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Post Posted: September 3rd, 2006, 3:48 pm

I noticed some problems with following test pages: http://www.foodstyling.be/testindex3.htm [url]http:www.foodstyling.be/navigation3.css[/url] I work on a mac and it seems that the menu on the page does'nt show up right with IE 6 & 7 with Windows, which is weird to reach 85 % of ...

Post Posted: August 4th, 2004, 6:41 am

Hello, I have a question that could be easy to answer, but I can't find the solution I made a little search engine on my site. It has list menu where you can choose a categoriy, then insert a ingredient in a textbox and it works just fine. Now I would like to be able to bypass the list element so yo...

Post Posted: March 23rd, 2004, 10:57 am

The static page is ok, what I want is to link the results I get from the query to go to to the detail pages. in fact I want to make a kind of site menu page with links to get spidered by search engines wich seems not possible with results coming from a search by a form.

Post Posted: March 22nd, 2004, 11:24 pm

thanks, and for the case you still wonder, the language is Dutch

Post Posted: March 22nd, 2004, 4:11 pm

I have a search engine on my site, which is dynamic with php/mysql. Now I wonder if I can generate pages that lists results from a query, for example when I search my database for a keyword, I get results pages with links to dynamic generated pages. for example when I search my database for "chocola...

Post Posted: February 26th, 2004, 4:42 am

Thanks, I got it

Post Posted: February 26th, 2004, 12:07 am

Could you please specify what whatever means with the code i posted? I am realy new in this php stuff. Thanks

Post Posted: February 25th, 2004, 11:37 pm

I have a form with 2 parts. The first is a dynamic generated list box to select a categorie. The second is a text field. Now I want to pass the result from the first part to the second one and i can't find how to make it work. This is the code for the form: [code] <form method="post" ac...

Post Posted: February 14th, 2004, 7:37 am

I try to get the results from this query to next page, using a pagination class, but I get this message and no results. How can I solve this ? This is the code: [code]$total_results = mysql_query_test("SELECT COUNT(recept) AS Num FROM recepten WHERE ingredienten LIKE '%$param1%' AND...

Post Posted: February 12th, 2004, 9:06 am

Thanks,
I thought the query was the problem for my next/previous problem, but it doesn't seem to be so. Can I ask you to check out the code on my next/previous post, because it start to drive me crazy.

Thanks in advance,

foodstyling

Post Posted: February 11th, 2004, 7:49 am

Hi, I tried a new script (Multiple pages from evolt.org) and i get the same results, no results for next page. I tested the queries and they stay empty, but in the browser's window I get this: localhost/test2/results.php?ingredienten=peper&catID=4&page=10&limit=10 this part of the code w...

Post Posted: February 11th, 2004, 12:39 am

I found a piece of code that I don't understand (the $query part after like): $numresults = mysql_query("SELECT * FROM your_table WHERE name LIKE '% ". $query . "%' "); In fact, I would like to use the elements of that query and replace them with my own elements. My query: $numresults = mysql_query ...

Post Posted: February 11th, 2004, 12:06 am

Sorry, I did not look at the lines before pasting them, but ir doesn't seem to help: still no next page!

Post Posted: February 10th, 2004, 4:41 pm

I did, and it gives me a pars error.

Post Posted: February 10th, 2004, 3:44 pm

It seems that i made a mistake , an d the next piece of code is missing. This is the part that doesn't work: [code]// Build Previous Link if($page>1) { $prev = ($page - 1); echo "<a href=\"".$_SERVER['PHP_SELF']."?ingredienten=$param1&ca...
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