Passing results from Gender Drop down to PHP

dixiebuyer

I have a project I am porting to the Web from stand alone Access and Word with VB. I have been pleased with LAMP so far, but this one part of the project is killing me!

Basically what I want to do is take the gender from a drop down box, and based on the selection, I would like to populate the drop down box for relation - i.e. male would return values such as son, father, uncle and female would return daughter, mother, aunt. Although it may be overkill, I want the values returned from a database.

So far in the body of the form I have the following:

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<form name = "personal" method="POST">
 
<div id="right"><div class="title">Family & Friends Information</div>
 
    <tr><td>Gender: </td><td><select id="famgenderid" name="famgender" default="Male">

                    <option value="Male">Male</option>

                    <option value="Female">Female</option>

                </select></td></tr>

    <tr><td>Relationship: </td><td>

                <?php    

                    $gender = "Male";//This is the part that needs to be fixed!!!!!!

                    selectSubset($connection, "tblRelationship" , "ctlRelationship", "ctlGender" , "relation", $gender); 
 
                 ?>

                </td></tr>

    </table>

</div>

And the php script is as follows:

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//Selects a Subset of values to place in a pulldown box based on value of prior box selected e.g. Male = brother, father

function selectSubset ($connection, $tableName , $attributeName1, $attributeName2 , $pulldownName, $gender)

    {

    $defaultWithinResultSet = FALSE;

    $selectSubset = "SELECT {$attributeName1}

                    FROM {$tableName}

                    WHERE {$attributeName2} = '{$gender}'";
 
    //Run selectSubset on the databaseName

    if(!($resultID = @ mysql_query ($selectSubset, $connection)))

        mysql_error();

    print "\n<select name=\"$pulldownName\">";

    //Retrieve each row from the Query

    while ($row = @ mysql_fetch_array($resultID))

        {

            $result = $row[$attributeName1];

            if (isset($defaultvalue) && $result == $defaultvalue)

                // Yes, show as selected

                print "\n\t<option selected value=\"{$result}\">{$result}";

            else

                // No, just show as an option

                print "\n\t<option value=\"{$result}\">{$result}";

            print "</option>";

        }

            print "\n</select>";

dixiebuyer

In the example above I have hard coded "Male" to get it to work one way. I am unsure how to pass the value from the famgender drop down box (Male or Female) to the variable $gender.

Atli

Hi.

Personally, I would use AJAX to do this, but this can easily be done using PHP.

The complication here is that PHP is server-side, which means that by the time your user even gets the chance to see the <select> drop-down, the PHP script will already have completed it's job.

What you need to do is have PHP submit the <select> box first, and hold of posting the rest of the code until the form has been submitted with a gender value before re-posting the data.

Consider this:

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<form name="genderForm" action="?" method="post">            

<?php

// Check if all the info has been submitted

if(isset($_POST['submitButton'])) 

{

    // Process the form

    echo "<h3>Complete!</h3>";

}
 
// Check if the gender has been selected

else if(isset($_POST['gender']) && $_POST['gender'] != "null") 

{

    // Print some gender specific content

    if($_POST['gender'] == "male") {

        echo 'Enter you wifes name: <input type="text" name="wifeName" /><br />';

    }

    else {

        echo 'Enter you husbands name: <input type="text" name="hubbyName" /><br />';

    }
 
    // And of course the submit button

    echo '<input type="submit" name="submitButton" value="submitted" />';

}
 
// Nothing has been sent.

else 

{

    // Show the gender <select> box.

    echo <<<HTML

Gender: <select name="gender" onchange="javascript: this.parentNode.submit();">

    <option value="null" selected="selected">--Select please--</option>

    <option value="female">Female</option>

    <option value="male">Male</option>

</select>            

HTML;

}

?>

</form>

This simply prints a gender <select> box. Once you select a gender, it automatically submits the form and loads the gender-specific form elements.
Once the submit button has been pressed, it should process the data.

It probably won't work exactly like you want yours to work, but it should at least point you in the right direction.

Let me know if something is unclear.

P.S.
Please use [code] tags when posting your code examples. (See How to ask a question)

[code] ... Code goes here... [/code]

Thank you.
Moderator

dixiebuyer

Thanks atli. That makes sense. I think my best bet is just to stay away from php for this function and use javascript so it is all client side. The need to access the mysql server just is not that great in this instance.

Passing results from Gender Drop down to PHP

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re: Passing results from Gender Drop down to PHP

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