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PHP - Is There A Way To Enter The Chosen Form Value, Without Having To Submit First?
Hello PHP world! I have this roll-down-select-menu working, reflecting a column in my database. I am able to retrieve the selected value after submit, that takes me to another code page. But I would like to use this value on the form page, that contains the roll-down-menu and the submit button. Is there a way to enter the chosen value into this variable $uan, without having to submit the page first? I tried to retrieve the value on this page with the POST lines at the button, but without luck. I think the value is only entered upon submit? Thanks in advance for your clever advice. Best regards Morris <?php $query = "SELECT * FROM Uniq_artist_name" ; $result = mysql_query($query); echo'<select name="uan">'; while($row = mysql_fetch_assoc( $result )) { echo '<option value="'.$row['uniqartistname'].'">'. $row['uniqartistname'] .'</option>'; } echo '</select>'; $uan = $_POST["uan"]; echo "$uan"; ?> Similar TutorialsHi guys. As I am new to PHP, I have a question. Necessarily, this need not be done in php, if you have something in html or js that can be done, let me know! Well, I have a form where you enter information like name, city, phone and product option. These product options include real estate and auto options. By clicking submit (to submit form), I need to direct the visitor to their chosen product option. For example, when I filled out the form I chose 'real estate'. When I clicked submit, it went to a page where the real estate-only content appeared. Note: I already have this code ready below and it already directs the information to the database. However, I don't know how to make him redirect to the customer, the product option he chose.
<form action="simulador.php" method="post" name="dados" id="dados" onSubmit="return validaform()">
<div class="col-md-5 esquerda">
Selecione o bem<br/>
<select name="tipo" type="text" class="contat3" placeholder="Selecione o bem">
<option value="Im�vel""
style="background-color: #fff;">Imóveis</option>
<option value="Autom�vel""
style="background-color: #fff;">Automóveis</option>
<option value="Moto" style="background-color: #fff;">Motos</option>
</select><br/>
Selecione o plano<br/>
<select name="plano" type="text" class="contat3">
<option value="Crédito""
style="background-color: #fff;">Crédito</option>
<option value="Parcela""
style="background-color: #fff;">Parcela</option>
</select><br/>
<input name="valorcon" type="text" id="valorcon" class="contat3" placeholder="Digite o valor" maxlength="1000" /><br/>
<input name="nomecon" type="text" id="nomecon" class="contat3" placeholder="Nome" maxlength="1000" /><br/>
<input name="telefone" type="text"
onkeypress="Mascara('TEL',this,event);"
type="text" id="telefone" class="contat3" placeholder="Telefone" />
<!---
<input name="tel2" type="text" id="tel2" onkeypress="Mascara('TEL',this,event);" /><br/>
--->
<input name="emailcon" type="text" id="emailcon" class="contat3" placeholder="E-mail" maxlength="1000" /><br/>
<input name="cidadecon" type="text" id="cidadecon" class="contat3" placeholder="Cidade" maxlength="1000" />
<br/><br/><br/>
<a id="enviar-form" class="button solid-color" href="#">Enviar</a>
<input type="submit" id="enviar-form-btn" style="display: none;" />
</div>
</form>
Hello! Before I go digging around for code or conversion to unix time stamp, I'm curious how many of you have a birthday entry on a form which is the least hassle for the person filling out the form. For example a text box and they can enter 12/1/1992 or 12/1/92 or 12-1-1992 or three text boxes for month, day and year and if so, require an 01 or a 08 or 4 characters for year. The second method with three boxes would easily convert to a date stamp entry on the other hand if there is a php conversion that would convert any format (12/1/1992, 12/01/92, etc) to a unix time stamp that would make it easier for the person filling out the form. Thanks in advance for any ideas or help. Mike The company I work for is a not-for-profit organization, we provide medical services for people who can not afford to go to the regular doctor. We need a way to keep track of our patients and their visits and a way to search the data. I found an html form, with name, address, phone number, zip code,,,,,, and made a few changes to fit our needs. But I do not know how to get that information into the database. I also need a way to make sure the information is not a duplicate. Since we are a not-for-profit organization, funds are kinda tight. That is why I want to go with php instead of buying medical software. Most of the medical software on the market cost more then we can