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Handling a radio button array in php

开发者 https://www.devze.com 2022-12-21 21:45 出处:网络
I have a bit of PHP code that is really taking a toll on me. To simplify, I have a table where three of the columns are foreign keys (event_type, accommodation_type and public_user). I am dynamically

I have a bit of PHP code that is really taking a toll on me. To simplify, I have a table where three of the columns are foreign keys (event_type, accommodation_type and public_user). I am dynamically creating a set of radio buttons and I have set the ID attribute on each input value correctly.

                        $result2 = mysql_query("SELECT * FROM event_type", $connection);
                        if(!$result2){
                            die("Database query failed: " . mysql_error());
                        }
                        $event = mysql_fetch_array($result2);   

                    echo "<input type='radio' name='event_type' value='";
                    echo $event["id"];
                    echo "' >&nbsp;&nbsp;";
                    echo $event['name'];
                    echo "</input><br /><br />";$result4 = mysql_query("SELECT * FROM accommodation_type", $connection);

                        if(!$result4){
                            die("Database query failed: " . mysql_error());
                        }
                        while($row = mysql_fetch_array($result4)){
                            if($row["id"] == 7 || $row["id"] == 8){
                            echo"<tr><td>";
                            echo $row["name"] ;
                            echo"</td><td>$";
                            echo $row["price"];
                            echo ".00</td><td>";
                            echo "<input type='radio' name='accommod开发者_开发知识库ation_type' value='";
           echo $row["id"];
                            echo "' />";
                            echo "</td></tr>";
                            }
                        }

I retrieved the correct id from the query. Thus after submitting the form with POST, I go on to do some minor validation and prepped the names from my post as follows:

    $event_type = trim(mysql_prep($_POST['event_type']));
    $accommodation_type= trim(mysql_prep($_POST['accommodation_type']));
    $public_user = trim(mysql_prep($_POST['public_user']));
    $comments = trim(mysql_prep($_POST['comments']));
    $grand_total = trim(mysql_prep($_POST['grand_total']));

I then proceeded to write insert statements to insert the data into the relevant tables. This requires two queries as follows.

        $query = "INSERT INTO event_registration (event_type, accommodation_type, public_user, comments, grand_total) VALUES ('{$event_type}','{$accommodation_type}','{$public_user}','{$comments}','{$grand_total}')";
        $query1 = "INSERT INTO additional_member (first_name, last_name, gender, age_group, public_user) VALUES ('{$first_name}','{$last_name}','{$gender}','{$age_group}','{$public_user}')";
        $result = mysql_query($query, $connection);
        $result1 = mysql_query($query1, $connection);

The query1 works as expected, however the first query fails to insert the data. It says undefined index at the lines where I have

    $event_type = trim(mysql_prep($_POST['event_type']));
    $accommodation_type= trim(mysql_prep($_POST['accommodation_type']));

I am not entirely sure where things went wrong. Everything seems to be set up correctly and the data from the form just refuses to be saved.

Any ideas ?


why not try reviewing the code from scratch and see if there are any syntax errors and then move on.


You are creating accommodation type

 echo "<input type='radio' name='accommodation_type' value='";
       echo $row["id"];

But not echoing event_type anywhere. Try echoing that too:

 echo "<input type='radio' name='accommodation_type' value='";
       echo $row["id"];

 echo "<input type='radio' name='event_type' value='";
       echo $row["whatever"];
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