[StrunkWriter]

I read all the earlier posts I could find regarding PHP/MySql drop down menus, but I still need help with my code. I'm creating a simple database (kind of a glorified blog really) that includes a form to upload information and a separate form to edit the information. It is the "edit" form I'm having an issue with.

If someone selects a row to edit, they are directed to a form that has contains their information and they can edit the information and clicking submit will update the table. My problem exists with the drop down menu in this form.

Inside the form the drop down selects categories which are kept in a separate table. But, because the categories come from a separate table, it doesn't show the category the user originally selected. I'm not explaining this well. So, here's the code (minus extraneous fields and formatting stuff):

PHP Code:

$story_select = $_GET['story_select'];



  if(isset($_POST['edit']))

{

$category = $_POST['category'];

$headline = $_POST['headline'];

$byline = $_POST['byline'];

$organization = $_POST['organization'];

$article = $_POST['article'];

$moreinfurl = $_POST['moreinfurl'];

$ReleaseDate = $_POST['ReleaseDate'];

$news_candidate = $_POST['news_candidate'];

mysql_query("UPDATE news_stories SET category = '$category', headline = '$headline', byline = '$byline', organization = '$organization', moreinfurl = '$moreinfurl', article = '$article', ReleaseDate = '$ReleaseDate', news_candidate = '$news_candidate' WHERE storynum = '$story_select'");



echo "story was updated!";

}

else

{



$query  = "SELECT * FROM news_stories WHERE storynum='$story_select'";

$result = mysql_query($query);

while($row = mysql_fetch_assoc($result))  {



echo  "<h1>Edit Article# " .$story_select ."</h1>";

echo "<form method=\"post\">";

echo "Category: <select name=\"category\">";



  $query2  = "SELECT * FROM news_categories";

  $result2 = mysql_query($query2);



  while($row2 = mysql_fetch_assoc($result2))  {

  echo "<option value=\"{$row2['catnum']}\">{$row2['catname']}</option><br>";

  }

  

  echo "</select>";

  echo "Headline (100 characters max): <input type=\"text\" name=\"headline\" value=\"{$row['headline']}\"> <br>";

  

 echo "<input name=\"edit\" type=\"submit\" id=\"edit\" value=\"Edit\">";

echo "</form>";

}



 } 

I hope I expressed this clearly. Any help will be appreciated.

[job0107]

We're going to need a little help here understanding what it is you want.

I am assuming the category select box is used to change the category you want the article to appear under?

And you want the news article category to appear selected in the dropdown box?

And I am assuming that $category is the value that would be found in $row2['catname']?

Then you could do it like this:

PHP Code:

<?php
$story_select = $_GET['story_select'];
if(isset($_POST['edit']))
{
 $category = $_POST['category'];
 $headline = $_POST['headline'];
 $byline = $_POST['byline'];
 $organization = $_POST['organization'];
 $article = $_POST['article'];
 $moreinfurl = $_POST['moreinfurl'];
 $ReleaseDate = $_POST['ReleaseDate'];
 $news_candidate = $_POST['news_candidate'];
 mysql_query("UPDATE news_stories SET category = '$category', headline = '$headline', byline = '$byline', organization = '$organization', moreinfurl = '$moreinfurl', article = '$article', ReleaseDate = '$ReleaseDate', news_candidate = '$news_candidate' WHERE storynum = '$story_select'");
 echo "story was updated!";
 }
else
{
 $query  = "SELECT * FROM news_stories WHERE storynum='$story_select'";
 $result = mysql_query($query);
 while($row = mysql_fetch_assoc($result))
 {
  echo  "<h1>Edit Article# " .$story_select ."</h1>
         <form method=\"post\">
         Category: <select name=\"category\">";
  $query2  = "SELECT * FROM news_categories";
  $result2 = mysql_query($query2);
  while($row2 = mysql_fetch_assoc($result2))
  {
   if($row2['catname']==$category)
   {
    echo "<option value=\"{$row2['catnum']}\" selected>{$row2['catname']}</option>";
    }
   else
   {
    echo "<option value=\"{$row2['catnum']}\">{$row2['catname']}</option>";
    }
   }
  echo "</select>
        Headline (100 characters max): <input type=\"text\" name=\"headline\" value=\"{$row['headline']}\"> <br>
       <input name=\"edit\" type=\"submit\" id=\"edit\" value=\"Edit\">
       </form>";
  }
 }
?>

Another problem I see is you don't have the other fields included in the form.
So when the form is submitted, only the category and headline are submitted.

