[RESOLVED] Radio Box Not Populating

cash09

I'm trying to create a drop down box that is populated with authors that are already in the database but the drop down box is not populating?

Here's the code

<?php 
//This page allows the administrator to add a book.

// Sets the page title and includes the PHP header.
$page_title = 'Add Book';
include ('./includes/header.php');

if (isset($_POST['submitted'])) { // Check if the form has been submitted.

require_once ('/mysqli_connect.php'); // Connect to the database.

} else { // Display the form.
	?>

<form enctype="multipart/form-data" action="test.php" method="post">

	<input type="hidden" name="MAX_FILE_SIZE" value="524288">

	<fieldset><legend>Fill out the form to add a book to the catalog:</legend>

    <p><b>Author:</b> 
	<p><input type="radio" name="author" value="existing" /> Existing
	<select name="existing"><option>Select One</option>
	<?php // Retrieve all the artists and add to the pull-down menu.
	$query = "SELECT author_id, CONCAT_WS(' ', first_name, last_name) AS name FROM author ORDER BY last_name, first_name ASC";		
	$result = mysqli_query ($dbc, $query);
	while ($row = mysqli_fetch_array ($result, MYSQLI_ASSOC)) {
		echo "<option value=\"{$row['author_id']}\">{$row['name']}</option>\n";
	}
	mysqli_close($dbc); // Close the database connection.
	?>
	</select></p>

</form>
<?php
}// End of main conditional.

// Include the PHP footer.
include ('./includes/footer.php');
?>

Any help would be greatly appreciated.

bradgrafelman

The only thing I see you trying to populate using information from a database is a SELECT dropdown, not a radio box.

cash09

Ooops... Yeah that is what i'm trying to populate. Sorry!

Pikachu2000

And what errors are you getting?

cash09

No errors from the code at all... The drop down box just doesn't have any options in it to select like it should do?

halojoy

As you can see here
http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_concat-ws
you should have 'single quotes' around all 3 arguments in CONCAT_WS()
Not only around separator: ' ' as you have.

Now, if you would use some MYSQLI ERROR CHECKING after each query
at least while in developing
then you would easily catch such syntax errors by yourself 😉
http://php.net/manual/en/mysqli.error.php

I hope I was right and I have not said TOO much 🙂
Besides, you write very good code!
It is pleasure to try to assist you.

        <?php 
		// Retrieve all the artists and add to the pull-down menu.

	// we put single quotes around all arguments in CONCAT_WS()
    $query = "SELECT author_id, CONCAT_WS(' ', 'first_name', 'last_name') 
			AS name FROM author ORDER BY last_name, first_name ASC";        
    $result = mysqli_query ($dbc, $query);

    // if the query fails and $result is = FALSE
    // let's see what MySQL have to report about it
    if(!$result){
    	 echo '<br />'. $query .'<br />'. mysqli_error($dbc) .'<br />';
    }

    while ($row = mysqli_fetch_array ($result, MYSQLI_ASSOC)) {
        echo "<option value=\"{$row['author_id']}\">{$row['name']}</option>\n";
    }
    mysqli_close($dbc); // Close the database connection.
    ?>

cash09

halojoy;10949217 wrote:
As you can see here
http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_concat-ws
you should have 'single quotes' around all 3 arguments in CONCAT_WS()
Not only around separator: ' ' as you have.

Now, if you would use some MYSQLI ERROR CHECKING after each query
at least while in developing
then you would easily catch such syntax errors by yourself 😉
http://php.net/manual/en/mysqli.error.php

I hope I was right and I have not said TOO much 🙂
Besides, you write very good code!
It is pleasure to try to assist you.
 <?php 
		// Retrieve all the artists and add to the pull-down menu.

	// we put single quotes around all arguments in CONCAT_WS()
 $query = "SELECT author_id, CONCAT_WS(' ', 'first_name', 'last_name') 
			AS name FROM author ORDER BY last_name, first_name ASC"; 
 $result = mysqli_query ($dbc, $query);

 // if the query fails and $result is = FALSE
 // let's see what MySQL have to report about it
 if(!$result){
 	 echo ' '. $query .' '. mysqli_error($dbc) .' ';
 }

 while ($row = mysqli_fetch_array ($result, MYSQLI_ASSOC)) {
 echo "<option value=\"{$row['author_id']}\">{$row['name']}</option>\n";
 }
 mysqli_close($dbc); // Close the database connection.
 ?>

There's still no error message being displayed even with the error checking? The sql query works in phpadmin fine as well?

bradgrafelman

halojoy;10949217 wrote:
As you can see here
http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_concat-ws
you should have 'single quotes' around all 3 arguments in CONCAT_WS()

Looks like first_name and last_name are columns in the table... in which case no, he shouldn't use single quotes around all 3 arguments - only the first.

EDIT:

cash09 wrote:
The sql query works in phpadmin fine as well?

Did you echo out the SQL query from within the PHP script to examine it visually? Did you copy and paste the query into something like phpMyAdmin to see if any rows were returned? If not... how do you know the query works?

