WDjoe

It's working now. I had to go line by line by line and follow every step of the code in extreme detail and analyze after comma and semi-colon and so forth. I had an "else" statement in my code where if any records were found to display the table and if no records were found to display a message. Unfortunately for me the message was nothing ( I had not yet written the message I wanted to display) and I had the "if" statement reversed, so when I ran the query it found records but because the "if" statement was reversed nothing was shown on the webpage. No problem, only spend 3 days trying to get it to work, lots of coffee and aspirin and a huge learning experience for me! ...lol Thank you for your help HartleySan for helping me out.

Hi, yes I did notice the comma and removed the comma at the end of the query (no difference). $q = "SELECT *, DATE_FORMAT(payment_date, '%M %D, %Y') AS formatted_date FROM item_payments WHERE username = 'benm' AND item_name='Invitations' " ; And this is what the database looks like: CREATE TABLE IF NOT EXISTS `item_payments` ( `payment_id` int(11) NOT NULL AUTO_INCREMENT, `item_name` varchar(30) DEFAULT NULL, `budget_id` int(50) DEFAULT NULL, `amount` decimal(10,2) DEFAULT NULL, `payment_date` date DEFAULT NULL, `username` varchar(50) DEFAULT NULL, `notes` varchar(200) NOT NULL, `invoice_number` varchar(50) NOT NULL, PRIMARY KEY (`payment_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=227 ; -- -- Dumping data for table `item_payments` -- INSERT INTO `item_payments` (`payment_id`, `item_name`, `budget_id`, `amount`, `payment_date`, `username`, `notes`, `invoice_number`) VALUES (226, 'Invitations', 728, '32.00', '2015-01-01', 'benm', '', 'abc123'), (225, 'Invitations', 728, '36.77', '2015-01-08', 'benm', ' test notes in invitations', 'test invoice number varchar ABCD12345');

$item_name = 'Invitations'; $q = "SELECT *, DATE_FORMAT(payment_date, '%M %D, %Y') AS formatted_date FROM item_payments WHERE username='$user' AND item_name='$item_name',";

Tried it and I get the following. But I still don't see what is wrong with it: #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$item_name="Invitations" ' at line 1

Ok, I have pounded my head long enough (literally hours working on this) and I just can't figure this one out. I am hoping someone can solve this issue for me. Hopefully I can explain myself well enough to be understood. I have a database table ("items") with a column named "item_name", inside this column I have "Invitations" as a category name. I am doing a simple query like this; -------------------------------------------------------------------------------------------------------------------------- $category = "Invitations" Query ="SELECT * FROM items WHERE username='$user' AND item_name='$category' "; -------------------------------------------------------------------------------------------------------------------------- The problem is it will not return the rows where item_name = "Invitations". I checked and rechecked making sure all letters are correct and that there are no spacing , etc. If I edit that column and replace the text ("Invitations") with numbers, i.e."1234" and change $category ="1234" it returns the rows exactly as I would expect. I tried changing the column type to VARCHAR, CHAR, TEXT. So what's going on?? Any ideas?

I figured it out and would like to share my solution. I knew it had to be a sessions issue but all the "session_start()" was correctly placed on all my pages. But in Chapter 12 there is a mention that you can put the "session_start()" in the php.ini folder. That's what I did. Just make a folder and name it php.ini and inside it just place "session_start ()" (without the quotation marks). Everything now works as it should. :-)

Hi, I am implementing the login script in chapter 12 but I don't understand something and hope someone can give me an answer. Every time I enter my credentials into the login form and press "Login" it automatically transfers me to my index.php page. Can anyone help please? Thanks

Hi every, can someone help me out with this problem that I have concerning zero's showing up in fields that I want to show blank if there is no value being called from the database. I read various topics on this issue but I just don't get it. To keep it simple lets say I have a form that has an optional FAX telephone number. If someone that fills in the form does not have a FAX number he or she will omit this field, but when the form is summited to the database a zero gets placed into the fax_telephone field in the database. I would like it to have nothing placed in the database. The way the fax number shows now has the fax number as "0", I would like it not to show anything at all. This is a very simplified query with everything else removed but the fax telephone field that I am using: $q = "INSERT INTO users (fax_telephone) VALUES ('$fax_telephone')";/code] On my database I have set up like this: [b]Field:[/b] fax_telephone [b]Type[/b]:INT [b]Lenngth[/b]:10 [b]Default:[/b] NULL [b]Collation:[/b] utf8_general_ci [b]Attributes:[/b] [b]Null:[/b] (CHECKED) Can anyone tell me what I am have to do to correct this? Many thanks in advance.

