whiskedaway Posted November 7, 2009 Share Posted November 7, 2009 I think I've made a silly mistake somewhere, but I can't find it, so any help would be appreciated! I've used if (isset) in other parts of my website and it works fine, but I can't get it working here. Basically, the PHP is meant to run a query on my database and return $staffID, $name, $address and $phone, and then is meant to be displayed in a table that is located in stafflook.inc. Code snippets: $cxn = mysqli_connect($host, $user, $password, $dbname) or die ("Connection failed."); $sql = "SELECT staffID, name, address, phone FROM Staff WHERE staffID = '$_POST[staffID]'"; $result = mysqli_query($cxn, $sql) or die ("Couldn't execute query."); $rowname = mysqli_fetch_assoc($result); extract ($rowname); $echoName = "<tr><td><strong>ID:</strong></td><td>$name</td></tr>"; $echoID = "<tr><td><strong>ID:</strong></td><td>$staffID</td></tr>"; $echoAddress = "<tr><td><strong>ID:</strong></td><td>$address</td></tr>"; $echoPhone = "<tr><td><strong>ID:</strong></td><td>$phone</td></tr>"; include("stafflook.inc"); stafflook.inc: <table width="250" border="0" align="center" cellpadding="2" cellspacing="2"> <?php if (isset($echoID)) { echo "$echoID";} if (isset($echoName)) { echo "$echoName";} if (isset($echoAddress)) { echo "$echoAddress";} if (isset($echoPhone)) { echo "$echoPhone";} ?> </table> If someone can see where I've made a mistake, it'd be greatly appreciated. Quote Link to comment Share on other sites More sharing options...
rajivgonsalves Posted November 7, 2009 Share Posted November 7, 2009 code looks okay what is the output your getting ? Quote Link to comment Share on other sites More sharing options...
keldorn Posted November 7, 2009 Share Posted November 7, 2009 use print_r($array); fro the array to make sure you getting what you expect from the database. Quote Link to comment Share on other sites More sharing options...
whiskedaway Posted November 8, 2009 Author Share Posted November 8, 2009 I'm actually getting no output at all, even with putting print_r($array);. I would assume that opening the page I'd get no output, but when I run the query, it should work. Here's the full code for both of the pages, hopefully it might be something that's not in the first code snippets: <?php session_start(); if (@$_SESSION['auth'] != "yes") { header("Location: management.php"); exit(); } include("credentials.inc"); switch (@$_POST['do']) { case "search": $cxn = mysqli_connect($host, $user, $password, $dbname) or die ("Connection failed."); $sql = "SELECT staffID, name, address, phone FROM Staff WHERE staffID = '$_POST[staffID]'"; $result = mysqli_query($cxn, $sql) or die ("Couldn't execute query."); $rowname = mysqli_fetch_assoc($result); extract ($rowname); $echoName = "<tr><td><strong>Name:</strong></td><td>$name</td></tr>"; $echoID = "<tr><td><strong>ID:</strong></td><td>$staffID</td></tr>"; $echoAddress = "<tr><td><strong>Address:</strong></td><td>$address</td></tr>"; $echoPhone = "<tr><td><strong>Phone:</strong></td><td>$phone</td></tr>"; print_r($array); include("stafflook.inc"); break; default: include("stafflook.inc"); } ?> stafflook.inc <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>Adelaide Books</title> <link href="style.css" rel="stylesheet" type="text/css" /> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> </head> <body> <div id="body"> <table id="main" cellspacing="0px"> <tr><td id="logo" colspan="3"> <img src="images/logo.gif" alt="Adelaide Books"/></td> <td class="space"> </td> <td class="right"> </td></tr> <tr><td class="left"> </td> <td class="space"> <form action="stafflookup.php" method="POST"> Staff ID: <input type="text" name="staffID" size="25" value=""><br/> <input type="submit" name="Search" value="Search"></form></td> <td id="text"> <br/> <br/> <br/> <table width="250" border="0" align="center" cellpadding="2" cellspacing="2"> <?php if (isset($echoID)) { echo "$echoID";} if (isset($echoName)) { echo "$echoName";} if (isset($echoAddress)) { echo "$echoAddress";} if (isset($echoPhone)) { echo "$echoPhone";} ?> <tr><td width="75px" align="right"> </td><td> </td></tr> </table> <br/> <br/> <br/> <br/><a href="management2.php"><img src="images/back.gif" alt="Back" border="0" /></a></td> <td class="space"> </td> <td class="right"> </td></tr> </table> </div> </body> </html> Quote Link to comment Share on other sites More sharing options...
MadTechie Posted November 8, 2009 Share Posted November 8, 2009 shouldn't <input type="submit" name="Search" value="Search"> be <input type="submit" name="do" value="Search"> OR switch (@$_POST['do']) { be switch (@$_POST['Search']) { As $_POST['do'] isn't in the form Quote Link to comment Share on other sites More sharing options...
whiskedaway Posted November 8, 2009 Author Share Posted November 8, 2009 Thank you! I added <input type="hidden" name="do" value="search"> and that fixed everything. I knew I'd done something really dumb! Quote Link to comment Share on other sites More sharing options...
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