Jay2391 Posted October 16, 2006 Share Posted October 16, 2006 I have a database and I have a relationship from one DB to another...now I want to put a drop down menu on my PHP page to select from that relationship...does anyone has some documentation on this... ??? Quote Link to comment Share on other sites More sharing options...
printf Posted October 16, 2006 Share Posted October 16, 2006 Are they different databases or just different tables in a single database?me! Quote Link to comment Share on other sites More sharing options...
Jay2391 Posted October 16, 2006 Author Share Posted October 16, 2006 one data base different tables so i guess the relationship is between tables... Quote Link to comment Share on other sites More sharing options...
craygo Posted October 16, 2006 Share Posted October 16, 2006 A little database structure would be cool and what fields you want to use.Ray Quote Link to comment Share on other sites More sharing options...
gijew Posted October 16, 2006 Share Posted October 16, 2006 I don't know your table structure but off the top of my head as long as you have a foreign key between the two (unique number that identifies the parent record in the child table) the query could be done several ways but most common would be looping through the two tables like this...[code]<?php$parent_query = mysql_query("SELECT id, name FROM parent_table ORDER BY name ASC");while ($parent= mysql_fetch_array($parent_query)) { $child_query = mysql_query("SELECT id, name FROM child_table WHERE foreign_key = $row[id] ORDER BY name ASC"); while ($child = mysql_fetch_array($child_query)) { // display your code }}?>[/code] Quote Link to comment Share on other sites More sharing options...
Jay2391 Posted October 17, 2006 Author Share Posted October 17, 2006 This is what i created .... what i need is that the states fields will show a dropdown menu and chosse from a table i have with all the states ... the table name is "states"hope this helps<html><head><title>Add Record</title></head><body><?php$user="root";$pass="12345";$self=$_SERVER['PHP_SELF'];$userid=$_POST['userid'];$name=$_POST['name'];$last=$_POST['last'];$city=$_POST['city'];$state=$_POST['state'];$country=$_POST['country'];$email=$_POST['email'];$gender=$_POST['gender'];$nickname=$_POST['nickname'];$age=$_POST['age'];?><form action="<?php echo( $self );?>" method="post">User ID: <input type="text" name="userid" size"30"><br>Name: <input type="text" name="name" size"30"><br>Last: <input type="text" name="last" size"30"><br>City: <input type="text" name="city" size"30"><br>State: <input type="text" name="state" size"30"><br>Country: <input type="text" name="country" size"30"><br>Email: <input type="text" name="email" size"30"><br>Gender: <input type="text" name="gender" size"30"><br>Nickname: <input type="text" name="nickname" size"30"><br>Age: <input type="text" name="age" size"30"><br><input type="submit" value="Submit"><br></form> <?php if($userid and $name and $last and $city and $state and $country and $email and $gender and $nickname and $age){ $dbh=mysql_connect("localhost", $user, $pass) or die ("Err:Conn"); if($dbh){ echo ("Congrats!!!"); $rs=mysql_select_db("natgal", $dbh) or die ("Err:Db"); $sql="INSERT INTO users( userid, name, last, city, state, country, email, gender, nickname, age) VALUES ($userid, \"$name\", \"$last\", \"$city\", \"$state\", \"$country\", \"$email\", \"$gender\", \"$nickname\", \"$age\")"; $rs=mysql_query($sql, $dbh); if($rs){ echo ("Record Added!!!"); } } }?></body></html> Quote Link to comment Share on other sites More sharing options...
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