PHP - Php Print Button Not Working
Hi - I designed a regular website using html. One page I used was on the previous site, and it had links to a database. Its for an online wedding registry. Problem is, on the page that pulls up with the couples list of wish list items, there is a "Print the Bliss List" link that is not working, right below the list. It is just seen as text on the page, not a button element. Can you look at the code and see why this isn't working or what I can do to fix it? I use Dreamweaver CS5 and am familiar with working in html, and css code, but not php. Here is the link to the page in question:
http://getinteriormo...s2.php?id=03313
thanks so much!
Similar TutorialsI am making a pos(point of sale) application using object oriented php. I have found thermal receipt printers support ESC/POS . i want that when i will press button to print directly , it will do automatically. anyone can help me please. Thanks in advance. so this is my code Code: [Select] <? $b=$_GET['b']; $m=$_GET['m']; $y=$_GET['y']; if ($b <> '' && $m <> '' && $y <> '') { $branch=$b; $leavemonth=$m; $leaveyear=$y; } else { $branch=$_POST['branch']; $leavemonth=$_POST['leavemonth']; $leaveyear=$_POST['leaveyear']; } $connection=mysql_connect("$server", "$username", "$password") or die("Could not establish connection"); mysql_select_db($database_name, $connection) or die ("Could not select database"); $query="SELECT * from tblworkgroup"; $result=mysql_query($query);?> Branch <select name="branch" > <option value="all">ALL Branches</option> <? while($row=mysql_fetch_array($result)) { ?> <option value= "<?=$row['WorkGroupID']?>" <? if ($branch==$row['WorkGroupID']){ echo 'selected'; } ?> ><?=$row['WorkGroupName']?></option> <? } ?> </select> </td> </tr> <br><br><br> Month <select name="leavemonth" > <option value = "1" <? if ($leavemonth==1){ echo 'selected'; } ?>>January</option> <option value = "2" <? if ($leavemonth==2){ echo 'selected'; } ?>>February</option> <option value = "3" <? if ($leavemonth==3){ echo 'selected'; } ?>>March</option> <option value = "4" <? if ($leavemonth==4){ echo 'selected'; } ?>>April</option> <option value = "5" <? if ($leavemonth==5){ echo 'selected'; } ?>>May</option> <option value = "6" <? if ($leavemonth==6){ echo 'selected'; } ?>>June</option> <option value = "7" <? if ($leavemonth==7){ echo 'selected'; } ?>>July</option> <option value = "8" <? if ($leavemonth==8){ echo 'selected'; } ?>>August</option> <option value = "9" <? if ($leavemonth==9){ echo 'selected'; } ?>>September</option> <option value = "10" <? if ($leavemonth==10){ echo 'selected'; } ?>>October</option> <option value = "11" <? if ($leavemonth==11){ echo 'selected'; } ?>>November</option> <option value = "12" <? if ($leavemonth==12){ echo 'selected'; } ?>>December</option> </select> Year <select name="leaveyear" > <option value = "2010" <? if ($leaveyear==2010) { echo 'selected';} ?>>2010</option> <option value = "2011" <? if ($leaveyear==2011) { echo 'selected';} ?>>2011</option> <option value = "2012" <? if ($leaveyear==2012) { echo 'selected';} ?>>2012</option> <option value = "2013" <? if ($leaveyear==2013) { echo 'selected';} ?>>2013</option> <option value = "2014" <? if ($leaveyear==2014) { echo 'selected';} ?>>2014</option> <option value = "2015" <? if ($leaveyear==2015) { echo 'selected';} ?>>2015</option> <option value = "2016" <? if ($leaveyear==2016) { echo 'selected';} ?>>2016</option> <option value = "2017" <? if ($leaveyear==2017) { echo 'selected';} ?>>2017</option> </select> <input type='submit' name='submit' value='Preview' Class="button" onclick='return validate()'> </div> </form> <br /><br /> <div id="box" valign="top"> <h3> <strong>Leave Application Details</strong>  </h3> <br><br> <FORM name="printbutton" method="post" align="left" action="report_print.php" target="_blank"> <input type="hidden" name="leavemonth" value="<?php echo $leavemonth; ?