shadiadiph Posted December 13, 2008 Share Posted December 13, 2008 I have been told this is a php problem and not so much of a javascript problem which is why i am posting it here in php help. This is almost working when i select an email it shows the name field but it shows more than one name on select as the php echo array returns more than 1 value how can i get it working to show the correct name only? The problem seems to be in defining the php echo "names" in the <script> area! <?PHP include("../checkadminlogin.php"); include("../../global/connection.php"); $sql = "select intNameID, email,fname, lname, place from tbladvisorsdetails where place='A' order by intNameID ASC"; $temps01=$DB_site->query($sql); while ($row=$DB_site->fetch_array($temps01)) { $selected = ($nameid==$row["intNameID"]) ? ' selected="selected"' : ''; $email_options .= " <option value=\"{$row['intNameID']}\"$selected>{$row['email']}</option>\n"; $js_names_array .= "names[{$row['intNameID']}] = '{$row['fname']} {$row['lname']}';\n"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>APA</title> <link rel="stylesheet" type="text/css" href="office.css" /> <script type="text/javascript"> var names = new Array(); <?php echo $js_names_array; ?> function changeName(value) { document.getElementById('USEDname').value = (names); } </script> </head> <body> Link to comment https://forums.phpfreaks.com/topic/136834-solved-php-and-javascript-array-problem/ Share on other sites More sharing options...
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