Inserting Dropdown List values in MYSQL Database

Question

So i was looking for solution for a couple of hours now.. my problem is that the data from the dropdown list (yearlevel) wont save in the database...

here's the full code of my work named qwe.php

<!DOCTYPE html>
<html>
<head>
</head>

<form method = "POST" name = "subjects" action ="qwe.php">

<?php
require_once('xcon.php');
$query = "SELECT yearlevel from yearlevel";
$result = mysql_query($query);

echo "Select Year Level: <select name ='yearlevel'>";
    while($row = mysql_fetch_array($result)){
        echo "<option value = '" . $row['yearlevel'] . "'>" . $row['yearlevel'] . "</option>";
    }
    echo "</select><br>";

?>

Multiple Choice: <input type = "text" name="MC"><br>
Identification: <input type = "text" name ="Identification"><br>

                 <input type = "submit" name = "confirm" value = "Confirm">
</form>


<?php

if(isset($_POST['confirm'])){

$MC = $_POST['MC'];
$Identification = $_POST['Identification'];
echo "<form method = 'POST' name = 'items' action ='qwe.php'>";
$items = 1;
$items2 = 1;

echo "<center>MULTIPLE CHOICE</center><br><br><br>";
    for ($x = 1; $x <= $MC; $x++) {

        echo "Question Number $items:"; echo "<input type = 'text' name = 'questions[]' style='width: 500px'><br><br>";
        echo "A. "; echo "<input type = 'text' name = 'ans1[]'>";
        echo "B. "; echo "<input type = 'text' name = 'ans2[]'><br>";
        echo "C. "; echo "<input type = 'text' name = 'ans3[]'>";
        echo "D. "; echo "<input type = 'text' name = 'ans4[]'><br>";
        echo "Correct Answer: "; echo "<input type = 'text' name ='correctans[]'><br><br>";
        $items++;

    }
echo "<center>IDENTIFICATION</center><br><br><br>";
    for ($y = 1; $y <= $Identification; $y++){
        echo "Question # $items2:"; echo "<input type = 'text' name = 'identification[]' style='width: 500px'><br>";
        echo "Answer: "; echo "<input type = 'text' name = 'identificationans[]'><br><br>";
        $items2++;
    }
        echo "<input type ='submit' name = 'save' value = 'Save'>";
        echo "</form>";
}

?>

<?php

    if(isset($_POST['save'])){

        $yearlvl = $_POST['yearlevel'];
        require_once('xcon.php');

        foreach ($_POST['questions'] as $key => $question){
            $ans1 = $_POST['ans1'][$key];
            $ans2 = $_POST['ans2'][$key];
            $ans3 = $_POST['ans3'][$key];
            $ans4 = $_POST['ans4'][$key];
            $correctans = $_POST['correctans'][$key];

            echo "<input type = 'hidden' value = '$question'>";
            echo "<input type = 'hidden' value = '$yearlvl'>";

            $query = "INSERT INTO mcq (mc_id, questions, ans1, ans2, ans3, ans4, correctans, yearlvl) 
                  VALUES ('NULL','$question','$ans1','$ans2','$ans3','$ans4','$correctans', '$yearlvl')";
        $result = mysql_query($query);

        }


        foreach($_POST['identification'] as $key => $identification){
            $identificationans = $_POST['identificationans'][$key];

            $query = "INSERT INTO identification (identification_id, identification_question, identification_answer, yearlvl)
                    VALUES ('NULL','$identification','$identificationans','$yearlvl')";
            $result = mysql_query($query);
        }

            if($result){
                echo 'Insert Success!<br>';
            }else{
            echo 'Error<br>';
            }    
    }

?>

Did i do anything wrong ? I really hope that you guys will be able to help me ! Thank you so much in advance !

Other than the obvious sermon against using mysql_* functions, what i have to say is print out the sql statement to see that it's what you expect it to be. Then run it in the mysql console. — e4c5
– e4c5, Commented Oct 13, 2015 at 10:46
i have changed the null part.. I dont know how to do that 'running in the mysql console' thing since i'm a beginner in php — Ger Mel
– Ger Mel, Commented Oct 13, 2015 at 15:00

Chirag Senjaliya · Accepted Answer · 2015-10-13 11:09:25Z

1

I think you should try this on ur last two query:

  $query = "INSERT INTO mcq (mc_id, questions, ans1, ans2, ans3, ans4, correctans, ) 
                          VALUES ('NULL','$question','$ans1','$ans2','$ans3','$ans4','$correctans')";


    $query = "INSERT INTO identification (identification_id, identification_question, identification_answer, )
                        VALUES ('NULL','$identification','$identificationans')";

answered Oct 13, 2015 at 11:09

Chirag Senjaliya

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3 Comments

Ger Mel Over a year ago

where did the yearlevel go ?

