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Ronald245

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Everything posted by Ronald245

  1. hi everyone, im new to these forums. as part of a project i'm working on i'm building a website for a large airline, i've had my ajax working perfectly but when i logged in on the site this morning it didnt update the <div> code on the page <div align="center"> <button class="buttonup" type="button" onclick="tempup()"></button> <script language="javascript"> function tempup() { $("#targettemp").load("targettemp.php"); $.ajax({ url: "tempup.php", //the page containing php script type: "POST", //request type }) $(document).ready(function () { $("#targettemp").load("targettemp.php"); }); } </script> <h4 class="target-temp panel-body" id="targettemp "> <?php require 'dbconnection.php'; $con; $dbcon; $temp = "SELECT `target temp` FROM Temperature WHERE homeid = 'BG0001' "; $query = mysql_query($temp); $results = mysql_result($query, 0); echo "$results ℃"; ?> </h4> <div align="center"> <button class="buttondown" type="button" onclick="tempdwn()"></button> <script language="javascript"> function tempdwn() { $("#targettemp").load("targettemp.php"); $.ajax({ url: "tempdwn.php", //the page containing php script type: "POST", //request type }) $(document).ready(function () { $("#targettemp").load("targettemp.php"); }); } </script> </div> </div> tempdwn.php <?php require 'dbconnection.php'; $con; $dbcon; $query = mysql_query("SELECT `target temp` FROM `Temperature` WHERE homeid = 'BG0001' "); $huidigetemp = mysql_result($query, 0); $newtemp = $huidigetemp - 1; mysql_query("UPDATE Temperature SET `target temp`= $newtemp WHERE homeid = 'BG0001' "); ?> targettemp.php: <?php require 'dbconnection.php'; $con; $dbcon; $temp = "SELECT `target temp` FROM Temperature WHERE homeid = 'BG0001' "; $query = mysql_query($temp); $results = mysql_result($query, 0); echo "$results ℃"; ?>
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