girl Posted January 3, 2013 Share Posted January 3, 2013 Hello people.Happy New Year.I started new year with learning php, but there is something I need, but still don know. I have a table in database with records, and I want to have links in a page: link 1link 2...link n when I click on link 1, it selects and outputs the record with id1, link 2-with id 2 and so on.Will you please advice me a way to do it.Thanks for advance. Link to comment Share on other sites More sharing options...
kanchatchai Posted January 3, 2013 Share Posted January 3, 2013 (edited) HAPPY NEW YEAR GIRL tableid, title, subject 1. show id as linkHere how to with databasehttp://www.w3schools...ysql_select.asp in file menu.phpwho link by code... echo '<a href="page.php?id='. $row['id'] .'">'.$row['title'].'</a>'; 2. after clicked php will accept with GEThttp://www.w3schools...php/php_get.asp in file page.php sql string...'select * from your_table_name where id='.$_GET['id'] and show record... echo '<p>'.$row['title'].'<br/>'; echo '<p>'.$row['subject'].'<br/>'; Edited January 3, 2013 by kanchatchai Link to comment Share on other sites More sharing options...
girl Posted January 4, 2013 Author Share Posted January 4, 2013 Thank You very much.You know I have tried it, but I don know why when in menu.php I write echo '<a href="page.php?id='. $row['id'] .'">'.$row['title'].'</a>';it echoes only the first record of database, though I have a lot there. Link to comment Share on other sites More sharing options...
birbal Posted January 4, 2013 Share Posted January 4, 2013 Post your code please. Link to comment Share on other sites More sharing options...
girl Posted January 5, 2013 Author Share Posted January 5, 2013 OK, here t is:menu.php <form action="page.php" method="get"> <?php include ("connect-db.php"); session_start(); mysql_select_db("db-name", $con); $id=mysql_query("SELECT * FROM tour_programs") or die($id."<br/><br/>".mysql_error() ); $row= mysql_fetch_array($id); echo '<a href="page.php?id='. $row['id'] .'">'.$row['tour_name'].'</a>'; echo "</br>"; ?> </form> page.php <?phpinclude ("connect-db.php");mysql_select_db("db-name", $con); $var=mysql_query("SELECT * FROM tour_programs where id='".$_GET["id"]."' ");$row= mysql_fetch_array($var);if($var === FALSE) { die(mysql_error()); // TODO: better error handling}echo '<p>'.$row['tour_name'].'<br/>';echo '<p>'.$row['description'].'<br/>'; ?> Link to comment Share on other sites More sharing options...
birbal Posted January 5, 2013 Share Posted January 5, 2013 you need to loop through (eg while loop) the mysql result set to get all the data. now it is just showing the first results. please check the tutorials regarding this. there is example of it. also in page.php you need to check the $var before you use it otherwise it will show errors anyway. 1 Link to comment Share on other sites More sharing options...
girl Posted January 5, 2013 Author Share Posted January 5, 2013 Yeah, thank you. I did just so, and it works.THanks Link to comment Share on other sites More sharing options...
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