PHP - Php String Variable In A Html Form Value Parameter ?
Hi
I struggle with something i'm trying to achieve. Maybe this is not the correct approach, i don't know. My goal is to attribute a string of a field of the form to a string variable in order to use this string variable in order to "auto fill" the field of the form previously completed. This is to do what is apparently called a sticky form... First the program attributes the string of a field of the form to a string variable : $EmitterFirstName = $_POST["EmitterFirstName"]; Then in the html form there is a field called "EmitterFirstName", and i want to use the string variable $EmitterFirstName in order to auto fill the field with the string previously typed by the user : <input type="text" name="EmitterFirstName" value="<?php $EmitterFirstName ?>"/> But it doesn't work as expected. Do you have any idea why or maybe an advice on how to code such a thing ? Thanks, Similar TutorialsI have a php string variable that is created by php code within an html form ($answer). I need to pass this string variable along with all the html form input data to another php script specified with the form "action" (post method). All the html form input data is coming thru fine but not the variable ($answer). How do I do this? Here is the php code for importing html form data at the script called in the form action: $languages = $_POST['languages']; $answergiven = $_POST['answergiven']; $problemanswer = $_POST['$answer']; 'languages' and 'answergiven' are form inputs and come thru fine. '$answer' does not get passed to the second script. How do I do this? Here is the php code within the first html form <?php // OPEN DATABASE $username="servics3_sample"; $password="sample"; $database="servics3_sample"; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); // GENERATE RANDOM PROBLEM NUMBER $probnum = (rand ( 1 , 9 )); echo $probnum; // RETRIEVE ANTI-SPAM PROBLEM $query="SELECT * FROM liasantispam WHERE `problem number` LIKE '%$probnum%' "; $result=mysql_query($query) or die("Error: ". mysql_error(). " with query ". $query); $firstnum=mysql_result($result,0,"first number"); $operator=mysql_result($result,0,"operator"); $secondnum=mysql_result($result,0,"second number"); $answer=mysql_result($result,0,"answer"); echo $firstnum," ",$operator," ",$secondnum," = "; mysql_close(); ?> Hi all I am in the process of creating a product page where the user can type in three initials into a form and then 'clicks add to basket'. The button doesn't use a form just a url: basket.php?action=add&id=1&qty=1 This adds the item into a basket using sessions rather than a database. The question is, how do I add three variables into my form so that they are sent via the URL and into the session? Wil I have to start again with a form? Here's the code: <td class="table-font-grey">Why not make your whip extra special by adding up to three elegantly engraved Kunstler script initials.</td> <td width="42" style="border: 1px solid grey;"><input name="initial_1" type="text" size="8" maxlength="1" height="40px" /></td> <td width="43" style="border: 1px solid grey;"><input name="initial_2" type="text" size="8" maxlength="1" height="40px" /></td> <td width="43" style="border: 1px solid grey;"><input name="initial_3" type="text" size="8" maxlength="1" height="40px" /></td> </tr> <tr> <td height="15"> </td> <td width="328" class="table-font"> </td> <td width="42" align="center" class="table-font">1</td> <td width="43" align="center" class="table-font">2</td> <td align="center" class="table-font">3</td> </tr> <tr> <td> </td> <td colspan="3" class="table-font"> </td> <td align="right" class="table-font"> </td> </tr> </table> <hr /> <p> </p> <table width="494" border="0"> <tr class="table-font"> <td width="263" height="24"> <a href="<?php echo "basket.php?action=add&id=1&qty=1" ?>" title="Add <?php echo $product['name'] ?