>"> <input type="hidden" name="leaveyear" value="<?php echo $leaveyear; ?>"> <input type="hidden" name="branch" value="<?php echo $branch;?>"> <Input type = "Submit" Name = "Submit1" Class="button" VALUE = "Print Leave Details"> </form> <table width="80%" align="center" > <thead> <tr> <th width="700px" align="left">Name</a></th> <th width="500px" align="center"></a>Application Date</th> <th width="500px" align="center"></a> </th> <th width="500x" align="right"></a>Reason</th> </tr> </thead> <tbody> <? /*echo "branch:".$branch.'<br />'; echo "leavemonth:".$leavemonth.'<br />'; echo "leaveyear:".$leaveyear.'<br />';*/ //-------------------------------------------------- if ($leavemonth =='' || $leaveyear =='' || $branch == '') { // one value missing, so don't display } else { // start display data $branchtracker = ''; if ($branch=='all'){ $branch = '%'; } $rs = mysql_query( " call vwleavereport('$leavemonth', '$leaveyear', $branch, 0);"); $query="SELECT *, MONTH(tblleaveapplication.DateFrom) as DFMonth, MONTH(tblleaveapplication.DateTo) as DTMonth, YEAR(tblleaveapplication.DateFrom) as DFYear FROM `tblleaveapplication` LEFT JOIN `tblemployee` ON tblleaveapplication.employeeid = tblemployee.id WHERE WorkGroupID LIKE '".$branch."' AND ( (MONTH(tblleaveapplication.DateFrom)=".$leavemonth." AND YEAR(tblleaveapplication.DateFrom)=".$leaveyear.") OR (MONTH(tblleaveapplication.DateTo)=".$leavemonth." AND YEAR(tblleaveapplication.DateFrom)=".$leaveyear.") ) ORDER BY WorkGroupID "; $result = mysql_query($query); while($row=mysql_fetch_array($result)) { if ($branchtracker<>$row['WorkGroupID']){ echo '<tr><td colspan=4 style="background-color: #e8e8e8; font-weight: bold;">'.$row['WorkGroupID'].'</td></tr>'; $branchtracker = $row['WorkGroupID']; } echo '<tr>'; echo '<td>'.$row['EmployeeName'].'</td>'; echo '<td>'.$row['DateFrom'].'</td>'; echo '<td>'.$row['DateTo'].'</td>'; echo '<td>'.$row['Purpose'].'</td>'; echo '</tr>'; } } ?> the problem is i want the print button only appear when there is output.. I apologize if I am in the wrong forum so I hope somebody can help. I have a function which sends an emails to client everytime an order is processed. They have asked me to add a print button in the emails so they could just print the email. I have tried using javascript and call the window.print function but when I click the button nothing happens. I know that it is possible to add a print button in an email but I can't figure out what I am doing wrong. Here is some of the code use to create the email using paypal ipn script. Code: [Select] $body = "<html>". "<head>". "<script type='text/javascript'>". "//<![CDATA[ function PrintEmail(){ window.print(); } //]]> ". "</script>". "</head>". "<body>". "<a href='javascript:window.print()'>Print Page</a><br />". "<input type='button' onclick='javascript:PrintEmail();' value='Print This Page' /><br /><b>Customer Name</b>: ".$name."<br />"; foreach($id_str_array as $key => $value){ $id_quantity_pair = explode(":", $value); $product_id = $id_quantity_pair[0]; // Get the product ID $cameo_color = $id_quantity_pair[1]; $product_quantity = $id_quantity_pair[2]; // Get the quantity $product_line1 = $id_quantity_pair[3]; // Get Line 1 $product_line2 = $id_quantity_pair[4]; // Get Line 2 $product_picture = $id_quantity_pair[5]; // Get picture $discount_code = $id_quantity_pair[6]; $body .= "<b>Product</b>: ".$product_id."<br />". "<b>Quantity</b>: ".$product_quantity."<br />". "<b>Engraving Line 1</b>: ".$product_line1."<br />". "<b>Engraving Line 2</b>: ".$product_line2."<br />". "<b>Your picture</b>: <img src='http://t3.sysguru.com/mcf/pictures/".$product_picture."' height='300' width='556' /><br />"; } $body .= "<b>Total</b>: $".$_POST['mc_gross']."