afford. I'am a computer tech here at the office, but my field is hardware, networking, and malware removal. PHP programming is new to me. So where do I start? How can I build a form with a drop down menu with the year, day, month? And then send that information to a database. If there is another/better option besides a drop down menu? If so, please tell me. The rest of the form is working fine, and all the inforamtion is going to the database - last name, first name, address, phone number, zip code, race,,,,,,,,,. I work for a not-for-profit medical office - we provide medical services to the local people who can not afford to go to a regular doctor. We need a simple way to keep track of patient information. One of the things we need to enter is the birthdate. Searching the forums provided a lot of results, but the words date and form are used so much, I can not find anything that helped me. I also looked through php.net, but can not find anything there. Hello everyone, I have a page with a short form with just a password field and a submit button. I also have php code setting variables depending on whether or not the password is correct. When the password is entered and the submit button is CLICKED, everything works as it should. However, when the password is entered and the ENTER button is pushed to submit the form, the form is reset and nothing else happens. Also, this only happens in I.E., it works fine with Firefox. I have plenty of other forms that I use in the same manner but they all work whether the submit button is clicked or the enter button is pushed to submit, regardless of what browser is used. The idea is to have this window close (it's opened as a pop-up) and the main window load the next page. Again, it works perfectly as long as the submit button is clicked but not if the enter button is pushed. Not sure if it has something to do with my php code or what. Any ideas on how to fix this? Here's my code: Code: [Select] <?php if (isset($_POST['submit'])) { $pword = md5($_POST['pword']); $q = mysql_query("SELECT * FROM `accounts` WHERE acct_number = '$acct' AND pword = '$pword'") or die (mysql_error()); $r = mysql_num_rows($q); // Checks to see if anything is in the db. if ($r == 1) { $student = "true"; } else { $student = "false"; } } else { $student = "unknown"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Student Verification...</title> </head> <!-- <body> tag is in PHP --> <?php if ($student == "true") { echo "<body onunload=\"opener.location=('signup.php?acct1=$acct')\">"; } else { echo "<body>"; } if ($student == "true") { echo "<div align='center'>"; echo "<p><font size='+2'>Passcode Correct!</font></p>"; echo "</div>"; echo "<div align='justify'>"; echo "<p>You will now be directed to the signup page.</p>"; echo "</div>"; echo "<div align='center'>"; echo "<p>Thank you.</p>"; echo "<input type='button' value='Continue to Signup...' onClick='window.close()'>"; echo "</div>"; echo "<meta http-equiv=\"REFRESH\" content=\"20;url=close.php\">"; } elseif ($student == "unknown") { echo "<div align='center'>"; echo "<table width='325' height='175' border='0' cellspacing='0' cellpadding='0'>"; echo "<tr>"; echo "<td width='325' align='center' valign='middle'>"; echo "<p><font size='+1'><strong>Welcome!</strong></font></p>"; echo "<p>Please enter the school passcode.</p>"; echo "<form name='form' method='POST' action='schoolpw.php'>"; echo "<p><input type='password' name='pword' size='30' maxlength='20' /></p>"; echo "<input type='submit' name='submit' id='submit' value='Submit' /> "; echo "<input type='button' name='cancel' value='Cancel' onClick='window.close()' />"; echo "</form>"; echo "</p>"; echo "</td>"; echo "</tr>"; echo "</table>"; echo "</div>"; } elseif ($student == "false") { echo "<div align='center'>"; echo "<table width='325' height='175' border='0' cellspacing='0' cellpadding='0'>"; echo "<tr>"; echo "<td width='325' align='center' valign='middle'>"; echo "<p><font size='+1'><strong>Welcome!</strong></font></p>"; echo "<p><font color='#FF0000'>Incorrect Passcode! Please try again.</font></p>"; echo "<form name='form' method='POST' action='schoolpw.php'>"; echo "<p><input type='password' name='pword' size='30' maxlength='20' /></p>"; echo "<input type='submit' name='submit' value='Submit' /> "; echo "<input type='button' name='cancel' value='Cancel' onClick='window.close()' />"; echo "</form>"; echo "</p>"; echo "</td>"; echo "</tr>"; echo "</table>"; echo "</div>"; } ?> </body> </html> Say I have an "Entries" table. I want to submit same multiple entries using a form submission. And If I have other queries submitted in the same form, I want those quarries to be submitted only once. Is that possible to do? Here's my code.
