techker Posted December 4, 2008 Share Posted December 4, 2008 hey guys im trying to do a select pic to insert the name in a database. the insert works but it does not insert the pic name only an id.. i can't seem to figure it out.. <? $dbh = mysql_connect("localhost","user","pass") or die("There was a problem with the database connection."); $dbs = mysql_select_db("db", $dbh) or die("There was a problem selecting the categories."); $type=$_POST['type']; $sql = "SELECT * FROM `gymball` WHERE `type` = '$type' "; $fileLIST=mysql_query($sql);?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>GymGraph</title> </head> <body> <p align="center"><u>Sélectionner UNE photo</u></p> <table width="346" border="0" align="center" cellpadding="0" cellspacing="0"> <tr><? while($row = mysql_fetch_array($fileLIST)) { ?> <td width="173"><?php echo '<a href="/Gymgraph/Gymgraph/gymball/'. $row['name'] .'" target="_blank"> <img src="/Gymgraph/Gymgraph/gymball/'. $row['name'] .'" border="0" alt="" width=115/> </a> <br />'; ?></td> <td width="173"><label> <input type="checkbox" name="select" id="pic" value='<? $row['name'] ?>'/> Selection de photo</label></td> </tr> <? } ?> </table> <p> <p> <label> <div align="center"> <select name="location" id="location"> <option value="pic1">1</option> <option value="pic2">2</option> <option value="pic3">3</option> <option value="pic4">4</option> <option value="pic5">5</option> <option value="pic6">6</option> <option value="pic7">7</option> <option value="pic8">8</option> </select> Emplacement de la photo </div> </label> </p> <label> <div align="center"> <input type="submit" name="submit" id="submit" value="Submit" /> </div> </label> <div align="center"> Envoyer la sélection </div> </p> </form> </body> </html> and this is the insert pic <?php $pic=$_POST['pic']; $location=$_POST['location']; mysql_connect("localhost", "user", "t") or die(mysql_error()) ; mysql_select_db("techker_gymgraphpics") or die(mysql_error()) ; //Writes the information to the database mysql_query("INSERT INTO $location (name) ". "VALUES ('$pic')"); //Tells you if its all ok $id= mysql_insert_id(); echo "<p>This file has the following Database ID: <b>$id</b>"; echo "You'll be redirected to Home Page after (4) Seconds"; echo "<meta http-equiv=Refresh content=4;url=gymball_select.php>"; ?> Quote Link to comment Share on other sites More sharing options...
revraz Posted December 4, 2008 Share Posted December 4, 2008 Use mysql_error() after your query to check for errors Are you saying this isn't working? mysql_query("INSERT INTO $location (name) ". "VALUES ('$pic')"); if so, echo $location and $pic to make sure the are what they should be. Quote Link to comment Share on other sites More sharing options...
MadTechie Posted December 4, 2008 Share Posted December 4, 2008 first off i ma not sure if you mean radio instead of check boxes but in anycase you have 2 problems 1. you requesting $_POST['pic']; which is the ID not the name.. the name is "select" 2. it will only take the last ticked value.. so to fix this.. change <input type="checkbox" name="select" id="pic" value='<? $row['name'] ?>'/> to <input type="checkbox" name="pic[]" id="pic" value='<?php $row['name'] ?>'/> and then change $pic=$_POST['pic']; to $pic=$_POST['pic'][0]; //assuming you only tick 1 box Quote Link to comment Share on other sites More sharing options...
gevans Posted December 4, 2008 Share Posted December 4, 2008 From what I can see, your html form isn't started you have the closing tag </form> but no opening tag - <form> Quote Link to comment Share on other sites More sharing options...
techker Posted December 4, 2008 Author Share Posted December 4, 2008 first off i ma not sure if you mean radio instead of check boxes but in anycase you have 2 problems 1. you requesting $_POST['pic']; which is the ID not the name.. the name is "select" 2. it will only take the last ticked value.. so to fix this.. change <input type="checkbox" name="select" id="pic" value='<? $row['name'] ?>'/> to <input type="checkbox" name="pic[]" id="pic" value='<?php $row['name'] ?>'/> and then change $pic=$_POST['pic']; to $pic=$_POST['pic'][0]; //assuming you only tick 1 box so what would be the value?i need the named of the image that is ticked. whats the [0] added? Quote Link to comment Share on other sites More sharing options...
techker Posted December 4, 2008 Author Share Posted December 4, 2008 ok i setled the prob by putting a hidden filed that echo the name of the pic. so in my add pic i just added the post for that field..and voila!thx.. but im still wondering why it does not work.. Quote Link to comment Share on other sites More sharing options...
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