</body></html>"; $subject = "Transaction Completed"; $mailer->SendPayPalInfo($subject, $body); I tried just calling the javascript print function with a link and aslo create a function and called the function from the button and neither of these methods work. here is the code which send the message out. Code: [Select] function SendPayPalInfo($subject, $body){ $headers = "From: " .EMAIL_FROM_ADDR. "\r\n"; $headers .= "Reply-To: ".EMAIL_FROM_ADDR."\r\n"; $headers .= "MIME-Version: 1.0\r\n"; $headers .= "Content-Type: text/html; charset=ISO-8859-1\r\n"; return mail(EMAIL_FROM_ADDR, $subject, $body, $headers); } Am I missing something in the headers of am I coding this wrong? Any help would be appreciated. heres the line for my button Code: [Select] echo "<tr><td colspan='2'><input type='submit' value='Add Reply' onClick=\"window.location = post_reply.php?cid=".$cid."&tid=".$tid."\" /><hr /> "; for some reason when i click on it in the site it does not do anything, but when i type the link in maunaly that the button should link to the page works. heres the code for the page that the button is on Code: [Select] <?php require("top.php"); ?> <div id='content'> <div id='homepageright'> <?php require("scripts/connect.php"); if($username){ $cid = intval($_GET['cid']); $tid = intval($_GET['tid']); $sql = "SELECT * FROM topics WHERE category_id = $cid AND id = $tid"; $res = mysql_query($sql) or die(mysql_error()); /*$cid = $_GET['cid']; $tid = $_GET['tid']; $sql = "SELECT * FROM topics WHERE category_id='".$cid."' AND id='".$tid."'"; $res = mysql_query($sql) or die(mysql_error());*/ if(mysql_num_rows($res) == 1){ echo "<table width='100%'>"; if($username){ echo "<tr><td colspan='2'><input type='submit' value='Add Reply' onClick=\"window.location = post_reply.php?cid=".$cid."&tid=".$tid."\" /><hr /> "; } While ($row = mysql_fetch_assoc($res)) { $sql2 = "SELECT * FROM post WHERE category_id='".$cid."' AND topic_id= '".$tid."'"; $res2 = mysql_query($sql2) or die(mysql_error()); while ($row2 = mysql_fetch_assoc($res2)) { echo "<tr><td valign='top' style='border: 1px solid #000000;'><div style='min-height: 125px; '>".$row['topic_title']."<br /> by ".$row2['post_creator']." - ".$row2['post_date']. "<hr /> ".$row2['post_content']."</div></td><td width='200' valign='top' align='center' style='border: 1px solid #000000;'>User Info Here</td></tr><tr><td colspan='2'><hr /></td></tr>"; } echo "</table>"; } } else{ echo "This Topic Does Not Exist."; echo "$sql"; } } else{ echo "You Must Be Logged In To Continue."; } ?> </div> <div id='homepageleft'> <?php ?> </div> </html> </body> Why the button "Enviar Emails" is not working? --> http://www.ibnshare.com/email/ (login: 123abc password: 123abc) Code: [Select] echo '<form action = "send.php" method = "post">'; echo '<strong>Assunto do Email</strong><br />'; echo $emailSubject; echo '<br />'; echo '<strong>Corpo do Email</strong><br />'; echo $emailBody; echo '<br /><br />'; echo '<input name = "send" type = "button" value = "Enviar Emails" />'; echo '</form>'; I am totally new in PHP, can anyone help me with this problem:
Login Button is not working after clicking this button this should print a echo message. <from action="" method="POST"> <input type="text" placeholder="username"> <br> <input type="password" placeholder="Password"><br> <input type="submit" name="login" value="Login"> </from> <?php if ( isset($_POST['login'])){ echo "hissssssssssssssssssss"; } ?>
I'm trying to use checkbox to choose multiple data and submit(button 'Send SMS') them and direct them to another page, which will send text SMS to the numbers chosen. But when I try to click on one checkbox, it automatically directs me to the next page, without having to click on the 'Send SMS' button. I tried to see which part is the problem, but I can't determine which one is wrong. Please help! Thank you.
