' . $row['origin'] . '</td>';
	echo '<td>' . $row['sw_key'] . '</td>';
            echo '<td><a href="edit.php?sw_id=' . $row['sw_id'] . '">Edit</a></td>';
            echo '<td><a href="delete.php?sw_id=' . $row['sw_id'] . '">Delete</a></td>';
            echo "</tr>"; 
    }
    // close table>
    echo "</table>";
 ?>

If I comment out the second drop down menu 'sw_name' and only implement the 'dept' drop down menu, the variable is successfully passed to the filtered_view.php and I only see records that have a dept field value that matches my drop down selection value. As soon as I implement the second drop down menu I get no results. If I statically place a value in the SELECT command on filter_view.php then everything works. I think that the $st variable just isn't being injected into the SELECT string.

Can anyone help me? As I said, I am self taught, so I am sure my code is horribly messy, and I will welcome any tips or recommendations. I have posted on other forums and recieved no attention in this issue. A friend of mine recommended this site as he has had great success working through problems with the help of this community.

I hope someone can help a self taught idiot like me figure this out. 🙂

Thanks so much in advance.

dalecosp

Welcome to PHPBuilder!

Please check out "Debugging 101" in my .sig or that of some other long-timers. It may help.

Here's one suggestion for your code:

<?php
//escape from the string when quoting array members/indices
echo "<element rel='foo' value='" . $array['index'] . "'>Label</element>"; // this
echo "<element rel='foo' value='$array[index]'>Label</element>"; // not this

That might have some bearing here, but I'm not sure.

The place to start is looking at the source code sent to your browser....

Derokorian

Actually dale both of your examples are equally valid. See http://php.net/manual/en/language.types.array.php#language.types.array.donts for more info, specifically the part showing using arrays in strings:

<?php
// Show all errors
error_reporting(E_ALL);

$arr = array('fruit' => 'apple', 'veggie' => 'carrot');

// Correct
print $arr['fruit'];  // apple
print $arr['veggie']; // carrot

// Incorrect.  This works but also throws a PHP error of level E_NOTICE because
// of an undefined constant named fruit
// 
// Notice: Use of undefined constant fruit - assumed 'fruit' in...
print $arr[fruit];    // apple

// This defines a constant to demonstrate what's going on.  The value 'veggie'
// is assigned to a constant named fruit.
define('fruit', 'veggie');

// Notice the difference now
print $arr['fruit'];  // apple
print $arr[fruit];    // carrot

// The following is okay, as it's inside a string. Constants are not looked for
// within strings, so no E_NOTICE occurs here
print "Hello $arr[fruit]";      // Hello apple

// With one exception: braces surrounding arrays within strings allows constants
// to be interpreted
print "Hello {$arr[fruit]}";    // Hello carrot
print "Hello {$arr['fruit']}";  // Hello apple

// This will not work, and will result in a parse error, such as:
// Parse error: parse error, expecting T_STRING' or T_VARIABLE' or T_NUM_STRING'
// This of course applies to using superglobals in strings as well
print "Hello $arr['fruit']";
print "Hello $_GET['foo']";

// Concatenation is another option
print "Hello " . $arr['fruit']; // Hello apple
?>

I do however notice you never close the select tag for the second drop down. Also your values should be encloses in quotes if any of the values may have spaces. Finally have you confirmed that there is a record in the DB with BOTH of the values in the drop downs present? Is it possible you are choosing options with nothing to find? I notice you don't ever check to see if any rows are actually returned before trying to use them, try checking if there are any rows and outputting something like "No results found." if there are no rows returned.