Chirag Senjaliya Over a year ago

it will not inserted!!

Ger Mel Over a year ago

oh, but what i want to happen is to insert the value in the dropdown list in my database..

mikie · Accepted Answer · 2015-10-14 06:34:19Z

try getting the value into a hidden input and then access it like this.

<!DOCTYPE html>
<html>
<head>
</head>

<form method = "POST" name = "subjects" action ="">

<?php
require_once('xcon.php');
$query = "SELECT yearlevel from yearlevel";
$result = mysql_query($query);

echo "Select Year Level: <select name ='yearlevel'>";
while($row = mysql_fetch_array($result)){
    echo "<option value = '" . $row['yearlevel'] . "'>" . $row['yearlevel'] . "</option>";
}
echo "</select><br>";

?>

Multiple Choice: <input type = "text" name="MC"><br>
Identification: <input type = "text" name ="Identification"><br>

<input type = "submit" name = "confirm" value = "Confirm">
</form>


<?php

if(isset($_POST['confirm'])){

$MC = $_POST['MC'];
$Identification = $_POST['Identification'];
echo "<form method = 'POST' name = 'items' action =''>";
$items = 1;
$items2 = 1;

$level = $_POST['yearlevel'];
echo'<input type="hidden" name="yearlevel2" value="'.$level.'">';
echo "<center>MULTIPLE CHOICE</center><br><br><br>";
for ($x = 1; $x <= $MC; $x++) {
echo "Question Number $items:"; echo "<input type = 'text' name = 'questions[]' style='width: 500px'><br><br>";
    echo "A. "; echo "<input type = 'text' name = 'ans1[]'>";
    echo "B. "; echo "<input type = 'text' name = 'ans2[]'><br>";
    echo "C. "; echo "<input type = 'text' name = 'ans3[]'>";
    echo "D. "; echo "<input type = 'text' name = 'ans4[]'><br>";
    echo "Correct Answer: "; echo "<input type = 'text' name ='correctans[]'><br><br>";
    $items++;

}
echo "<center>IDENTIFICATION</center><br><br><br>";
for ($y = 1; $y <= $Identification; $y++){
    echo "Question # $items2:"; echo "<input type = 'text' name = 'identification[]' style='width: 500px'><br>";
    echo "Answer: "; echo "<input type = 'text' name = 'identificationans[]'><br><br>";
    $items2++;
}
    echo "<input type ='submit' name = 'save' value = 'Save'>";
    echo "</form>";
}

?>


<?php

if(isset($_POST['save'])){

    $yearlvl = $_POST['yearlevel2'];

    echo $yearlvl;
    exit();
    require_once('xcon.php');

    foreach ($_POST['questions'] as $key => $question){
        $ans1 = $_POST['ans1'][$key];
        $ans2 = $_POST['ans2'][$key];
        $ans3 = $_POST['ans3'][$key];
        $ans4 = $_POST['ans4'][$key];
        $correctans = $_POST['correctans'][$key];

        echo "<input type = 'hidden' value = '$question'>";
        echo "<input type = 'hidden' value = '$yearlvl'>";

        $query = "INSERT INTO mcq (mc_id, questions, ans1, ans2, ans3, ans4, correctans, yearlvl) 
              VALUES ('NULL','$question','$ans1','$ans2','$ans3','$ans4','$correctans', '$yearlvl')";
    $result = mysql_query($query);

    }


    foreach($_POST['identification'] as $key => $identification){
        $identificationans = $_POST['identificationans'][$key];

        $query = "INSERT INTO identification (identification_id, identification_question, identification_answer, yearlvl)
                VALUES ('NULL','$identification','$identificationans','$yearlvl')";
        $result = mysql_query($query);
    }

        if($result){
            echo 'Insert Success!<br>';
        }else{
        echo 'Error<br>';
        }    
}

?>

I've edited my answer to show the entire code. You must make sure that the hidden input field is within a form, in your case it should be within the second form. Let me know if this works for you.

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