> to Cart"> <a href="<?php echo WEB_URL."basket.php?action=add&id=1&qty=1" ?>" title="Add <?php echo $product['name'] ?> to Cart"> <img src="<?php echo WEB_URL."images/product-add-button.gif" ?>" alt="Buy <?php echo $product['name'] ?>" /></a> Many thanks Pete apparently I am doing this wrong. I want my string results from a form that searches DB content to appear within HTML table, tr, td tags. I get the results fine, but the HTML part isn't appearing. How should I be doing this? <?php $string = ''; $result = mysql_query($sql); /// This is the execution if (mysql_num_rows($result) > 0){ while($row = mysql_fetch_object($result)){ echo "<table>"; echo "<tr>"; $string .= "<td>".$row->last_name."</td> "; $string .= "<td>".$row->first_name."</td>"; $string .= "<td>".$row->employee_id."</td>"; $string .= "<td>".$row->title."</b>"; $string .= "<td>".$row->territory."</td>"; $string .= "<td>".$row->district."</td>"; $string .= "<td>".$row->Phase1A_Score."</td>"; $string .= "<td>".$row->Phase1B_Score."</td>"; $string .= "<td>".$row->Phase1_Average."</td>"; $string .= "<td>".$row->Phase1A_HS_Exam."</td>"; $string .= "<td>".$row->Phase1A_HS_Exam_RT."</td>"; $string .= "<td>".$row->Phase1B_HS_Exam."</td>"; $string .= "<td>".$row->Phase1B_HS_Exam_RT."</td>"; $string .= "<td>".$row->Class_Date."</td>"; $string .= "<td>".$row->Awards."</td>"; $string .= "<br/>\n"; echo "</tr>"; echo "</table>"; } }else{ $string = "No matches found!"; } echo $string; ?> Hi, I have a Flash contact form that sends its name, e-mail, and message variables to a PHP script. The script works fine if the message is under ~700 characters, but anything more and the script won't post the message at all. An e-mail with just the name and e-mail address will come through to my inbox (which is embarrassing as a potential client sent me a contact form e-mail about a development job ). Is there some way to check the maximum character length for a string variable? Or to set its length to something higher? This is my php. Thank you!! <? $senderName = $_POST['userName']; $senderEmail = $_POST['userEmail']; $senderMessage = $_POST['userMsg']; $senderName = stripslashes($senderName); $senderMessage = stripslashes($senderMessage); $senderMessage = strip_tags($senderMessage, '<p><br>'); $to = "kevin@kevinburkeportfolio.com"; $from = "email@kevinburkeportfolio.com"; $subject = "E-mail from $senderName"; $message = "<p><font size='2.5' face='Helvetica Neue, Helvetica, Sans Serif'><b>FROM:</b> $senderName <br><br> <b>E-MAIL:</b> <a href='mailto:$senderEmail'>$senderEmail</a> <br><br> <b>MESSAGE:</b> $senderMessage</p>"; $headers = "From: $from\r\n"; $headers = "MIME-Version: 1.0\n"; $headers .= 'Content-type: text/html; charset=iso-8859-1' . "\r\n"; $to = "$to"; mail($to, $subject, $message, $headers); $my_msg = "Thanks $senderName, your message has been sent."; print "return_msg=$my_msg"; exit(); ?> Hey all, When someone clicks a link, link looks like this: http://site/homes/edit/1?image=true Now on the same page, I have another link, which posts to the edit method as well, but no query string passed: http://site/homes/edit/1 I want to check if image key is set to true and basd on that, determine which view to render for the user, that is, whether to display the edit form to edit post or the image form to edit image. But by using the get array to try to grab query string: Code: [Select] if($_GET["image"]=="true"){ I get the following error: Message: Undefined index: image It appears that $_GET is not successfully grabbing the parameter. Thanks for response. I have a simple form and need to convert it to XML so I can send it to my payment gateway. I have the XML data types (?) they are expecting, but do not have a clue of how to build the string. Can someone help me out? Thanks, Debbie Warning: mysql_query() expects parameter 1 to be string, resource given in C:\wamp\www\mariyano\profile.php on line 37 Help me to solve this problem I just want to display the datas in table Thanks in advance hi, I need a code that switch the background of a stylesheet according to url parameter. I am using this (it wont work): <div class="wFormContainer" style="width:880px; background-image: url(<?