PHP/MYSQL Could not display data from MYSQL in Drop Down Box

Question

I am having a problem displaying data from MYSQL into drop down box.

Output:

<html>
<body>
    <form name=displayQuestion>
        Survey Categories : 
        <select name="surveyCategory">
        <option> Choose Survey Category </option>
        <?php
            $surveyQuery = "SELECT survey_id, survey_name FROM surveys";
            $result = mysql_query($surveyQuery) or die (mysql_error());
            while($menu=mysql_fetch_assoc($result)){
                echo "<option value=$menu[survey_id]>$menu[survey_name]</option>";                  
            }
        ?>
    </select>
    </form>
</body>
</html>

First off, mysql_query has been deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. That said, where are you establishing your connection to the db? Add that to your question. — hanleyhansen
– hanleyhansen, Commented Dec 15, 2013 at 5:34

virmaior · Accepted Answer · 2013-12-15 07:19:37Z

1

no reason to use that ugly formatting:

echo '<option value="' . $menu['survey_id'] . '">' . $menu['survey_name'] . '</option>';

edited Dec 15, 2013 at 7:19

answered Dec 15, 2013 at 5:35

virmaior

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2 Comments

Ignacio Ocampo Over a year ago

Access field value with simple quotes $menu['survey_name']

Ignacio Ocampo Over a year ago

Your code have syntax error: '<option value="'> . should be '<option value="' .

Ignacio Ocampo · Accepted Answer · 2013-12-15 05:34:04Z

1

You need put your option value within quotes and access array with simple quotes like as:

echo "<option value=\"$menu['survey_id']\">$menu['survey_name']</option>";

answered Dec 15, 2013 at 5:34

Ignacio Ocampo

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