如何从数据库创建选择选项？

Question

I'm working on a form that has 4 different select elements from 2 tables of a database. I haven't done anything like this and I don't really know how to do it.

I have a table called "students" from I need "name" and "class" and a table called "books" from I need "writer" "title" ... all is one select element and all has more than 2 option values.

I've tried with only one sql query and one select but it shows only one option on the site, wether it has about 6 values in the database.

My code:

$sql = "SELECT class
        FROM students";
$result = mysql_query($sql);


while ($row = mysql_fetch_assoc($result)) {
    $select_class = "<option value={$row['class']}>{$row['class']}</option>";
}

<select id="class" name="class">
    <?php print $select_class; ?>
</select>

How would it be correct?

Answer 1

try this

$sql = "SELECT class FROM students";
$result = mysql_query($sql);

echo '<select id="class" name="class">';

while ($row = mysql_fetch_assoc($result)) {
    echo "<option value={$row['class']}>{$row['class']}</option>";
}

echo '</select>';

Answer 2

Changing this:

$select_class = "<option value={$row['class']}>{$row['class']}</option>";

to this:

$select_class .= "<option value={$row['class']}>{$row['class']}</option>";

might solve your problem.

Right now you are constantly resetting the value of $select_class, instead of adding to it. The .= assignment should help you get around this.

As always, be sure to up-vote any StackOverflow answers you find useful.

Answer 3

You are overwriting $select_class on each while() loop. You need to concatenate $select_class . Change to $select_class .=

$select_class = "";
while ($row = mysql_fetch_assoc($result)) {
    $select_class .= "<option value={$row['class']}>{$row['class']}</option>";
}

Answer 4

Try this

        <?php
                        $dbhandle = mysql_connect($hostname, $username, $password) or die("can't connect");
                        $table = "students";
                        $sql = "SELECT * FROM students";
                        $result = mysql_query($sql, $dbhandle);
                        mysql_data_seek($result, 0);
        ?>
        <form>                
        <select class="dropdown" name="dropdown">           
    <?php 
    if(mysql_num_rows($result) > 0){
        while($row = mysql_fetch_array($result)) {
            echo '<option value="' . $row['class'] . '">' . $row['class'] . '</option>';
            }
         }
    ?>
</select>
</form>