(empty($_GET['iswoman'])) ? 'l_banner1.jpg' : 'l_banner1_man.jpg';?>);background-color:#FFFFFF;background-repeat: no-repeat;margin:auto;"> please help This is my code $file = "E:/wamp/www/Project/Changes/".$LineNo.".txt"; if(file_exists($file)) { fopen($file,'a') or die("can't open file"); fwrite($file,$Username); } else { fopen($file,'w'); fwrite($file,$Username); } I have a strange error from the code below. I have run the query with the resulting values in PHPMyAdmin and it works fine. The code is: $query = "SELECT * FROM sn_matches WHERE (P1_ID='".$P1_ID."' AND P2_ID='".$P2_ID."') OR (P1_ID='".$P2_ID."' AND P2_ID='".$P1_ID."')"; while($row = mysql_fetch_assoc($query)){ The error is: Quote Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in /customers/ronflorax.com/ronflorax.com/httpd.www/snookerstats/h2h.php on line 21 Line 21 is the one which has mysql_fetch_assoc on it. Any help would be much appreciated! Hi, I'm getting this error: Warning: mktime() expects parameter 5 to be long, string given in /home/user/public_html/include/functions.php on line 58 The code is: if($Date_Format == 1) #For dd-mm-yyyy { $Date_Output = date("d-m-Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 2) #For dd/mm/yyyy { $Date_Output = date("d/m/Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 3) #For mm-dd-yyyy { $Date_Output = date("m-d-Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 4) #For mm/dd/yyyy { $Date_Output = date("m/d/Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 5) #For Mon Day YYYY { $Date_Output = date("M D Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 6) #For Month Day YYYY { $Date_Output = date("F D Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 7) #For Date"nd" Month YYYY { $Date_Output = date("d\\t\h F Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 8) #For Month Day YYYY { $Date_Output = date("F d Y",mktime(0,0,0,$Month,$Day,$Year)); } elseif($Date_Format == 9) #For Month Day YYYY { $Date_Output = date("d/m/Y H:i",mktime($DateDispsplittimehr,$DateDispsplittimemin,0,$Month,$Day,$Year)); // Line 58 } return $Date_Output; } else { return NULL; Line 58 says $Date_Output = date("d/m/Y H:i",mktime($DateDispsplittimehr,$DateDispsplittimemin,0,$Month,$Day,$Year)); I really appreciate any help on this. Here is the error: Code: [Select] Warning: image_type_to_extension() expects parameter 1 to be long, string given in C:\wamp\www\Login\inc\edit-profile.php on line 23 Here is the code for line 23: Code: [Select] $type = image_type_to_extension($_FILES['pic']['type'], true); Here is the code for the whole page: Code: [Select] <?php if(!$userb) { login('?p=edit-profile'); die(); } if(isset($_POST['submit'])) { $mottof = mysql_real_escape_string(bb($_POST['motto'])); $aboutf = mysql_real_escape_string(bb($_POST['about'])); $interesrsf = mysql_real_escape_string(bb($_POST['interests'])); $booksf = mysql_real_escape_string(bb($_POST['books'])); $authorsf = mysql_real_escape_string(bb($_POST['author'])); $randf = mysql_real_escape_string(bb($_POST['random'])); if($_FILES['pic']['type'] == 'image/gif' || $_FILES['pic']['type'] == 'image/jpeg' || $_FILES['pic']['type'] =-'image/pjpeg' || $_FILES['pic']['type'] == 'image/png') { if($_FILES['pic']['size'] < 8388608) { if($_FILES['pic']['name'] != '') { //mkdir('/images/user_images/' .$user_log); $type = image_type_to_extension($_FILES['pic']['type'], true); move_uploaded_file($_FILES['pic']['tmp_name'], 'images/user_images/' .$user_log .'pp' .$type); $path = 'images/user_images/' .$user_log .'pp' .