Populate html <select> with array data from mysql in PHP

Question

I can see the query returning results, but I can't seem to be able to put them into a html dropdown box. Also, the dropdown box has just as many entries as the query returns, but THEY ARE ALL WHITE SPACES. HOWEVER, the page source shows correct option values such as

<option value="3 John"></option>

<option value="Jude"></option>

<option value="Revelation"></option>

Can somebody help me out? Why dont they actually show in the dropdown box?

<html>
<?php
    //Connect to the database
    $mysqli = new mysqli("localhost", "root", "", "bible");

    //Return an error if we have a connection issue
    if ($mysqli->connect_error) {
        die('Connect Error (' . $mysqli->connect_errno . ') '
                . $mysqli->connect_error);
        }

    //Query the database for the results we want
    $query = $mysqli->query("select distinct bname as Name from kjv limit 1");

    //Create an array  of objects for each returned row
    while($array[] = $query->fetch_object());

    array_pop($array);

    //Print out the array results
    print_r($array);
    ?>
    <h3>Dropdown Demo Starts Here</h3>
    <select name="the_name">
    <?php foreach($array as $option) : ?>
        <option value="<?php echo $option->Name; ?>"></option>
    </select>
        <?php endforeach; ?>

Reshil · Accepted Answer · 2013-09-22 05:13:21Z

4

Try This

<select name="the_name">
<?php foreach($array as $option) : ?>
        <option value="<?php echo $option['Name']; ?>"><?php echo $option['Name']; ?></option>
<?php endforeach; ?>
</select>

answered Sep 22, 2013 at 5:13

Reshil

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3 Comments

Vicky Thakor Over a year ago

You are right he is printing values in attribute. @Slava Basat print it within option tag.

Slava Basat Over a year ago

thank you but print what within option tag? the <select> tag?

andreithegiant Over a year ago

A minor quibble but the code will work without the <?php endforeach; ?> at the end. I suspect @Reshil is carrying it over from another language like vba where it is in fact necessary to end a foreach loop with it.

Tim Wax · Accepted Answer · 2013-09-22 05:28:35Z

1

After the query is executed use the while loop to add the options to select

$query = $mysqli->query("select distinct bname as Name from kjv limit 1"); ?>

<select>
    <?php while($option = $query->fetch_object()){ ?>
        <option><?php echo $option->Name; ?></option>
    <?php } ?>
</select>

Not sure what the array_pop is doing in the code

answered Sep 22, 2013 at 5:28

Tim Wax

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2 Comments

Slava Basat Over a year ago

Hello I will try this now. Array pop removes the last empty record from the query results. I saw it on youtube otherwise I would never put it there

Slava Basat Over a year ago

YOU ARE A G. THIS WORKED RESPECT MAN

Slava Basat · Accepted Answer · 2013-09-22 16:19:45Z

1

AS TIM WAX SAID THIS IS THE SOLUTION

$query = $mysqli->query("select distinct bname as Name from kjv limit 1"); ?>

<select>
    <?php while($option = $query->fetch_object()){ ?>
        <option><?php echo $option->Name; ?></option>
    <?php } ?>
</select>

answered Sep 22, 2013 at 16:19

Slava Basat

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Comments

Vicky Thakor · Accepted Answer · 2013-09-22 05:11:32Z

0

<select name="the_name">
<?php foreach($array as $option) : ?>
        <option value="<?php echo $option->Name; ?>"></option>
<?php endforeach; ?>
</select>

You ended your loop in a way that it also create <select> tag again and again. Change it and try again. I don't know much about .php but it could be a problem in showing your dropdown box.

answered Sep 22, 2013 at 5:11

Vicky Thakor

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2 Comments

Slava Basat Over a year ago

You mean like this? <select name="the_name"> <?php foreach($array as $option) : ?> <option value="<?php echo $option->Name; ?>"></select></option> <?php endforeach; ?> If so, it still gives me empty data.

Vicky Thakor Over a year ago

stackoverflow.com/questions/16746244/… Check out the thread

Bobski · Accepted Answer · 2014-02-16 05:45:53Z

0

here is mine .. im a beginner but it works for me,

    $query = $mysqli->query("SELECT * FROM  `student_type_db`"); //table of student type

    echo "<select>";
    while($row = $query->fetch_array()){
         echo "<option>";
         echo $row['student_type'] . " - " . $row['student_description'];
         echo "</option>"; 
    }
    echo "</select>";

   // student type = 1 | student description = regular
   // output : 1 - regular

answered Feb 16, 2014 at 5:45

Bobski

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