<script type="text/javascript"> function checkedAll (frm1) {var aa= document.getElementById('frm1'); if (checked == false) { checked = true } else { checked = false }for (var i =0; i < aa.elements.length; i++){ aa.elements[i].checked = checked;} } </script> <TABLE width="1106" border="1" align="center" cellpadding="5" cellspacing="2" > <tr> <td colspan="9"> <form id ="frm1" action="sms_latecomers.php"> <input type='checkbox' name='checkall' onclick='checkedAll(frm1);' value=""> <input type="submit" name="Submit" id="button" value="SMS Reminder" style="background:#FFCC33"/> </td> </tr> <TR bgcolor="#996699"> <?php $count=0; ?> <TH width="32"></TH> <TH width="37"> <span class="style1"> <div align="center">No. </div></span></TH> <th width="101"><span class="style1">SUPERVISOR ID</span></th> <th width="101"><span class="style1">EMPLOYEE ID</span></th> <th width="153"><span class="style1">EMPLOYEE NAME</span></th> <th width="124"><span class="style1">DATE</span></th> <th width="153"><span class="style1">LATE CHECK IN</span></th> </TR> <?php while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { ?> <TR> <TD><input type="checkbox" name="chk1" value="<?php echo $row['supervisor_telno']; ?>" onClick="window.open('sms_latecomers.php?employee_id=<?php echo $row['employee_id']; ?>&supervisor_telno=<?php echo $row['supervisor_telno']; ?>');"></TD> <TD height="66">[bad html removed]<input type="checkbox" name="chk1" >--> <?php $count=$count+1; print($count);?></TD> <TD><div align="center"><?php echo $row['supervisor_id']; ?></div></TD> <TD><div align="center"><?php echo $row['employee_id']; ?></div></TD> <TD><div align="center"><?php echo $row['employee_name']; ?></div></TD> <TD><div align="center"><?php echo $row['DATE']; ?></div></TD> <TD><div align="center"><?php echo $row['TIME']; ?></div></TD> </TR> <?php } ?>[bad html removed]</form>--> </TABLE> Hey guys! I've looked through this code over and over today, and I still haven;t found where my error is. Here is my entire code: Code: [Select] <?php session_start(); include("config536.php"); ?> <html> <head> <link rel="stylesheet" type="text/css" href="style.css" /> </head> <?php if(!isset($_SESSION['username'])) { echo "<banner></banner><nav>$shownavbar</nav><ubar><a href=login.php>Login</a> or <a href=register.php>Register</a></ubar><content><center><font size=6>Error!</font><br><br>You are not Logged In! Please <a href=login.php>Login</a> or <a href=register.php>Register</a> to Continue!</center></content>"; } if(isset($_SESSION['username'])) { echo "<nav>$shownavbar</nav><ubar>$ubear</ubar><content><center><font size=6>Quest Agency</font><br><br>"; $startjob = $_POST['submit']; $jobq = "SELECT * FROM jobs WHERE username='$showusername'"; $job = mysql_query($jobq); $jobnr = mysql_num_rows($job); if($jobnr == "0") { ?> <form action="<?php echo "$PHP_SELF"; ?>" method="POST"> <input type="submit" name="submit" value="Start Job"></form> <?php } if(isset($startjob)) { $initemidq = "SELECT * FROM items ORDER BY RAND() LIMIT 1"; $initemid = mysql_query($initemidq); while($ir = mysql_fetch_array($initemid)) { $ids = $ir['itemid']; } mysql_query("INSERT INTO jobs (username, item, time, completed) VALUES ('$showusername', '$ids', 'None', 'No')"); $wegq = "SELECT * FROM items WHERE itemid='$ids'"; $weg = mysql_query($wegq); while($wg = mysql_fetch_array($weg)) { $im = $wg['image']; $nm = $wg['name']; $id = $wg['itemid']; } $_SESSION['theid'] = $id; $yeshere = $_SESSION['theid']; echo "<font color=green>Success! You have started this Job!