populating dropdown list from mysql - php/html

Question

Trying to populate a dropdown list from my database - connection is ok and in my mind the code I have should work, but currently getting a blank dropdown... Have looked at PHP- Fetch from database and store in drop down menu html as well as other tutorials but alas no luck so far!

code is as follows...

<?php
//get constants for database
require("database.php");

// Opens a connection to a MySQL server
$connection = mysqli_connect ($server, $username, $password);
if (!$connection){  
    die('Not connected : ' . mysqli_error());
}

// Set the active MySQL database
$db_selected = mysqli_select_db($connection, $database);
if (!$db_selected) {
    die ('Can\'t use db : ' . mysqli_error($connection));
}

$query = "SELECT * FROM route WHERE 1";
$result = mysqli_query($connection, $query);


echo '<select name="list" style="width:400px;">';

while($r = mysqli_fetch_assoc($result)){
    echo "<option value=".$r['alt']."</option>"; 
}

echo '</select>';
?>

Your option-tag is broken. Should be something like echo "<option value='".$r['alt']."'>[Text in dropdown here]</option>"; — Qirel
– Qirel, Commented Sep 29, 2015 at 12:37

Pupil · Accepted Answer · 2015-09-29 12:39:13Z

3

<option> tag is broken.

Corrected code:

while($r = mysqli_fetch_assoc($result)){
  echo '<option value="'.$r['alt'].'">'.$r['alt'].'</option>'; 
}

Note: You can use single and double quotes either, but, they should be properly closed.

answered Sep 29, 2015 at 12:39

Pupil

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Comments

Pang · Accepted Answer · 2016-03-02 01:10:52Z

0

Please have a look on this Select Dropdown Syntax

Syntax For Select Dropdown

<select>
    <option value="volvo">Volvo</option>
    <option value="saab">Saab</option>
</select>

You have missed a closing of value in option.

<?php
//get constants for database
require("database.php");

// Opens a connection to a MySQL server
$connection = mysqli_connect ($server, $username, $password);
if (!$connection){  
die('Not connected : ' . mysqli_error());
}

// Set the active MySQL database
$db_selected = mysqli_select_db($connection, $database);
if (!$db_selected) {
    die ('Can\'t use db : ' . mysqli_error($connection));
}

$query = "SELECT * FROM route WHERE 1";
$result = mysqli_query($connection, $query);

//Changes From Here

?>
<select name="list" style="width:400px;">
    <?
    while($r = mysqli_fetch_assoc($result))
    {?>
        <option value="<?echo $r['alt'];?>"><?echo $r['alt'];?></option>
    <?}?>
</select>

edited Mar 2, 2016 at 1:10

Pang

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answered Sep 29, 2015 at 12:41

Nana Partykar

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3 Comments

Jay Blanchard Over a year ago

Why should the OP try this? A good answer will always have an explanation of what was done and why it was done that way, not only for the OP but for future visitors to SO.

Nana Partykar Over a year ago

Ok @JayBlanchard. Sorry. Explaining. Editing.

CodeGodie Over a year ago

Why are you including the entire code? Just include the areas you fixed.

lakshman · Accepted Answer · 2015-09-29 12:44:02Z

-1

<?php
        //get constants for database
        require("database.php");

        // Opens a connection to a MySQL server
        $connection = mysqli_connect ($server, $username, $password);
        if (!$connection){  
            die('Not connected : ' . mysqli_error());
            }

        // Set the active MySQL database
        $db_selected = mysqli_select_db($connection, $database);
        if (!$db_selected) {
            die ('Can\'t use db : ' . mysqli_error($connection));
        }

        $query = "SELECT * FROM route WHERE 1";
        $result = mysqli_query($connection, $query);


            echo '<select name="list" style="width:400px;">';

                while($r = mysqli_fetch_assoc($result)){
                    echo "<option value=".$r['alt'].">".$r['alt']."</option>"; 
                }

            echo '</select>';
            ?>

so if your select condition is right/hidden by you then you will have to add a display value between option tag. As per the code you have assigned values in option value tag but not given the values to be displayed. Try the above

edited Sep 29, 2015 at 12:44

answered Sep 29, 2015 at 12:42

lakshman

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1 Comment

CodeGodie Over a year ago

Why are you including the entire code? Just include the areas you fixed.

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