Redlightpacket

I have looked up on the web for some good tutorials, so I haven't really found any "SIMPLE ONES". I'm looking for a good simple tutorial on how to make your page continue and continually load data when you scroll down the page. Will need ajax to pull some data from the database. I looked at 9Lessions tutorial on page load data jquery, but it's pretty complex. I tried to make it work in an example and it didn't work. But, all I'm looking for is a very simple tutorial on how to make a page load data jquery. Thanks..............

Any help would be nice.....really looking for the holy grail of jquery debugging.......like the best html editor is Adobe Dreamweaver.......A compiler would be nice for Jquery if it was easy to do....Thanks for the help, Chris

Try this, this way works for me $query = "SELECT * FROM Cards"; $result = mysql_query($query) or die("Query failed: " . mysql_error()); while ($row = mysql_fetch_assoc($result)) { //Prints the id numbers in the table. The id numbers go from 1 and forever //You will need to set a primary key and it will be the primary id number in your mysql table echo $row['id']; //Prints row numbers }

Not really sure what you are talking about a little bit. But this is where I would start. Hopefuly this might help you some way. <?PHP $query = "SELECT * FROM Cards"; $result = mysql_query($query) or die("Query failed: " . mysql_error()); while ($row = mysql_fetch_assoc($result)) { $Cards_ID[1] = $row['Cards_ID']; $Cards_MemberID[1] = $row['Cards_MemberID']; $Cards_Card[1] = $row['Cards_Card']; $Cards_DateTime[1] = $row['Cards_DateTime']; } Array ( [889557275612] => Array ( [0] => Array ( [Cards_ID] => 51 [Cards_MemberID] => 889557275612 [Cards_Card] => 308292114653 [Cards_DateTime] => 1969-12-31 18:00:00 ) [1] => Array ( [Cards_ID] => 51 [Cards_MemberID] => 889557275612 [Cards_Card] => 308292114653 [Cards_DateTime] => 1969-12-31 18:00:00 ) ?>

Excuse me, GET[] ---------->>> GET

I believe you need to check out the GET[].

I tell you this, I had a problem with my broswer not doing sessions right and somebody on here told me to change your session settings in your php.ini file. I think you need to enable sessions and cookies settings.

I'll try that, but I'm not sure it will work, but will give it a try any way

I'm trying to make a profile page like on social networks and I can't get the id=333334444 to stay at the top of the page or in the query string when I send post a message. I want the id=3333334444 to stay in the query string all the time when i'm on the certain user profile page, for example like when I post a message on the profile page. This stuff has been a little tricky to figure out. But I need a little documentation on this subject or something that will help me. Thanks a lot, ~~~~Chris Clark

I am really puzzled how to work with these. I need a good information source that covers this topic broadly. I know the basics how query strings work and how to use them, but I need more information how to use them.

Here is my error message I get. 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 '[1] = 'listen'' at line 1 Some how it won't write data in the first and next and next fields in my database table. //Trying to make this data go into the database table. //SET $column=array("listen","date","message"); $value=array("listen","date","message"); //WHERE $column2=array("video","listen","date"); $value2=array("video","listen","date"); $counts = 1; while($counts < 4) { $counts = 0 + $counts; $n = 1; $n2 = 1; $n = 0 + $n; $n2= 0 + $n2; $result = mysql_query("UPDATE _0_sermon_data_file SET $column[$n] = '$value[$n]' WHERE column2[$n2] = '$value2[$n2]'") or die(mysql_error()); } ///////////////////////////////////////////////////////////////////////////////////////////////////////////

I need this script to be basic so I can understand it. Thanks, Chris

Some how the line below will not work. I'm trying to make the current date input into the database when I insert data into the database. $query = "CREATE TABLE IF NOT EXISTS _0_calendar_data(id SMALLINT NOT NULL AUTO_INCREMENT PRIMARY KEY, news TEXT(1200), date INT(30), date2 INT(30))"; //The one right here needs attention $result = "INSERT INTO _0_calendar_data(id, news, date, date2) VALUES (NULL, '$news', 'YEAR: Auto CURDATE()', CURDATE())";

I really need a book that has some pagnation examples in it. I can't figure out how to get the pagnation examples working on the internet. I would highly appreciate anybody's help on this matter. Thanks, Chris

Ok, Please read more into this question than it is acually it is. What I want to do is make my sentence on the screen break line by line in a table when I have data to display. I have written a little code to try to do this, but it seems to hard and complicated. What I was wondering does cakePHP provide a function that does this. If not can you give me some code or another framework that could cut out all of the work to reinvent the wheel. I would dearly appreciate your help.

I want to build a netvibes like web page, but I need some examples or some help to get me started. Anybody have any advice how I can do this?

I would like to know how they made the fql. Did they use XML - rpc ? Any body Have any clues how this was done

I figured it out, but thanks for the time to look at my post.

I need some help trying to get my UPDATE to work the right way for this particular piece of code. I am really confused here. $array = array('p', $user); $com = join("", $array); //test me echo $com; $query = "CREATE TABLE IF NOT EXISTS $com(id SMALLINT NOT NULL AUTO_INCREMENT PRIMARY KEY, bio VARCHAR(450), idea VARCHAR(450), location VARCHAR(50))"; $table1 = mysql_query($query) or die("Query failed: ".mysql_error()); //$query = "UPDATE $com SET idea = $idea, bio = $bio, location = $location, WHERE id = "1""; //Here is my update statement, when I run the script, it says there is an error in this line. $result = mysql_query("UPDATE $com SET bio = '$bio', idea = '$idea' location = '$location' WHERE id = '1'") or die(mysql_error());

I am trying to learn PHP. I am starting simple. I'm wondering what kinds of simple projects do I need to do to build my skills in PHP. I know you need to learn how to build a forum. But what other kinds of simple projects could help me learn php better.

I am looking for a place to find some web application code that is simple so it can help me get a better understanding how to make web apps. Thanks, Chris Clark

I changed this around a little bit and it worked I changed file:// to http:// echo "http://localhost/".$row['picture']."\"></td><td>"; but thanks for the help everybody

I'll try what yall are talking about

I'm really not sure what you are doing, but if you are a beginner programming in PHP and have a pc. I suggest you get WAMP to run your php scripts on. I hope this helps.

Here is my code right here. I need some help with displaying the image in the code. I might be having a bug. $query = "SELECT * FROM picture_tb"; $result = mysql_query($query) or die("Query failed: " . mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo'<table border="1"> <tr> <td>';echo $row["id"];echo'</td> <td><IMG SRC="';echo'file:///C:/wamp/www/';$row['picture'];echo'">';echo'</td> <td>';echo $row["title"]; echo'</td> <td>';echo $row["desc"]; echo'</td> </tr> </table> '; } ////////////////////////////////////////////////////////code]