</font><br><br>Please bring me this item: <b>$nm</b><br><br><img src=/images/items/$im><br><br><br>"; } if($jobnr == "1") { $yeshere = $_SESSION['theid']; $finish = $_POST['finish']; $quit = $_POST['quit']; $okgq = "SELECT * FROM items WHERE itemid='$yeshere'"; $ok = mysql_query($okgq); while($ya = mysql_fetch_array($ok)) { $okname = $ya['name']; $okid = $ya['itemid']; $okimage = $ya['image']; } $yeshere = $_SESSION['theid']; echo "Where is my <b>$okname</b>?<br><br><img src=/images/items/$okimage><br><br><br>"; ?> <form action="<?php echo "$PHP_SELF"; ?>" method="POST"> <input type="submit" name="finish" value="I have the Item"><br><br> <input type="submit" name="quit" value="Quit"></form> <?php } } if(isset($finish)) { $yeshere = $_SESSION['theid']; $cinq = "SELECT * FROM uitems WHERE theitemid='$_SESSION[theid]'"; $cin = mysql_query($cinq); $connr = mysql_num_rows($cin); if($connr != "0") { mysql_query("DELETE FROM uitems WHERE username='$showusername' AND theitemid='$yeshere' LIMIT 1"); mysql_query("UPDATE users SET jobs=jobs+1 WHERE username='$showusername'"); mysql_query("UPDATE users SET credits=credits+320 WHERE username='$showusername'"); mysql_query("DELETE FROM jobs WHERE username='$showusername'"); echo "<font color=green>Success! You have completed this job! You have been given <b>320</b> credits as an award. Thank You!</font>"; } else { echo "<font color=red>Error! You do not have my item!</font>"; } if(isset($quit)) { mysql_query("DELETE FROM jobs WHERE username='$showusername'"); echo "<font color=green>Success! You have quit this quest.</font>"; } $yeshere = $_SESSION['theid']; } ?> The variable for the button is: $quit = $_post and I want the quit button to work. Does anybody know why it will not work? Thank you in advance! I trying to make it so i can in put the data from a database into a theme i got from theme forest but im having few problems. the theme and page http://warp.nazwa.pl/dc/innovation/portfolio.html The NEXT button on the portfolio see how it slides across and works nicely here but on my page http://www.jigsawsoulmedia.com/root/Files/jigsawsoulmedia.com/_public/template.php its stopped working since I've place my php code in there and i can't see why not, everything look right to be from the html and php. Could one you lovely people see what you can spot wrong. page code with php below <!-- PAGE CONTENT HERE, PORTFOLIO LIST AND RIGHT SIDE BAR --> <div id="portoflioHeaderContainer"> <!-- PAGE TITLE AND NAVIGATION TREE --> <h1 class="commonPageTitle">Portfolio</h1> <div id="navigationTreeContainer"> <a href="index.html" class="prev">Home</a> \ <a class="current">Portfolio</a> </div> <!-- navigationTreeContainer --> <!-- SHORT TEXT THAT DESCRIBE PAGE CONTENT --> <p class="commonIntroductionText"> Sed ut perspiciatis unde omnis iste natus error sit voluptatem accusantium doloremque laudantium, totam rem aperiam, eaque ipsa quae. Nemo enim ipsam voluptatem quia <span class="spanBold">voluptas sit aspernatur</span> aut odit aut fugit, sed quia consequuntur magni dolores eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet consectetur, adipisci velit. </p> <!-- PORTFOLIO STATISTICS, CURRENT/ALL PAGE AND CURRENT/ALL PROJECT NUMBER --> <div class="portfolioStatisticsContainer"> <div class="pageStatsWrapper"> <span id="pageNumber"></span><span id="pageCount"></span> </div> <div class="imageStatsWrapper"> <span id="hoveredImageIndex">Project: 1/</span><span id="numberOfImages">0</span> </div> </div> <!