cffood Posted April 16, 2013 Share Posted April 16, 2013 I have a form that posts into the database but not the username. I have for the form <td><input name="username" type="text" disabled="disabled" class="inputbox" value="<?php echo $row['username'];?>" size="45" /></td> to fetch the username from the database so they don't have to insert it as I will be making another code to echo the username's inputs so they can track what they have inserted. but its not inserting the username back in to the database. any while will do Quote Link to comment https://forums.phpfreaks.com/topic/277013-not-adding-into-database/ Share on other sites More sharing options...
Jessica Posted April 16, 2013 Share Posted April 16, 2013 code? Quote Link to comment https://forums.phpfreaks.com/topic/277013-not-adding-into-database/#findComment-1425093 Share on other sites More sharing options...
cffood Posted April 16, 2013 Author Share Posted April 16, 2013 <?php define("_VALID_PHP", true); require_once("init.php"); $row = $user->getUserData(); ?> <?php include("header.php");?> <?php if($user->checkMembership('1,6')): ?> <!-- ============= CONTENT AREA STARTS HERE ============== --> <div id="content"> <!-- SLIDER STARTS HERE --> <div id="slider" class="slider2"> <div > </div> <h2 class="slider-head">Track your Weight</h2><br /><br /> <div id="whats-hot" > <p> Insert your daily Weight in here. The best time to weigh yourself is within the first hour as you wake up. </p> </div> <!-- end of news div --><!-- end of container div --> <div class="w-pet-border"> <script language="javascript"> function calculateWeight(fld) { if(fld.name=="weight_lb" || fld.name=="lose_lb") { // calculate in kg if(isNaN(fld.value) || fld.value=="") w=0; else w=fld.value; wKg=Math.round(w*0.453*10)/10; if(fld.name=="weight_lb") fld.form.weight_kg.value=wKg; else fld.form.lose_kg.value=wKg; } else { // calculate in lbs if(isNaN(fld.value) || fld.value=="") w=0; else w=fld.value; wP=Math.round(w*2.2); if(fld.name=='weight_kg') fld.form.weight_lb.value=wP; else fld.form.lose_lb.value=wP; } } </script> <table width="300" border="0" align="center" cellpadding="0" cellspacing="1"> <tr> <td><form name="form1" method="post" action="weighttrack_db.php"> <table width="100%" border="0" cellspacing="1" cellpadding="3"> <tr> <td><table width="100%" border="0" cellspacing="1" cellpadding="3"> <tr> <td colspan="3"><strong>Insert Data Into mySQL Database </strong></td> </tr> <tr> <th width="180"><strong>Username:</strong></th> <td> </td> <td><input name="username" type="text" disabled="disabled" class="inputbox" value="<?php echo $row['username'];?>" size="45" /></td> </tr> <tr> <td>Lastname</td> <td>:</td> <td><label>Your weight:</label> <br /> <input type="text" name="pounds" size="4" onkeyup="calculateWeight(this);" value="<?=$_SESSION["calc_weight_lb"]?>" /> lbs <b>OR</b> <input type="text" name="kilo" size="4" onkeyup="calculateWeight(this);" value="<?=$_SESSION["calc_weight_kg"]?>" /> kg</td> </tr> <tr> <td>Email</td> <td>:</td> <td><input id="date" name="date" value="<?php echo date("M j, Y - g:i"); ?>"/></td> </tr> <tr> <td colspan="3" align="center"><input type="submit" name="Submit" value="Submit" /></td> </tr> </table></td> </tr> </table> </form> </td> </tr> </table> </div></div><br /><br /><br /> <!-- ============= CONTAINER AREA ENDS HERE ============== --> <?php else: ?> <h2>Please <a href="index.php">Login</a> to view this page. or if you are logged in please weight 24 hour or less so Admin can change your Membership </h2> <?php endif; ?> <?php include("footer.php");?> Quote Link to comment https://forums.phpfreaks.com/topic/277013-not-adding-into-database/#findComment-1425095 Share on other sites More sharing options...
cffood Posted April 16, 2013 Author Share Posted April 16, 2013 if I take disabled="disabled" of the input then it wokrs fine..but it means they can change their username. which means if they change it they will never see what they inputted Quote Link to comment https://forums.phpfreaks.com/topic/277013-not-adding-into-database/#findComment-1425096 Share on other sites More sharing options...
Jessica Posted April 16, 2013 Share Posted April 16, 2013 There is nothing in your code about inserting into a database. You know, you can just... Echo the text. You don't have to put it in an input box. Wtf is your question? Quote Link to comment https://forums.phpfreaks.com/topic/277013-not-adding-into-database/#findComment-1425097 Share on other sites More sharing options...
DavidAM Posted April 16, 2013 Share Posted April 16, 2013 if I take disabled="disabled" of the input then it wokrs fine..but it means they can change their username. which means if they change it they will never see what they inputtedDisabled INPUT elements are NOT posted with the form. They are disabled. You can use READONLY to display an input that cannot be typed in, it WILL be posted with the form. That being said, it is VERY VERY easy for a user to change the value in ANY INPUT element -- even READONLY and HIDDEN and, yes, DISABLED -- before posting the form. If you have a value that needs to survive from one "page" to another with no chance of being modified by the user, you need to look into sessions Quote Link to comment https://forums.phpfreaks.com/topic/277013-not-adding-into-database/#findComment-1425099 Share on other sites More sharing options...
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