$type; } else { $path = 'images/user_images/user.jpg';} } $q2 = "UPDATE login_info SET profile_picture = '$path' WHERE user='$user_log' "; mysql_query($q2) or die(mysql_error()); } $query = " UPDATE login_info SET motto = '$mottof', about = '$aboutf', fav_books = '$booksf', fav_authors = '$authorsf', interests = '$interesrsf', random = '$randf' WHERE user='$user_log' "; mysql_query($query); echo '<h3> Profile Updated </h3>'; } $q = "SELECT * FROM login_info WHERE user='$user_log' "; $s = mysql_query($q) or die(mysql_error()); $r = mysql_fetch_assoc($s); $motto = $r['motto']; $about = $r['about']; $books= $r['fav_books']; $authors = $r['fav_authors']; $rand = $r['random']; $interests = $r['Interests']; if(isset($_GET['act'])){$red = '?p=' .$rpage ;} else {$red='?p=edit_profile';} echo" <form action='?p=edit-profile' enctype='multipart/form-data' method='post' > <label>Motto: </label> <input type='text' name='motto' value='$motto' placeholder='Motto' /> <label> About :</label> <textarea name='about' cols='40' rows='10'> $about</textarea> <label> Interests: </label> <textarea name='interests' cols='40' rows='10'> $interests </textarea> <label> Favorite Books: </label> <textarea name='books' cols='40' rows='10'> $books </textarea> <label> Favorite Authors </label> <textarea name='author' cols='40' rows='10'> $authors </textarea> <label> Random: </label> <textarea name='random' cols='40' rows='10'> $rand </textarea> "; ?> <script type='text/javascript'> function if_checked() { if($('#cur').is(':checked')) { $('#pic').attr('disabled', true); } else { $('#pic').removeAttr('disabled'); } } </script> <label> Profile Pictu </label> <input type='checkbox' name= 'current' id='cur' onchange="if_checked();" checked /> <label for='current' style='display:inline;'> Current profile picture </label> <br /> <input type='file' id='pic' name='pic' disabled="disabled" /> <br /> <input type='submit' value ='Submit' name='submit' /> </form> Ive used the function before and I havent encountered this error. Im not sure what it means. Hey, all. I'm new to php/mysql. I'm getting the warning: Quote mysql_fetch_array() expects parameter 1 to be resource, string...line 39 with reference to the mysql_fetch_array function in the following code: mysql_select_db("calculators", $con); $query = "SELECT * FROM calculator WHERE Abbreviation = '" . $_GET['abbreviation'] . "'"; $result = mysql_query($query, $con); while($row = mysql_fetch_array($result)) { //some code... } The code runs fine, but I'd like to know why I'm getting the warning. I've looked at similar post on this forum and other forums and can't quite get a straight answer. Thanks a bunch, Davis PHP Warning: file_exists() expects parameter 1 to be string, array given in /home/mysite/public_html/display.mysite.com/wp-content/themes/mytheme/form.php on line 9 PHP Warning: file_exists() expects parameter 1 to be string, array given in /home/mysite/public_html/display.mysite.com/wp-content/themes/mytheme/form.php on line 13I have a form that is having some issues and in the error log I see a bunch of these warnings. What is wrong with this?? I numbered the offending lines below (lines 9 & 13) Thanks for any help on this. function CheckExistance($VUrl) { /*9*/ if ( file_exists($VUrl) ) echo $VUrl; /*13*/ elseif ( file_exists(str_replace('www.display.mysite.com','www.display.com', $VUrl)) ) echo str_replace('www.display.mysite.com','www.mysite.com', $VUrl); else echo str_replace('www.mysite.com','www.display.mysite.com', $VUrl); } Edited by damion, 02 July 2014 - 06:01 PM. Hi all I have a field in mySQL table called dimensions. It has the double quote in in for inches - " When I echo the result from the mySQL query on the item page (Customer facing) it's fine. However, I have built a form so that the administrator can edit the dimensions in the admin panel and when I echo it out in to the form field it stops when it gets to the double quotes? Pete Hi, I am trying to make some adjustments to uploadify.php which comes with the latest version of uploadify (3.0 beta), so that it works with a session variable that stores the login username and adds it to the path for uploads. Here is uploadify.php as it currently looks: Code: [Select] <?php session_name("MyLogin"); session_start(); $targetFolder = '/songs/' . $_SESSION['name']; // Relative to the root if (!empty($_FILES)) { $tempFile = $_FILES['Filedata']['tmp_name']; $targetPath = $_SERVER['DOCUMENT_ROOT'] . $targetFolder; $targetFile = rtrim($targetPath,'/') .'