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 ""; ` — Qirel, Sep 29 '15 at 12:37

score 3 · Accepted Answer · answered Sep 29 '15 at 12:39

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 '15 at 12:39

Pupil

23,834
6
44
66

score 0 · Answer 2 · edited Mar 02 '16 at 01:10

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 02 '16 at 01:10

Pang

9,564
146
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122

answered Sep 29 '15 at 12:41

Nana Partykar

10,556
10
48
77

2

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. – Jay Blanchard Sep 29 '15 at 12:41
Ok @JayBlanchard. Sorry. Explaining. Editing. – Nana Partykar Sep 29 '15 at 12:42
1

Why are you including the entire code? Just include the areas you fixed. – CodeGodie Sep 29 '15 at 12:44

lakshman · Answer 3 · 2015-09-29T12:44:02.797

<?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

Why are you including the entire code? Just include the areas you fixed. — CodeGodie, Sep 29 '15 at 12:44

populating dropdown list from mysql - php/html

3 Answers3