barney0o0 Posted November 22, 2013 Share Posted November 22, 2013 HI Chaps, im losing sleep over this...with the below code, the info is been added to the database, however each record is added twice (with the same values)..i keep looking at it but cant see the problem...any ideas?..thanks in advance <?php if(isset($_POST['Submit'])) { $i=0; $ndim= $_POST["val3"]; foreach($_POST['idt'] as $idt) { $query = "INSERT INTO testdims (proid, dimid, dimvalue) VALUES ('$utid', '$idt', '$ndim[$i]')"; mysql_query($query) or die('Error, query failed : ' . mysql_error()); echo $query; $result_update = mysql_query($query); $i++; } } ?> <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post"> <?php $result = mysql_query("SELECT dims.dimid, dims.dimdesc, dimlinks.prodid, dimlinks.dimlink FROM dimlinks LEFT JOIN dims ON dimlinks.dimlink = dims.dimid WHERE dimlinks.prodid = '$utid'"); while($row=mysql_fetch_array($result)) { ?> <div class="field"> <label><?php echo $row['dimdesc']; ?></label> <input name="val3[]" value="" /> <input name="idt[]" type="hidden" value="<?php echo $row['dimid']; ?>" /> </div> <?php } ?> Quote Link to comment Share on other sites More sharing options...
Solution requinix Posted November 22, 2013 Solution Share Posted November 22, 2013 (edited) Well, doing the query twice could certainly cause that. mysql_query($query) or die('Error, query failed : ' . mysql_error()); echo $query; $result_update = mysql_query($query); Edited November 22, 2013 by requinix Quote Link to comment Share on other sites More sharing options...
barney0o0 Posted November 22, 2013 Author Share Posted November 22, 2013 thanks....it was so obvious that i didn't see it Quote Link to comment Share on other sites More sharing options...
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