mc136 Posted February 10, 2007 Share Posted February 10, 2007 Hi I have the following piece of code which enters all data into a table and if all data entered will show green image or blue (and text box) if not. My problem is with creating a form basically don't know where to start. Ive tried but will only update the last text box. Any help please!! ??? while($Array = mysql_fetch_array($result, MYSQL_ASSOC)){ $arr[] = $Array; } echo "<pre style=\"text-align: left; font-size: 10px;\">"; print_r( $arr ); echo "</pre>"; foreach($arr as $key => $value){ echo "<tr><td><H1>" . $value['qName'] . "</td>"; foreach($value as $chk => $num){ if(preg_match("/^chk/",$chk) && $num >= "0"){ echo "<td><p>" . $num . "</td>"; }elseif(preg_match("/^chk/",$chk) && $num == ""){ echo "<td><input type='text' name='check1' size='1'></td>"; } } //----------test-works--------- if($value[chk1] == ""){ print "<td align='center'><img src=\"blue.jpg\" width=\"15\" height=\"15\" alt=\"description of the photo\"></img></td>"; }elseif($value[chk2] == ""){ print "<td align='center'><img src=\"blue.jpg\" width=\"15\" height=\"15\" alt=\"description of the photo\"></img></td>"; }else{ print "<td align='center'><img src=\"green.jpg\" width=\"15\" height=\"15\" alt=\"description of the photo\"></img></td>"; } //----------end test------ } Quote Link to comment Share on other sites More sharing options...
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