shenagsandnags Posted February 7, 2011 Share Posted February 7, 2011 hey all. i previously posted about how i lost all my data so im trying to recreate my form again. after not many replies from my desperate call for help i tried to figure it out on my own, i spent hours and hours going through this forum and others but im just running into brick walls because of my minor knowledge of php (though i do have to say i am picking it up a little!) my form is short and simple so this shouldnt be too hard but this is what i have come up with. my tables ---------------- Movies: ID(P_id),Title,Category (F_K),URL Categories: ID (P_Id),Category All i am needing my form to do is enter records into my movies table, therefore it goes like this: Title: "textbox" URL: "textbox" Category: "dropdown menu" "submit" now the thing here is that the dropdown menu needs to be populated with data from Category.Categories table so i can add that info into category in the movies table. so pretty much it will be sending the ID from Categories table into Category in the Movies and yes they both have the same datatype. so here is what i have so far. my form.php <html> <form id="form1" name="Update" method="post" action="add.php"> <label> First Name: <input type="text" name="Title" id="textfield" /> </label> <br /> <label> Last Name: <input type="text" name="URL" id="textfield2" /> </label> <select name='Category'> <?php $con = mysql_connect("localhost", "root", ""); if (!$con) { die('Could not connect to DB: ' . mysql_error() ); } mysql_select_db ("mydb", $con); $query = "SELECT Category from Catgories"; $result = mysql_query($query) OR DIE ("There was an error .mysql_error()); while ($line = mysql_fetch_array($result,MYSQL_ASSOC)){ $categoryid = $line["playerid"]; $category = $line["category"]; echo "<option value='$categoryid'>$category</option>\n"; } php?> </select> <input name="" type="submit" value="send" /> </form> </html> and then my add.php <?php $con = mysql_connect("localhost", "root", ""); if (!$con) { die('Could not connect to DB: ' . mysql_error() ); } mysql_select_db ("mydb", $con); $sql="INSERT INTO movies (title, url) VALUES ('$_POST[title]','$_POST[url]')"; if (!mysql_query($sql,$con)) { die ('Error: ' . mysql_error()); } echo "Record added"; so first things first, although i have tested the code its throwing me a variable error i think on line 30 so i figured that instead of focusing on just the error i should just have everything looked over since i dont have anyone i can turn to about this stuff to make sure i didnt forget any " or { etc.. i am also not sure if the while loop is even needed in form.php, i kind of taylored this code to my own needs. THEN i have the add.php, granted that by some miracle that the form.php is correct i dont know what else i would need to add to it for my dropdown. i really need this bad and i hope someone can help me you know. if anyone has any ideas for a more simple code please, enlighten me but just give me some good examples because i really am a beginner. thanks alot. Quote Link to comment https://forums.phpfreaks.com/topic/226972-need-a-dropdown-menu-for-a-form-that-can-be-populated-by-one-of-my-tables/ Share on other sites More sharing options...
atrum Posted February 7, 2011 Share Posted February 7, 2011 That should work. but I may have missed something. <html> <form id="form1" name="Update" method="post" action="add.php"> <label> First Name: <input type="text" name="Title" id="textfield" /> </label> <br /> <label> Last Name: <input type="text" name="URL" id="textfield2" /> </label> <select name='Category'> <?php $con = mysql_connect("localhost", "root", ""); if (!$con) { die('Could not connect to DB: ' . mysql_error() ); } mysql_select_db ("mydb", $con); $query = "SELECT Category from Catgories"; $result = mysql_query($query) OR DIE ("There was an error" .mysql_error()); while ($line = mysql_fetch_array($result,MYSQL_ASSOC)){ $categoryid = $line["playerid"]; $category = $line["category"]; for($i = 0;$i<count($categoryid);$i++){ echo "<option value='$categoryid[$i]'>$category[$i]</option>"; } } php?> </select> <input name="" type="submit" value="send" /> </form> </html> Quote Link to comment https://forums.phpfreaks.com/topic/226972-need-a-dropdown-menu-for-a-form-that-can-be-populated-by-one-of-my-tables/#findComment-1171056 Share on other sites More sharing options...
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