gevans Posted December 6, 2008 Share Posted December 6, 2008 That will check if id is checked, and if it is only show that one title Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 ok one last piece to edit  the index.php where it displays available surveys  ===================================  Heres the code We need to make it so if it is in the complete_surveys then dont show that survey  <td><font size="6"> Surveys Available </font></td> <td> <?php mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM survey"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<a href=\"survey.php?id={$row['id']}\">{$row['title']}</a><br>\n"; } ?> </td></tr> </table> Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 First of all we need to do some work on survey.php to run an sql query  You need to say  insert into the second table survey id and user id  let me know how u get on! Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 Heres The Code I have to Do This  mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); mysql_select_db("sex1800_loginbux") or die(mysql_error()); $member = mysql_real_escape_string($_SESSION['userid']); $survey = mysql_real_escape_string($_GET['id']); mysql_query("INSERT INTO completed_surveys (member_id, survey_id) VALUES('$member', '$survey') ") or die(mysql_error());  Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 All we need to do is edit this code to only show surveys that havent been completed   <td><font size="6"> Surveys Available </font></td> <td> <?php mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM survey"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<a href=\"survey.php?id={$row['id']}\">{$row['title']}</a><br>\n"; } ?> </td></tr> </table> Quote Link to comment Share on other sites More sharing options...
premiso Posted December 6, 2008 Share Posted December 6, 2008 How do you know that a survey has not been completed? Answer that then add that identifier to the survey query in a WHERE clause and you got your answer. Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 Thats alot of confusing stuff you just  I know its completed if its in the member id and survey id is in completed_surveys table  but i dont know how to make it so if its in the table to tell the survey to be hidden  Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 change this; Â </tr> <?php mysql_connect("localhost", "sex1800_admin", "****") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM surveys"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href="survey.php?id={$row['id']}">{$row['title']}</a></td></tr> "; } Â to this; Â </tr> <?php mysql_connect("localhost", "sex1800_admin", "****") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM surveys JOIN completed_surveys ON surveys.id!=completed_surveys.survey_id AND memberid={$_SESSION['userid']}"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href="survey.php?id={$row['id']}">{$row['title']}</a></td></tr> "; } Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 Now i get this Error  Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home3/sex1800/public_html/epicbux/index.php on line 95  Heres lines 85 - 100  <tr> <td><font size="6"> Surveys Available </font></td> <td> <?php mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM surveys JOIN completed_surveys ON surveys.id!=completed_surveys.survey_id AND memberid={$_SESSION['userid']}"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href="survey.php?id={$row['id']}">{$row['title']}</a></td></tr>"; } ?> </td></tr> </table> Quote Link to comment Share on other sites More sharing options...
premiso Posted December 6, 2008 Share Posted December 6, 2008 echo "<tr><td><a href="survey.php?id={$row['id']}">{$row['title']}</a></td></tr>"; should be echo "<tr><td><a href=\"survey.php?id={$row['id']}\">{$row['title']}</a></td></tr>"; Â You have to escape " inside of ". Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 <tr> <td><font size="6"> Surveys Available </font></td> <td> <?php mysql_connect("localhost", "sex1800_admin", "****") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM surveys JOIN completed_surveys ON surveys.id!=completed_surveys.survey_id AND memberid={$_SESSION['userid']}"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href="survey.php?id={$row['id']}">{$row['title']}</a></td></tr>"; } ?> </td></tr> </table> Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 now it displays the page but where it should display the surveys it displays  Unknown column 'df1063585e41cc161b358347877f900c' in 'on clause'  Heres the code though i fixed it to match my tables  <?php mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM survey JOIN completed_surveys ON survey.id!=completed_surveys.survey_id AND member_id={$_SESSION['userid']}"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href=\"survey.php?id={$row['id']}\">{$row['title']}</a></td></tr>"; } ?> Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 did you get the same error after doing that? Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 yea  Tables:  survey *id *title *adlink  completed_surveys *member_id *survey_id Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 put this in for an error check; Â <?php mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM survey JOIN completed_surveys ON survey.id!=completed_surveys.survey_id AND member_id={$_SESSION['userid']}"; echo $query; die(); $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href=\"survey.php?id={$row['id']}\">{$row['title']}</a></td></tr>"; } ?> Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 it says  SELECT * FROM survey JOIN completed_surveys ON survey.id!=completed_surveys.survey_id AND member_id=df1063585e41cc161b358347877f900c <-- this is the problem Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 Can you look into the completed_surveys table and see what data you currently have in there? Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 member_id - survey_id 0 - 1 0 - 2 0 - 1 Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 ok, having a little error  didn't realise that the member_id was hashed.  in the table completed_surveys change member_id from INT to VARCHAR 32 Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 what should i put for the mysql code  instead of this <?php mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM survey JOIN completed_surveys ON survey.id!=completed_surveys.survey_id AND member_id={$_SESSION['userid']}"; echo $query; die(); $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href=\"survey.php?id={$row['id']}\">{$row['title']}</a></td></tr>"; } ?> Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 ok, having a little error  didn't realise that the member_id was hashed.  in the table completed_surveys change member_id from INT to VARCHAR 32   did you do that??  If you did delete all the entries in the table and it should work fine Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 yes i did and deleted all that data in it so and i open the index page and still get this  SELECT * FROM survey JOIN completed_surveys ON survey.id!=completed_surveys.survey_id AND member_id=324a6953910123430fa8e36e055d3c41 Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 ok, change this  <?php mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM survey JOIN completed_surveys ON survey.id!=completed_surveys.survey_id AND member_id={$_SESSION['userid']}"; echo $query; die(); $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href=\"survey.php?id={$row['id']}\">{$row['title']}</a></td></tr>"; } ?>  back to this  <?php mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM survey JOIN completed_surveys ON survey.id!=completed_surveys.survey_id AND member_id={$_SESSION['userid']}"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href=\"survey.php?id={$row['id']}\">{$row['title']}</a></td></tr>"; } ?> Quote Link to comment Share on other sites More sharing options...
jnerotrix Posted December 6, 2008 Author Share Posted December 6, 2008 Now it says  Unknown column '8190ef16bc01b56546027a349806ad2a' in 'on clause' Quote Link to comment Share on other sites More sharing options...
gevans Posted December 6, 2008 Share Posted December 6, 2008 try this; Â <?php mysql_connect("localhost", "sex1800_admin", "12921993") or die(mysql_error()); //add you password mysql_select_db("sex1800_loginbux") or die(mysql_error()); $query = "SELECT * FROM survey JOIN completed_surveys ON survey.id!=completed_surveys.survey_id WHERE member_id={$_SESSION['userid']}"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo "<tr><td><a href=\"survey.php?id={$row['id']}\">{$row['title']}</a></td></tr>"; } ?> Quote Link to comment Share on other sites More sharing options...
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