Html/Php adding a drop down with mysql query

Question

I am trying to design a website for my project and I got confused in the following.If I want to add a dropdown menu I do it with the following code.

<select>
    <option value="January"<?php if ($row[month] == 'January') echo ' selected="selected"'; ?>>January</option>
    <option value="February"<?php if ($row[year] == 'February') echo ' selected="selected"'; ?>>February</option>
</select>

But what I want to do is following.I have a database and bunch of data which I will query with a select statement.And I want to give all of them as an option to do this drop-down menu.How can I accomplish it?To simplify the question assume I have a query

$sql= "select Name FROM Course WHERE courseName='".$something."'"; 
$resulting=mysqli_query($connection,$sql);
$resultarray=mysqli_fetch_array($resulting);

How can I accomplish this?

Edit:Without a dropdown menu I would normally create a form in this way.

<form action="restaurantAction.php" method="post">
Restaurant Name:<input type="text" name="deleteRestaurant">
<input type="submit">
</form>

How can I apply this to a drop down menu based on the answer of Minh

foreach array item create an <option>

u_mulder
– u_mulder

2016-05-07 14:16:40 +00:00
Commented May 7, 2016 at 14:16 — u_mulder
– u_mulder, Commented May 7, 2016 at 14:16

Minh Thành · Accepted Answer · 2016-05-07 22:44:54Z

2

You can write like this

<select>
<?php foreach ($resultarray as $key => $value) {?>
    <option value="<?php echo $value?>"><?php echo $value?></option>
<?php } ?>
</select>

For your update

<form action="restaurantAction.php" method="post">
Restaurant Name:<input type="text" name="deleteRestaurant">
<select name="restaurantName">
<?php foreach ($resultarray as $key => $value) {?>
    <option value="<?php echo $value?>"><?php echo $value?></option>
<?php } ?>
</select>
<input type="submit">
</form>

edited May 7, 2016 at 22:44

answered May 7, 2016 at 14:16

Minh Thành

1439 bronze badges

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2 Comments

Minh Thành Over a year ago

Mask this to right answer if it is solutions :p. Thanks

Curriculaa Over a year ago

Yeah it works.I also upvoted the other answers but I already used this one so I ll accept this as an answer.I still have one little question.How can I turn this into a form.I will edit question to show the format

user6023290 · Accepted Answer · 2016-05-07 14:29:11Z

1

<?php
// List filtered items
// Fetch data as associated array
<select>
while ($row = mysqli_fetch_assoc($resulting)) : ?>
<option value="<?php echo $resulting['game_name']?>"><?php echo $resulting['game_name']?></option>// $resulting['game_name'], or however called that field in the DB
</select>

answered May 7, 2016 at 14:29

user6023290

Comments

Dharmik Patel · Accepted Answer · 2016-05-07 14:43:41Z

1

      <?php                    
         $result=mysqli_query($con,"select Name FROM Resturant WHERE ResturantName='".$something."'");
         echo '<form action="restaurantAction.php" method="post">'
         echo 'Resutarant Name:<select name="ResturantName">';
            while ($row = mysqli_fetch_array($result)) {
               echo '<option value="'.$row['Name'].'">'.$row['Name'].'</option>';
              }
         echo '</select>';
         echo '<input type="submit">';
         echo '</form>';
      ?>

Please Check this

edited May 7, 2016 at 14:43

answered May 7, 2016 at 14:30

Dharmik Patel

1016 bronze badges

1 Comment

Curriculaa Over a year ago

What does this line accomplish?echo 'Resutarant Name:<select name="ResturantName">'; .Also there is a typo in Resutarant Name but it is not a big deal.Does this work without writing that line and using an already fetched array?.So is it ok if I just omit the first 3 lines and apply the code starting from while statement.It seems like ok to me

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