Search:

You can also use cURL what would solve the problem with data base access.
It shouldn't be very complicated I think - Depends of the structure of the source page.

Check your PM ;)

As I have said in your different thread, it must be because of value names. Your mobile version of web page is like www.webpage/mobile/link_here.html or something like that I don't remember now.. So...

just type in google search API key. First result:
https://developers.google.com/maps/signup

There choose the Google Maps API JavaScript v3 and do rest step by step ;)

Is your browser has a javascript turned on ? Because it is working on my computer without any issues. So maybe, if the sitemap is working there is a problem with how you named the links. I have...

You have to get a API key from google to use a maps. Than you will have do something like:


<script src="http://maps.googleapis.com/maps/api/js?key=YOUR KEY IS HERE&sensor=FALSE">
</script>
...

It is all depend of what result you want to get. For example you can get data from database by PHP and then show 10 results on the page and make a button like: show more, after click on that user...

I think it is not good idea to store pictures in database, better would be to store just a photo names. However, you have to do it similar to that:




if(isset($_POST['button_pic'])){
...

<div id="drop-down">
<form name="form">
<SELECT name="guidelinks" onChange="window.location=document.form.guidelinks.options[document.form.guidelinks.selectedIndex].value">
...

What you mean by visualization ? Graphic charts or what ?

Hi,
you have to create html form and then pass the data from the to your query,
It will look similar to this:
HTML FORM



<form action="" method="POST">
<select name="field">
Choose a field:

Sorry, i dont know how to edit the post, but I have made a wrong there, it should be:

SELECT SUBSTRING(mytextarea,1,200) FROM `my_table` WHERE `user_id`=[user_id]

Hi,
just use substring function in you query, it will look like:

SELECT SUBSTRING(mytextarea,1,200) as `mytextarea` FROM `my_table` WHERE `user_id`=[user_id]

It should be something like that,...