sops21

One thing I noticed is that you have { on line 5 and } on line 8. I don't know if these are supposed to be part of a if statement or something but they should not be there alone. Are you sure that your query is returning anything? You …

You can do something like this User accesses [url]http://www.somewebsite.com/category/baseball[/url] .htaccess [CODE] RewriteEngine on RewriteRule ^ category/([^/.]*)/?([^/.]*)/?$ /displaycategory.php?category =$1 [L] [/code] Then on the displaycategory.php page do a simple get [code] $category = $_GET['category']; [/code] You could then use $category to lookup the information that you need to display. In my …

You are never getting the file name. To get the file name you need to use [CODE=php]$filename = $_FILES["Filedata"]["name"];[/CODE]

How do you know they are not using Oauth?

Try this: [code] RewriteEngine On rewriteRule ^([a-zA-Z0-9_-]+)$ index.php?m=$1 [/code]

I think your going to have to better explain what you are trying to do. You have IMAGE1.jpg and you want to make that JPGIMAGE1.jpg?

Zencart does not have this feature by default and there is no such plugin for amazon listed.

Do you have access to cron on the server? With cron you can schedule it execute a file every 12hrs. That file can run the mysql query that you want.

[CODE][/CODE]If you want to keep the same method that you are using for adding a record, you can do something like this: [CODE=php] echo "<tr><td>" . $row['id'] . "</td><td>" . $row['authorname'] . "</td><td>" . $row['publishyear'] . "</td><td>" . $row['booktitle'] . "</td><td>" . $row['edition'] . "</td><td>" . $row['place'] . "</td><td>" . …

Did you modify anything before you got this error?

Does is have to by using PHP? Depending on the webhost you might be able to setup a cron job to do that backup... Create a file called backup [CODE] #!/bin/sh mysqldump -uusername -ppassword --opt tablename > /path/tofile/location/youwant/mysqldumpfile.sql [/CODE] And then schedule cron to run and execute the script [CODE]0 …

I've used this before. $id is the the value of the primary key of the row that you just inserted. [CODE=php]$link = mysql_connect(DB_HOST, USER, PASSWORD) or die(mysql_error()); mysql_select_db(DATBASENAME) or die(mysql_error()); $query = "query here"; mysql_query($query) $id = mysql_insert_id($link); [/CODE]

I would do it all in wordpress. You can create a page in wordpress and then set that page as the homepage. This way you don't have to create a home page(html, css) and then also create the theme for the blog.

I have a select box with information about people in it. When you click on their name in the select box, I'm trying to display information (Name, Address, Height, Weight) about them inside of a <div>. Here is what I have so far: [CODE]<form> <SELECT NAME="list1" MULTIPLE SIZE=20 class="set_width" id='developer'> …