/'. $_FILES['Filedata']['name']; // Validate the file type $fileTypes = array('m4a','mp3','flac','ogg'); // File extensions $fileParts = pathinfo($_FILES['Filedata']['name']); if (in_array($fileParts['extension'],$fileTypes)) { move_uploaded_file($tempFile,$targetFile); echo '1'; } else { echo 'Invalid file type.'; } } echo $targetFolder; ?> I added Code: [Select] echo $targetFolder; at the bottom so that I could make sure that the string returned was correct, and it is, i.e. '/songs/nick'. For some reason though, uploads are not going to the correct folder, i.e. the username folder, but instead are going to the parent folder 'songs'. The folder for username exists, with correct permissions, and when I manually enter Code: [Select] $targetFolder = '/songs/nick';all works fine. Which strikes me as rather strange. I have limited experience of using php, but wonder how if the correct string is returned by the session variable, the upload works differently than with the manually entered string. Any help would be much appreciated. It's the last issue with a website that was due to go live 2 days ago! Thanks, Nick if($pid != "") { $bname = $_REQUEST['bname']; $btitle = $_REQUEST['btitle']; $btags = $_REQUEST['btags']; $bdesc = $_REQUEST['bdesc']; $btext = $_REQUEST['btext']; $bimg = $_REQUEST['bimg']; $bimgalt = $_REQUEST['bimgalt']; $data = ""; if($bname!="") { $data = $data." 'bname' => ".$bname.", "; } if($btitle!="") { $data = $data."'btitle' => ".$btitle.", "; } if($btags!="") { $data = $data."'btags' => ".$btags.", "; } if($bdesc!="") { $data = $data."'bdesc' => ".$bdesc.", "; } if($btext!="") { $data = $data."'btext' => ".$btext.", "; } if($bimg!="") { $data = $data."'bimg' => ".$bimg.", "; } if($bimgalt!="") { $data = $data."'bimgalt' => ".$bimgalt.", "; } $data = $data."'pid ' =>". $pid.", " ; $data = "[".$data."]"; $build = ""; if($bname!="") { $build = $build." page_name = :bname,"; } if($btitle!="") { $build = $build." page_title = :btitle,"; } if($btags!="") { $build = $build." page_tags = :btags,"; } if($bdesc!="") { $build = $build." page_desc = :bdesc,"; } if($btext!="") { $build = $build." page_text = :btext,"; } if($bimg!="") { $build = $build." page_img = :bimg,"; } if($bimgalt!="") { $build = $build." page_imgalt = :bimgalt,"; } $build = $build." page_id = :pid"; $sql = "UPDATE pages SET ".$build." WHERE page_id=:pid"; echo $sql."<br /><br />"; echo $data."<br /><br />"; $stmt= $pdo->prepare($sql); $stmt->execute($data); Result of echo SQL and Data
UPDATE pages SET page_name = :bname, page_title = :btitle, page_tags = :btags, page_desc = :bdesc, page_text = :btext, page_img = :bimg, page_imgalt = :bimgalt, page_id = :pid WHERE page_id=:pid Market Net is the place for you to find those goodies you saw on the markets, buy them online and support small local businesses. Watch this space for more information. Test #D hvac , 'bimg' => supportlocal.jpg, 'bimgalt' => Support Local Small Businesses, 'pid ' =>106, ] Yet I am getting said error message. What am I doing wrong? Hi all I am in the process of creating a product page where the user can type in three initials into a form and then 'clicks add to basket'. The button doesn't use a form just a url: basket.php?action=add&id=1&qty=1 This adds the item into a basket using sessions rather than a database. The question is, how do I add three variables into my form so that they are sent via the URL and into the session? Wil I have to start again with a form? Here's the code: <td class="table-font-grey">Why not make your whip extra special by adding up to three elegantly engraved Kunstler script initials.