Issues with getting value of currently selected option in a dropdown

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November 8th, 2012, 12:36 AM

crzybldthrwr

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PHP-General - Issues with getting value of currently selected option in a dropdown

Ok So I have 2 tables, one with Assignment names (Homework1, Homework2, etc) and a table with Student grades for each assignment (So mary might have entries for HW1, HW2 and Joe would also have entries for HW1 and HW2)

Currently I'm using php to grab the list of homeworks from the first table and shove them into a dropdown, no problem there.

But what I want to do is then grab that selected dropdown value, and use it in an SQL query to narrow grade results to JUST the selected Assignment

PHP Code:


		
			
<?

            $HWname = "SELECT hwname FROM Homeworks ORDER BY hwname ASC";

            $HWnameResult = mysql_query($HWname) or die(mysql_error());

            $table = "SELECT Student,NumCorrect,NumTotal,Score FROM StudentGrades WHERE Class='{$_POST[hwname]}' ORDER BY Student ASC";

            $tableResult = mysql_query($table) or die(mysql_error());

            

            //Dropdown for Assignment select 

            $dropdown = "<select name='Homeworks'";

            while($row = mysql_fetch_assoc($HWnameResult)) 

            {

              $dropdown .= "\r\n<option value='{$row['hwname']}'>{$row['hwname']}</option>";

            }

            $dropdown .= "\r\n</select>";

            echo $dropdown;

            ?>

            

            <table id="ViewGradesTable" class="table table-bordered table-condensed table-hover">

            

            <thead class="table-head">

                <?

                echo "

                <tr>

                <th>Student Name</th>

                <th>Correct</th>

                <th>Total</th>

                <th>Score</th>

                </tr>"; ?>



            </thead>



            <?

            //Table of Student HW grades for selected Assignment

            while($row = mysql_fetch_array($tableResult))

              {

              echo "<tr>";

              echo "<td> </td>";

              echo "<td> </td>";

              echo "<td> </td>";

              echo "<td> </td>";

              echo "</tr>";

              }

            echo "</table>";

        

            mysql_close($con);

            ?>

Left the connection to the database out as it's not part of the problem

November 8th, 2012, 05:49 AM

Axweildr

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Code:

 $dropdown = "<select name='Homeworks'>";

Comments on this post

ManiacDan agrees: Good eye

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November 8th, 2012, 03:14 PM

crzybldthrwr

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I saw that shortly after I posted this, fixed
still having an issue

November 8th, 2012, 03:30 PM

requinix

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And where's the code where you try to grab the value?

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November 8th, 2012, 03:31 PM

ManiacDan

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Is there more to this application? For instance, is there a properly-formed <form> tag and a properly formed <input type="submit"> button? If you have those, does the form's action have code which retrieves the value of your drop-down and draws the appropriate page?

Your "issue" seems to be that you're not done writing this application.

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November 8th, 2012, 03:47 PM

crzybldthrwr

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Here's my updated code, I added a button to submit, and I use the SQL query that grabs the HW name as the action.

PHP Code:


		
			
<?



            $HWname = "SELECT hwname FROM Homeworks ORDER BY hwname ASC";

            $HWnameResult = mysql_query($HWname) or die(mysql_error());

            $table = "SELECT Student,NumCorrect,NumTotal,Score FROM StudentGrades WHERE Class=???? ORDER BY Student ASC";

            ?>

            

            <form method="post" action="<? $tableResult = mysql_query($table) or die(mysql_error()); ?>">

            <?

            echo $_POST['Homeworks'];



            //Dropdown for Assignment select 

            $dropdown = "<select name='Homeworks' id='Homeworks'>";

            while($row = mysql_fetch_assoc($HWnameResult)) 

            {

              $dropdown .= "\r\n<option value='{$row['hwname']}'>{$row['hwname']}</option>";

            }

            $dropdown .= "\r\n</select>";

            echo $dropdown;

            ?>

            <input type="submit" id="submitbutton" name="Go" value="Check" />

            </form>

            

            <!--Table of Student HW grades for selected Assignment-->

            <table id="ViewGradesTable" class="table table-bordered table-condensed table-hover">

            <thead class="table-head">

                <?

                echo "

                <tr>

                <th>Student Name</th>

                <th>Correct</th>

                <th>Total</th>

                <th>Score</th>

                </tr>"; ?>

            </thead>



            <?

            while($row = mysql_fetch_array($tableResult))

              {

              echo "<tr>";

              echo "<td>{$row['Student']} </td>";

              echo "<td>{$row['NumCorrect']} </td>";

              echo "<td>{$row['NumTotal']}</td>";

              echo "<td>{$row['Score']}</td>";

              echo "</tr>";

              }

            echo "</table>";

        

            mysql_close($con);

            ?>

My issue right now is in the $table variable, I don't know what to put after Class= so that it grabs the submitted value of the dropdown box and uses it in the query

November 8th, 2012, 03:49 PM

crzybldthrwr

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I did the "echo $_POST['Homeworks']; " to make sure that it's grabbing info correctly and it is. If I select Homework1 and hit Check, "Homework1" pops up on the page and vice versa for anything else I select. I just can't figure out how to take that text and throw it into the second query

November 8th, 2012, 10:10 PM

ManiacDan

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PHP Code:


		
			
 $table = "SELECT Student,NumCorrect,NumTotal,Score FROM StudentGrades WHERE Class='" . mysql_real_escape_string($_POST['Homeworks']) . "' ORDER BY Student ASC";

PHP-General - Issues with getting value of currently selected option in a dropdown

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