-- portfolioStatisticsContainer --> </div> <!-- portoflioHeaderContainer --> <div id="portfolioContainer"> <div class="portfolioPage"> <?php $result = "SELECT * FROM portfolio"; $result = mysql_query ($result) or die (mysql_error()); $i=0; while($row = mysql_fetch_assoc($result)) { if($i==2) $divclass = 'portfolioProjectWrapper borderWhite'; else $divclass = 'portfolioProjectWrapper borderGray'; echo ' <div class="'.$divclass.'"> <a href="portfolioPage.html" class="image asyncImgLoad" title="img/'.$row['image290x290'].'"></a> <p class="imageDesc">'.$row['image290x290_phot'].'</p> <h3 class="title">'.$row['title'].'</h3> <p class="subtitle">'.$row['subtitle'].'</p> <p class="desc"> '.substr($row['description'], 0, 1200).' <a href="portfolioPage.html" class="commonLink">Read more</a> </p> </div> '; if($i==2) $i=0; else $i++; } ?> </div><!-- portfolioPage --> </div> <!-- portfolioContainer --> <!-- PORTFOLIO CONTROL PANEL --> <div id="portfolioControlPanel"> <div id="portfolioPrevPageBtn">Prev page</div> <div id="portfolioNextPageBtn">Next page</div> </div> <!-- portfolioControlPanel --> <div class="clearBoth"></div> Thanks for taking a look! Right ive got a user profile that i want a add friend button but i coded a little something what i fort wud work but no luck <?php session_start(); include "includes/db_connect.php"; include "includes/functions.php"; include"includes/smile.php"; logincheck(); $username=$_SESSION['username']; $viewuser=$_GET['viewuser']; $fetch=mysql_fetch_object(mysql_query("SELECT * FROM users WHERE username='$viewuser'")); if (!$fetch){ echo "No such user"; $totalf = mysql_num_rows(mysql_query("SELECT * FROM friends WHERE username = '$viewuser' AND active='1'")); $invite_text="<div>$username Has Sent You A Friend Request<br> <input name=Yes_Accept type=submit id=yes value=Accept Invite class=abutton> <input name=No_accept type=submit value=Decline Invite class=abutton></div><input type=hidden name=invite_id value=$bar2>"; if (($_GET['fri'])){ $exicst=mysql_query("SELECT * FROM users WHERE username='$viewuser'"); $nums=mysql_num_rows($exicst); $adding=mysql_fetch_object($exicst); $already=mysql_num_rows(mysql_query("SELECT * FROM friends WHERE type='Friend' AND person='$viewuser' AND username='$username'")); if ($already != "0"){ echo "<center><font color=orange><b><br>This user is already your friend.<br><br></font>"; }elseif ($already == "0"){ mysql_query("INSERT INTO `friends` ( `id` , `username` , `person` , `type` , `active`) VALUES ( '', '$username', '$viewuser', 'Friend' , '0' )"); mysql_query("INSERT INTO `friends` ( `id` , `username` , `person` , `type` , `active`) VALUES ( '', '$viewuser', '$username', 'Friend' , '0' )"); mysql_query("INSERT INTO `inbox` ( `id` , `to` , `from` , `message` , `subject` , `date` , `read`) VALUES ( '', '$viewuser', '$username', '$invite_text' , 'Friend Request' , '$date' , '0' )"); $bar2=mysql_insert_id(); echo "<center><font color=orange><br>Your Friend Invitation Was Sent To $viewuser<br><br></font>"; exit(); } }} ?