Dropdown list displays no value

Question

I have a Book details form where we can insert the book details.
One of the field in this form is "Category_Id". The value of this field is coming from database.
So I want to display all the category id that is stored in a table name " book_catagory ".
I have tried to write code which I have done it earlier also but here no values are displayed in the form. Please help me to find my mistake. My code snippet is as follow:

<td>Enter Category ID: </td>
<td>
    <?php
        $select_query = "select Category_Id from book_catagory";
        $select_query_run = mysql_query($select_query);

        echo "<select name='category_id' id='category_id'>";

        while ($select_query_array = mysql_fetch_array($select_query_run)) {
            echo "<option value='" . htmlspecialchars($select_query_array['Category_Id']) . "' >" .
 htmlspecialchars($select_query_array["Category_Id"]) . " </option>";
        }

        echo "</select>";
    ?>
</td>

This code appears to have been written in the 19th century. Come on; modern code for modern times. — Strawberry, Nov 24 '15 at 07:29
Great work @Vucko - now it actually looks like he did actually do this work rather than copy paste this from another site. Infact this question/code has been asked over and over again in SO. http://stackoverflow.com/questions/24839519/php-form-dropdown-containing-options-based-on-sql-data — Shaun, Nov 24 '15 at 07:34

Kevin · Answer 1 · 2015-11-24T08:40:58.033

I can recommend you to establish the connection via the object orientated approach.This way to connect to database server is since PHP 5.5.0 deprecated.[1] Your query seems to be correct. You parse your SQL results with mysql_fetch_array(), which return a numeric and associated array.[2] If you prefer to parse your database result by associated attribute name only you should use the PHP method mysql_fetch_assoc()[3].

You can alter your code by following to fix your issue:

<td>Enter Category ID: </td>
<td>
    <?php
        $select_query = "select Category_Id from book_catagory";
        $select_query_run = mysql_query($select_query);

        echo "<select name='category_id' id='category_id'>";

        while ($select_query_array = mysql_fetch_array($select_query_run), MYSQL_NUM) {
            echo "<option value='".htmlspecialchars($select_query_array[0])."'>" .htmlspecialchars($select_query_array[0]) . " </option>";
        }

        echo "</select>";
    ?>
</td>

References:
[1] http://php.net/manual/de/mysqli.summary.php
[2] http://php.net/manual/de/function.mysql-fetch-array.php
[3] http://php.net/manual/de/function.mysql-fetch-assoc.php

Did you have had send the sql statement direct in the database to check which results should been shown? Alternatively you can add an integer value to count while loops: `int i = 0; while() {} echo 'While loops:'.i;` — Kevin, Nov 24 '15 at 08:29

score 0 · Accepted Answer · answered Nov 24 '15 at 08:55

You should try like this :

    <td>Enter Category ID: </td>
    <td>
        <?php
            $select_query = "select Category_Id from book_catagory";
            $select_query_run = mysql_query($select_query);

            echo "<select name='category_id' id='category_id'>";

            while ($select_query_array=mysql_fetch_array($select_query_run)) {
     ?>
<option value=' <?php echo htmlspecialchars($select_query_array['Category_Id']) ?>' >
    <?php echo htmlspecialchars($select_query_array['Category_Id']) ?></option>
            }
    <?php
            echo "</select>";
        ?>
    </td>

Also I would say that you should use mysqli. And you can also use mysql_fetch_asssoc if you need an associative array.

Dropdown list displays no value

2 Answers2