zed420 Posted November 24, 2008 Share Posted November 24, 2008 Hi All Can someone please point out what am I doing wrong here, the query should display one set of results but its showing me THREE same results even thou there is only ONE set of results in the database so the insert query is fine it hasn't inserted the duplicate data. Some help will appreciated. Thanks function normalJob(){ $id = $_GET['id']; $query = "SELECT DISTINCT job_tb.job_id,job_tb.dateTime,job_tb.cust_name,job_tb.cust_address,job_tb.des,job_tb.typeOfbooking, user.id FROM job_tb,user WHERE job_tb.user_id = '$id'"; $result = mysql_query($query)or die(mysql_error()); ?><div class="smallerText"> <b>These are the jobs you have booked so far. Please click on Job ID to view further details about the job</b> <TABLE BORDER=0 WIDTH=100% CELLSPACING=3 CELLPADDING=3 ALIGN=CENTER bgcolor="#CCCCCC"> <TR bgcolor="#FFFFCC" align="center"> <td width="10%"><font color=red><b>Job No.</b></font></TD> <td width="15%"><font color=red><b>Date/Time</b></font></TD> <td width="15%"><font color=red><b>Name</b></font></TD> <td width="25%"><font color=red><b>Pick up Address</b></font></TD> <td width="20%"><font color=red><b>Destination</b></font></TD> <td width="20%"><font color=red><b>Booking Type</b></font></TD> </tr> <? while ($row = mysql_fetch_array($result)) { extract($row); echo "<tr > <td> <a href=\"javascript:open_window('detailNormal.php?job_id=$job_id');\">" . $row['job_id'] . "</a></td> <td>" . $row['dateTime'] . "</td> <td>" . $row['cust_name'] . "</td> <td>" . $row['cust_address'] . "</td> <td>" . $row['des'] . "</td> <td>" . $row['typeOfbooking'] . "</td> </tr>"; } ?></TABLE></div><? } normalJob(); Thanks Zed Quote Link to comment Share on other sites More sharing options...
mtoynbee Posted November 24, 2008 Share Posted November 24, 2008 Looks like your query would benefit from a JOIN. SELECT DISTINCT job_tb.job_id,job_tb.dateTime,job_tb.cust_name,job_tb.cust_address,job_tb.des,job_tb.typeOfbooking, user.id FROM job_tb LEFT OUTER JOIN user ON job_tb.user_id = user.id WHERE job_tb.user_id = ".int_val($id)." With the current query you are bound to get a resultset for each user Quote Link to comment Share on other sites More sharing options...
zed420 Posted November 24, 2008 Author Share Posted November 24, 2008 Thanks for quick reply, this is the error I'm getting Fatal error: Call to undefined function int_val() in C:\xampp\htdocs\t_line_ph\memberProAdmin.php on line 127 Quote Link to comment Share on other sites More sharing options...
mtoynbee Posted November 24, 2008 Share Posted November 24, 2008 Ah Try intval() instead Quote Link to comment Share on other sites More sharing options...
zed420 Posted November 25, 2008 Author Share Posted November 25, 2008 Thanks mtoynbee it worked. thank you Zed Quote Link to comment Share on other sites More sharing options...
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