Populate dropdown list get data from different MySQL tables

Question

I need your help to solve a silly problem. I have 2 tables in my database (contents and categories).

I have populated my MySQL table called "categories", and now I want to see in a form the old category stored in the database while I modify it to a new one.

Unfortunately what I wrote shows only the list of the categories get from the database table.

<select name="PostedCat">
    <?php 

    $query_category = "SELECT * FROM categ";
    $result_category =  mysql_query($query_categ) or die (mysql_error());


    while($categ = mysql_fetch_assoc($result_category)){        
    ?>
        <option value="<?php echo $categ['cat_title']; ?>" ><?php echo $categ['cat_title']; ?></option>
    <?php
    }                       
    ?>
</select>

With this code I can see the categories stored in the database, but how can I get the "old" selected one? The stored one?

Hope in some help, but I'm blind at the moment.

Thank you in advance.

What "Old category" ? selected by what ? stored where ? the category of contents ? — Manse
– Manse, Commented Jun 12, 2012 at 9:57
Is the contents of the categories database going to change? If not, it would be best to just have an array of the categories rather than a db. — max_
– max_, Commented Jun 12, 2012 at 18:02
Yes, the categories could be added/modified/deleted as for the content. — Mark
– Mark, Commented Jun 12, 2012 at 18:26

Eugen Rieck · Accepted Answer · 2012-06-12 09:59:03Z

2

Assuming your old category is in $oldcat, just do

$query_category = "SELECT * FROM categ";
$result_category =  mysql_query($query_categ) or die (mysql_error());


while($categ = mysql_fetch_assoc($result_category)){        
?>
    <option value="<?php echo $categ['cat_title']; if ($categ['cat_title']==$oldcat) echo '" selected="true'; ?>" ><?php echo $categ['cat_title']; ?></option>
<?php
}                       
?>

answered Jun 12, 2012 at 9:59

Eugen Rieck

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2 Comments

Eugen Rieck Over a year ago

I might be able to help you, If you give more info ... echo "" inside the loop and post teh output

Mark Over a year ago

Thank you, I moving back home from work and I will add some more code and details. Thank you again for the kindness

Mark · Accepted Answer · 2012-06-12 18:00:23Z

So, thank you again for the kindness. Here is what I did.

I have a page with the list of all my contents and near each one of those I have and "Edit" button. When I click it, I go to a new page with a filled form that takes data directly from MySQL and place quite everything in the correct field. In the function.php I have:

function getPost($id) {
    $id = (int) $id;
    $query = mysql_query("SELECT * FROM contents WHERE id = '$id'") or die (mysql_error());

    return mysql_fetch_array($query);
}

In the edit.php I have this for example:

<?php $editedpost = getPost($_GET['id']); ?>
<form action="editingPost.php" method="post">
<table>
    <tr>
        <td><label for="ContentTitle">Title</label></td>
        <td><input type="text" name="ContentTitle" value="<?php echo $editedpost['content_title']; ?>" /></td>
    </tr>

    <tr>
        <td><label for="ContentCategory">Category</label></td>
        <td>
            <select name="ContentCategory">
                <?php 
                $query_category = "SELECT * FROM categ";
                $result_category =  mysql_query($query_categ) or die (mysql_error());

                while($categ = mysql_fetch_assoc($result_category)){        
                ?>
                    <option value="<?php echo $categ['cat_title']; ?>" >Here I would like to see the data stored in the database, that I choosed before.</option>
                <?php
                }                                   
                ?>
            </select>
        </td>
    </tr>

    <tr>
        <td colspan="2"><input type="submit" name="submit_post_new" /></td>
        <td><input type="hidden" name="id" value="<?php echo $_GET['id']; ?>" /></td>
    </tr>
</table>
</form>

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