Weedpacket

Just for the sake of argument, does using

printf('<option value="&#37;s">%s</option>'."\n", $row['author_id'], $row['name']);

make any difference (just to cut down on the amount of quote-nesting going on)?

cash09

bradgrafelman;10949220 wrote:
Looks like first_name and last_name are columns in the table... in which case no, he shouldn't use single quotes around all 3 arguments - only the first.

EDIT: Did you echo out the SQL query from within the PHP script to examine it visually? Did you copy and paste the query into something like phpMyAdmin to see if any rows were returned? If not... how do you know the query works?

When I echo out the query it displays the same query as is in the code?

Also I have tried the query in phpMyAdmin and it returns the author_id and also the authors first and last names as one field as it should? I can't see why its not populating the drop down box?

cash09

Weedpacket;10949223 wrote:
Just for the sake of argument, does using
printf('<option value="%s">%s</option>'."\n", $row['author_id'], $row['name']); 
make any difference (just to cut down on the amount of quote-nesting going on)?

This doesn't seem to affect the script at all... The drop down box stays empty?

bradgrafelman

After the if() statement halojoy suggested, try adding this:

echo "\n<!-- # of rows found: " . mysqli_num_rows($result) . "-->\n";

Then, view the HTML source of the page in your browser. Do you see the comment? If so, how many rows does it say were found? Is there anything else between the <select>..</select> tags?

cash09

bradgrafelman;10949288 wrote:
After the if() statement halojoy suggested, try adding this:
echo "\n\n";
Then, view the HTML source of the page in your browser. Do you see the comment? If so, how many rows does it say were found? Is there anything else between the <select>..</select> tags?

Okay I put that in as well so now my code looks like this:

<?php
//This page allows the administrator to add a book.

// Sets the page title and includes the PHP header.
$page_title = 'Add Book';
include ('./includes/header.php');

if (isset($_POST['submitted'])) { // Check if the form has been submitted.

require_once ('/mysqli_connect.php'); // Connect to the database.

} else { // Display the form.
    ?>

<form enctype="multipart/form-data" action="test.php" method="post">

    <input type="hidden" name="MAX_FILE_SIZE" value="524288">

    <fieldset><legend>Fill out the form to add a book to the catalog:</legend>

    <p><b>Author:</b>
    <p><input type="radio" name="author" value="existing" /> Existing
    <select name="existing"><option>Select One</option>
    <?php
    // Retrieve all the artists and add to the pull-down menu.

    // we put single quotes around all arguments in CONCAT_WS()
    $query = "SELECT author_id, CONCAT_WS(' ', 'first_name', 'last_name')
            AS name FROM author ORDER BY last_name, first_name ASC";        
    $result = mysqli_query ($dbc, $query);

    // if the query fails and $result is = FALSE
    // let's see what MySQL have to report about it
    if(!$result){
         echo '<br />'. $query .'<br />'. mysqli_error($dbc) .'<br />';
    }
	 echo "\n<!-- # of rows found: " . mysqli_num_rows($result) . "-->\n"; 

    while ($row = mysqli_fetch_array ($result, MYSQLI_ASSOC)) {
        echo "<option value=\"{$row['author_id']}\">{$row['name']}</option>\n";
    }
    mysqli_close($dbc); // Close the database connection.
    ?>
    </select></p>

</form>
<?php
}// End of main conditional.

// Include the PHP footer.
include ('./includes/footer.php');
?>

When I view the html source of the page it shows these errors:

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\xampplite\htdocs\website\test.php on line 30

Warning: mysqli_error() expects parameter 1 to be mysqli, null given in C:\xampplite\htdocs\website\test.php on line 35 
 SELECT author_id, CONCAT_WS(' ', 'first_name', 'last_name')
AS name FROM author ORDER BY last_name, first_name ASC 
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in C:\xampplite\htdocs\website\test.php on line 37

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in C:\xampplite\htdocs\website\test.php on line 39 
 
Warning: mysqli_close() expects parameter 1 to be mysqli, null given in C:\xampplite\htdocs\website\test.php on line 42

Hope this is what you meant?

bradgrafelman

Well there's your problem - as the error message states, $dbc is null (or not defined).

cash09

I don't understand? Its the same mysqli_connect script that I've used to connect to the mysql database for all the other scripts I have and they have never come back as null or not defined?

bradgrafelman

But you never include it in this script:

if (isset($_POST['submitted'])) { // Check if the form has been submitted. 

require_once ('/mysqli_connect.php'); // Connect to the database. 

} else { // Display the form.

You only include the file if $_POST['submitted'] is set.

cash09

I can not thank you enough bradgrafelman. Silly mistakes always let me down. Thank you for being patient with me 🙂

bradgrafelman

Does that mean it's working now? If so, don't forget to mark this thread resolved (using the link on the Thread Tools menu).

cash09

Yeah its working now. I'll mark the thread resolved now. Thanks again!