carlostau

Hello there. Normally your hosting should provide you with the FTP login and password. And your database connection too Most of the sites use PHPMyAdmin. A simple tool that lets you create a database or just paste an SQL dump. An SQL dump is achieved by using whatever GUI you use to run your queries locally, once you have the .sql file, you can open it (it is a text file), copy its entire content and paste in an SQL window within PHPMYAdmin. that will create an exact duplicate of your datatbase on the server. I created a site on one of the many free php servers wondering around (php1h.com) you might want to give it a try to see how it works Hope it helps! Demian [quote author=mysty link=topic=104099.msg415071#msg415071 date=1155494867] I have developed most of my first dynamic site using my local computer as my server.  I have gotten a hosting package and am confused as to how this works with this new remote server (my host ISP). When in manage sites, I think I have the local info correct.  1) The remote info and testing server are both set to FTP access.  Is this correct?  What are the other settings supposed to be? 2) How do I get my database uploaded to the server?  Am I supposed to create a 'db' folder on my site and then upload from there?  Sorry, this is not clear to me. [/quote]

HI! Select your Dropdown, click on ListValues and add one Item Label called "All" or whatever you want to call it, now for the "All" thing to work, you have to assign a '%' Value (without the quotes) that way, when the form passes it, I will pass it as % and will work as a wildcard on your query. Lastly you have to modify the recordset query in your results page to use 'like' instead of '=' EXAMPLE: Select trainer, colour from your_table where trainer='trainer' and colour like 'colour' Both 'triner' and 'colour' will be replaced with what you had in the form Works like a charm! Good luck Demian [quote author=mysq_hell link=topic=84179.msg339614#msg339614 date=1138144253] Hi, Managed to Solve Problem One... Added Static menu item and Blanked out the Select value equal to Box.. Can anyone help with the SQL for problem 2, The Static Menu Options have a value of 0, Which Means if the Form is Submitted without Changing the Default value Nothing Happens... Paul. [/quote]

I have noticed that you have like a duplicated echo row there (highlghted in bold now). My row goes like: <a href="updatelist.php?recordID=<?php echo $row_rs_home['id']; ?>" without having anything else. It might be a Dreamweaver bug that left the line with two echoes. I would keep just the first part like the one I show here Then on the edit page all you need to do is make sure to create a Recordset that has the URL.Variable and that the name of that variable is 'recordID' and not 'pagename_con' Hope it helps! Demian [quote author=Theora link=topic=103745.msg413358#msg413358 date=1155231768] I'm trying to set up something that should be relatively simple; one page where the user selects the item they would like to update which passes them on to a second page where the record they selected is available for editing. I attempted to follow the instructions at livedocs.macromedia.com, but the second page isn't bringing up the selected record. This is what I have: Page 1: [code]<table width="300" border="0" align="center" cellpadding="5">   <tr>     <td>Which page would you like to edit? </td>   </tr>   <?php do { ?>     <tr>       <[b]td><a href="edit.php?recordID=<?php echo $row_rsContent['pagename_con']; ?>"><?php echo $row_rsContent['pagename_con']; ?></a></td>     </tr>[/b]     <?php } while ($row_rsContent = mysql_fetch_assoc($rsContent)); ?> </table>[/code] This part seems to work fine. When the next page comes up the url ends with ?recordID=main (for example). The code for the second page: [code] <?php require_once('Connections/content.php'); ?> <?php $colname_rsEdit = "-1"; if (isset($_GET['id'])) {   $colname_rsEdit = (get_magic_quotes_gpc()) ? $_GET['id'] : addslashes($_GET['id']); } mysql_select_db($database_content, $content); $query_rsEdit = sprintf("SELECT primary_con, pagename_con, pagecontent_con FROM content WHERE pagename_con = '%s'", $colname_rsEdit); $rsEdit = mysql_query($query_rsEdit, $content) or die(mysql_error()); $row_rsEdit = mysql_fetch_assoc($rsEdit); $totalRows_rsEdit = mysql_num_rows($rsEdit); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1" /> <title>Untitled Document</title> </head> <body> <form id="updating" name="updating" method="post" action="">   <table width="500" border="0" align="center" cellpadding="15" cellspacing="0">     <tr>       <td>Updating <?php echo ucfirst($row_rsEdit['pagename_con']); ?></td>     </tr>     <tr>       <td><textarea name="textarea" cols="55" rows="20"><?php echo $row_rsEdit['pagecontent_con']; ?></textarea></td>     </tr>     <tr>       <td><input name="Update" type="submit" id="Update" value="Submit" /></td>     </tr>   </table> </form> </body> </html> <?php mysql_free_result($rsEdit); ?> [/code] It appears (to my untrained eye) that there is a query to select three things where the page name equals the variable passed from the previous page. It also appears that the large text area is trying to call in the contents of "pagecontent." I've tried both the quick (using the wizard) and step-by-step methods given on the Macromedia site and get the same results. Can you see what I'm doing incorrectly here? Thanks! [/quote]