> <a href=?fri=Yes>Add Friend +</a> It just adds a blank person and comes back with No Such User and Your Friend Invitation Was Sent To I think ive put some things in the wrong place to be honest but as im not a pro i easily miss things This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=356816.0 This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=307745.0 Hi, there is probably a straightforward solution to my problem, but I can't quite work it out. I use the following PHP line to check whether the submit button in an html form was pressed or not: Code: [Select] if (isset($_POST['buttonname'])==FALSE) {do something} If the html form is contained within a PHP script as follows: Code: [Select] <?php print" <html> <body> <form action=check.php method=POST //form here </form> </body> </html> "; ?> then the check works fine. But if the html form is contained within an external html file that is called upon in an iframe as follows: Code: [Select] <?php print" <html> <body> <iframe src ='form.htm'></iframe> </body> </html> "; ?> then the check doesn't work and it always thinks the submit button wasn't pressed. How can I perform this check when the submit button is in an external html file that is called upon in an iframe? Any help would be much appreciated! Thanks. I need to have several sets of radio buttons, each of which will be used to assess criteria, e.g. endurance, strength, posture, etc. I've made a function, as follows: Code: [Select] <?php function makeRadio($min, $max, $lbl) { echo "<div class='inline_label'>".$lbl."</div>"; for ($i = $min; $i <= $max; $i++) { echo "<input type='radio' name='".$lbl."' value='".$i."' id='".$lbl.$i."_".($i-1)."'"; if(isset($_POST[$lbl])) { echo " checked='checked'"; } echo "/>".$i." "; } echo "<br />"; } I'm calling the function like this: Code: [Select] <form id="form1" name="form1" method="post" action="" > <?php makeRadio(1,10, 'Endurance'); makeRadio(1,10, 'Strength') ?> <p><input type="submit" name="submit" id="submit" value="Submit" /></p> </form> On form submission, however, I can't get the buttons to be sticky? Where am I going wrong in my function code? TIA Friends I am new to php and i have to submit my coursework in php by 3rd dec, I stuck at one place where i have to upload multiple photo and one can see all the photo he has uploaded and can edit or delete that photo so i have done uploading now i am showing those pics in table by running loop and generating tr and td but now i have two buttons with each row edit and delete now when i clicked on one delete or edit that pic should be delete or give text box to edit description of pic, Please help me how to do that....... Hello.
i am totally new to php and just started to learn now. i just dont understand why the following code is not printing the username that i enter on the page.
Please note that the code itself is saved with the name "basicForm.php".
Thanks.
<html> Code: [Select] $query ="SELECT oneID FROM table WHERE table.PersonID = 'game.PlayerA'" ; $result = mysql_query($query); $row = mysql_fetch_array($result); $oneID = $row[0]; [code] If I then echo "$oneID" why does it not print anything? $result echos resource7 Hello, How to customize print page in php ? I had a page, but I need to print it like an invoice look page. Thanks in advance Is there a way to dynamically print the url of a web page once it loads? If so, how? This is for metadata purposes. Thanks! I spent some time "Googling" today but all I found was how to create PDFs on the fly. What I wonder is if it is possible to print text on an existing PDF. And if so, could I get some pointer where to obtain this knowledge? Simply put. I have a PDF form that is not yet filled out. I would like to be able to look up an address and other information in a database, and use that information to print on the existing PDF file. Any help is much appreciated. |