</td> <td width="42" style="border: 1px solid grey;"><input name="initial_1" type="text" size="8" maxlength="1" height="40px" /></td> <td width="43" style="border: 1px solid grey;"><input name="initial_2" type="text" size="8" maxlength="1" height="40px" /></td> <td width="43" style="border: 1px solid grey;"><input name="initial_3" type="text" size="8" maxlength="1" height="40px" /></td> </tr> <tr> <td height="15"> </td> <td width="328" class="table-font"> </td> <td width="42" align="center" class="table-font">1</td> <td width="43" align="center" class="table-font">2</td> <td align="center" class="table-font">3</td> </tr> <tr> <td> </td> <td colspan="3" class="table-font"> </td> <td align="right" class="table-font"> </td> </tr> </table> <hr /> <p> </p> <table width="494" border="0"> <tr class="table-font"> <td width="263" height="24"> <a href="<?php echo "basket.php?action=add&id=1&qty=1" ?>" title="Add <?php echo $product['name'] ?> to Cart"> <a href="<?php echo WEB_URL."basket.php?action=add&id=1&qty=1" ?>" title="Add <?php echo $product['name'] ?> to Cart"> <img src="<?php echo WEB_URL."images/product-add-button.gif" ?>" alt="Buy <?php echo $product['name'] ?>" /></a> Many thanks Pete I am experimenting around with PDO, I have a DB handle class and instancing it in a function. The problem is when I want to bind elements in a query, for example I bind a parameter but I can only do it using a varible outside of the object. Code: [Select] <?php $database = new Database("localhost", "fry", "root", ""); $database->set_table("usertest"); $database->set_query("SELECT * FROM usertest WHERE user_name = :user_name"); $value = "matt"; $database->prepare_query(); $database->bind("parameter", ":user_name", $value ,PDO::PARAM_STR, 5); $database->execute(); while($result = $database->fetch()) { echo $result['user_name'].'<br />'; } ?> Is this bad OOP practice? I wanted to put $value = "matt"; into the Database class somehow here is the Database class Code: [Select] <?php class Database{ public $hostname; public $database; public $username; public $password; public $connection; public $prepare; public $query; public $table; public $fetch; public $bind_var; public $bind_val; function __construct($hostname, $database, $username, $password) { $this->hostname = $hostname; $this->database = $database; $this->username = $username; $this->password = $password; $this->Database_connection(); } public function Database_connection() { try { $this->connection = new PDO('mysql:host='.$this->hostname.';dbname='.$this->database, $this->username, $this->password); } catch (PDOException $e) { echo 'Connection failed: ' . $e->getMessage(); } } public function set_query($query) { $this->query = $query; } public function get_query() { return $this->query; } public function set_table($table) { $this->table = $table; } public function get_table() { return $this->table; } public function set_bind_var($var) { $bind_var = $var; } public function set_bind_val($val) { $bind_val = $val; } public function get_bind_var() { return $this->bind_var; } public function get_bind_val() { return $this->bind_val; } public function bind($bindtype, $val, $var, $pdo, $num) { if ($bindtype == "parameter") { $result = $this->prepare->bindParam($val,$var,$pdo); } return $result; } public function prepare_query() { $this->prepare = $this->connection->prepare($this->get_query()); } public function execute() { $this->prepare->execute(); } public function fetch() { return $this->prepare->fetch(); } } ?> Yet agian sorry for the sloppy code, this is mainly an experiment, would be very appriciated if someone could help me out with this problem. |