These fields aren't included in the form:

PHP Code:

$byline = $_POST['byline'];
$organization = $_POST['organization'];
$article = $_POST['article'];
$moreinfurl = $_POST['moreinfurl'];
$ReleaseDate = $_POST['ReleaseDate'];
$news_candidate = $_POST['news_candidate']; 

And this is the update query:

PHP Code:

mysql_query("UPDATE news_stories SET category = '$category', headline = '$headline', byline = '$byline', organization = '$organization', moreinfurl = '$moreinfurl', article = '$article', ReleaseDate = '$ReleaseDate', news_candidate = '$news_candidate' WHERE storynum = '$story_select'"); 

So when you submit the form and the table is updated, the excluded fields will contain NULL values and any data that was in them will be lost or the update query won't function at all.

And you aren't checking to see if all the variables have any values before you update the table.
You are only checking to see if the form was submitted.
So if the user were to erase a value in a field then that field would be empty and the data in the table would be over written with a NULL value.

You also should make sure any value submitted doesn't contain and data that could be used to harm your database.

[StrunkWriter]

Yah, I cut some of the code out because I didn't want to bore you with un-needed details. I figured after the form field for "headline" y'all could see I know how create a basic form -- so I just cut the rest out when I was posting.

I want the old values to show in every field, so the end user can see what exists before they decide to change it. If they don't edit a field, it will "update" the same information back into the table.

But it's the showing old values part I'm having issue with. How can I have the drop down contain all the options, but have it's "default" value show as the information already contained in the main table?

I guess another possibility is to show the whole form filled out and put an "edit" button next to each field. When the user clicks the edit button they will get a "miniform" to change that particular field -- without changing the other data. What do you think? Is this a "safer" solution for my user or a bigger pain in the butt for them?

[job0107]

You have got a lot of ideas, you just need to pick one.
I showed you how to select the item in the dropdown box when the form information matches the database information.

Look, you are loading the data from the record fetched from the news_stories table
and I am assuming you are loading the form inputs with this data.

And you are querying the news_categories table and using the results to populate the select box.

I know the news_stories table contains a category field.
And the news_categories table contains a catname field.
And I am assuming that the value in the category field
came from the catname field. So as you can see, I am comparing
the category field with the catname field and
marking that select option as selected, which selects it when displayed.

Now as long as you are loading the form from the database and you are updating the database every time you submit the form.
You just need to reload the form from the database every time you submit the form.

PHP Code:

<?php
$story_select=!empty($_GET["story_select"])?$_GET["story_select"]:"";
if(isset($_POST['edit']))
{
 $category = $_POST["category"];
 $headline = $_POST["headline"];
 $byline = $_POST["byline"];
 $organization = $_POST["organization"];
 $article = $_POST["article"];
 $moreinfurl = $_POST["moreinfurl"];
 $ReleaseDate = $_POST["ReleaseDate"];
 $news_candidate = $_POST["news_candidate"];
 mysql_query("UPDATE news_stories SET category = '$category', headline = '$headline', byline = '$byline', organization = '$organization', moreinfurl = '$moreinfurl', article = '$article', ReleaseDate = '$ReleaseDate', news_candidate = '$news_candidate' WHERE storynum = '$story_select'");
 echo "story was updated!<p>";
 }
if($story_select)
{
 $result = mysql_query("SELECT * FROM news_stories WHERE storynum='$story_select'");
 while($row = mysql_fetch_assoc($result))
 {
  echo  "<h1>Edit Article# ".$story_select."</h1>
         <form action='?story_select=".$story_select."' method='post'>
         Category: <select name='category'>";
  $result2 = mysql_query("SELECT * FROM news_categories";
  while($row2 = mysql_fetch_assoc($result2))
  {
   if($row2["catname"]==$row["category"])
   {
    echo "<option value=\"{$row2['catnum']}\" selected>{$row2['catname']}</option>";
    }
   else
   {
    echo "<option value=\"{$row2['catnum']}\">{$row2['catname']}</option>";
    }
   }
  echo "</select>
        Headline (100 characters max): <input type='text' name='headline' value=\"{$row['headline']}\"> <br>
       <input name='edit' type='submit' id='edit' value='Edit'>
       </form>";
  }
 }
else{header("Location: login.php");}
?>