In the top of the footer.html page of the CMS script what file is the footer.html referring to that is listed as <!--<?php include ('./includes/comments.inc.php'); ?>--> I don't see any explanation about this line of code in the book nor is it in the includes folder. If I erase this line the pages gets scrambled so I would image it is important.

Hi Stuart, thank you for your help. I have tried different combinations with no luck. I noticed something though, I only had one table selected (categories) I think I would need also to select the other table (pages) so a comparison could be made. But when I did add the other table I got multiple choices of the same item in the drop down menu. That is I have three of everything in the drop-down menu. And even that did not get the category selected. Here is the updated code I used thinking it would work but didn't. <select name="category_id"<?php if (array_key_exists('category', $add_page_errors)) echo ' class="error"'; ?>> <option>Select One</option> <?php // Retrieve all the categories and add to the pull-down menu: $q = "SELECT categories.id, categories.category, pages.category_id FROM categories, pages ORDER BY category ASC"; $r = mysqli_query ($dbc, $q); while ($row = mysqli_fetch_array ($r, MYSQLI_NUM)) { echo "<option value=\"$row[0]\""; // Check for stickyness: if (isset($_POST['categories.id']) && ($_POST['pages.category_id'] == $row[0]) ) echo ' selected="selected"'; echo ">$row[1]</option>\n"; } ?> </select> This is how the two table are set up : TABLE 'categories' ('id', 'category') TABLE 'pages' (id, category_id, title, description, content, date_created) Any thoughts? thanks again.

I am wondering if anyone can help me out with this. On the CMS script I now have the capability to edit any page or category. I have a drop-down menu where the administrator can change the category that the page will be placed under. My problem is when I edit a page I would like the category drop-down menu to have the category of the page that it is under (from the database) already selected. I copied the code from the "add_page.php" file to my new "edit_page.php" file but when I go to edit the page the category is not selected only the text "Select One" is. Though I am able to click on a category from the drop-down menu to choose a category I would like the one that page is classified under to show automatically. If I forget to enter a category when I edit a page a "0" is automatically placed under the "category_id" column in the database and the page is unable to be viewed. Here is the drop-down menu code that retrieves the categories from the database. <p><label for="category"><strong>Category</strong></label><br /> <select name="category_id"<?php if (array_key_exists('category', $add_page_errors)) echo ' class="error"'; ?>> <option>Select One</option> <?php // Retrieve all the categories and add to the pull-down menu: $q = "SELECT id, category FROM categories ORDER BY category ASC"; $r = mysqli_query ($dbc, $q); while ($row = mysqli_fetch_array ($r, MYSQLI_NUM)) { echo "<option value=\"$row[0]\""; // Check for stickyness: if (isset($_POST['category_id']) ) echo ' selected="selected"'; echo ">$row[1]</option>\n"; } ?> </select><?php if (array_key_exists('category', $add_page_errors)) echo ' <span class="error">' . $add_page_errors['category'] . '</span>'; ?></p> Many thanks for any help in advance.

Hi Larry, thank you for your reply. That did the trick :-)

I am working to on the first script in the book "Effortless E-Commerce with PHP and MySQL" and have it working the way I want it to work.I am customizing it to fit my particular needs. I have added pages where the administrator can also add or delete pages and categories from the database. I also made it where only the administrator can add the users to the database (no one can register). There is only one problem I am having and it's driving me nuts and I am hoping that someone can help me. First, I have added the TinyMCE html editor per instructions in the book ( I love this feature) and it works as it should, the problem I have is how do I edit a page that is already in the database so when I call it to be edited the original text from this page is automatically placed in the TinyMCE editor for easy editing?