if(isset($_POST['submit'])) {
$entries = 10;
$id = 55;
$name = 'Smith';
$insert = $db->prepare("INSERT INTO entries(id, name) VALUES(:id, :name)");
$insert->bindParam(':id', $id);
$insert->bindParam(':name', $name);
$result_insert = $insert->execute();
if($result_insert == false) {
echo 'Fail';
} else {
echo 'Success';
}
}
?>
<form action="" method="post">
<input type="submit" name="submit" value="SUBMIT" />
</form>
Edited January 13, 2019 by imgrooot Hi. Pretty straight forward I guess but as the name suggests am a newbie. I have a form that requires the user to enter certain parameters. If the values are blank it submits to itself and loads the error messages. What I want to do is create PHP code that submits the form to a different url. What I thought was create two forms (the second with hidden fields replicating the first form), each form having a different url in the action"" code. What I cant work out is the PHP IF ELSE code to submit form 2 if Form1 is is validated correctly. This is the PHP code relevant to the form validation. Help? <?php //If form was submitted if ($_POST['submitted']==1) { $errormsg = ""; //Initialize errors if ($_POST[width]){ $title = $_POST[width]; //If title was entered } else{ $errormsg = "Please enter width"; } if ($_POST[drop]){ $textentry = $_POST[drop]; //If comment was entered } else{ if ($errormsg){ //If there is already an error, add next error $errormsg = $errormsg . " & content"; }else{ $errormsg = "Please enter drop"; } } } if ($errormsg){ //If any errors display them echo "<div class=\"box red\">$errormsg</div>"; } //If all fields present if ($title && $textentry){ //Do something echo 'THIS IS WHERE I WANT THE CODE TO SUBMIT FORM 2 or SUBMIT FORM 1 TO A DIFFERENT URL'; } ?> I have developed a code for a login and seems to work well (No syntax error according to https://phpcodechecker.com/ but when I enter a username and a password in the login form, I get an error HTTP 500. I think that everything is ok in the code but obviously there is something that I am not thinking about. The code (excluding db connection): $id="''"; $username = $_POST['username']; $password = md5($_POST['password']); $func = "SELECT contrasena FROM users WHERE username='$username'"; $realpassask = $conn->query($func); $realpassaskres = $realpassask->fetch_assoc(); $realpass= $realpassaskres[contrasena]; $func2 = "SELECT bloqueado FROM users WHERE username='$username'"; $blockedask = $conn->query($func2); $blockedres = $blockedask->fetch_assoc(); $bloqueado = $blockedres[bloqueado];
//Login if(!empty($username)) { // Check the email with database Hello, first time poster.. I've looked the web over for a long time and can't figure this one out. - Below is basic code that successfully checks MySQL for a match and displays result. I was debugging and forced the "height" and "width" to be 24 and 36 to make sure that wasn't the problem. That's good.. - I'd like to give the user ability to select width and height from a form.. and have it do an onchange this.form.submit so the form can be changing as fields are altered (thus the onchange interaction) - In a normal coding environment I've done this numerous times with no "Page cannot be displayed" problems. It would simply change one select-option value at a time til they get down the form and click submit... but in WordPress I'm having trouble making even ONE single onchange work! - I've implemented the plugins they offer which allows you to "copy+paste" your php code directly into their wysiwyg editor. That works with basic tests like my first bullet point above. - I've copied and pasted the wordpress url (including the little ?page_id=123) into the form "action" url... that didn't work... tried forcing it into an <option value=""> tag.. didn't work. I'm just not sure. I've obviously put xx's in place of private info.. Why does this form give me Page Cannot Be Displayed in WordPress every time? It won't do anything no matter how simple.. using onchange.. Code.. $con = mysql_connect("xxxx.xxxxxxx.com","xxxxxx","xxxxx"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("xxxxxx", $con); $myprodwidth=24; $myprodheight=36; $result = mysql_query("SELECT * FROM product_sizes WHERE prodwidth='$myprodwidth' and prodheight='$myprodheight'"); while($row = mysql_fetch_array($result)) { echo $row['prodprice']; } mysql_close($con); <form method="post" action=""> <select name="myheight" onchange="this.form.submit();"> <option selected="selected" value="">select height</option> <option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">36</option> <option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">48</option> </select> I have a form where there are various fields one can edit and replace images as well. The problem is when one tries to replace the image they choose. It only replaces the path of the file on column "imageurl1" on the MySQL database and not on the one that was chosen. I'm also getting a "Invalid argument supplied for foreach()" Here is the script Code: [Select] <?php session_start(); include('SimpleImage.php'); $image = new SimpleImage(); //error_reporting(E_ALL); // image upload folder $image_folder = 'images/classified/'; // fieldnames in form $all_file_fields = array('image1', 'image2' ,'image3', 'image4', 'image5', 'image6', 'image7', 'image8', 'image9', 'image10', 'image11', 'image12'); // allowed filetypes $file_types = array('jpg','gif','png'); // max filesize 5mb $max_size = 5000000; //echo'<pre>';print_r($_FILES);exit; $time = time(); $count = 1; foreach($all_file_fields as $fieldname){ if($_FILES[$fieldname]['name'] != ''){ $type = substr($_FILES[$fieldname]['name'], -3, 3); // check filetype if(in_array(strtolower($type), $file_types)){ //check filesize if($_FILES[$fieldname]['size']>$max_size){ $error = "File too big. Max filesize is ".$max_size." MB"; }else{ // new filename $filename = str_replace(' ','',$myusername).'_'.$time.'_'.$count.'.'.$type; // move/upload file $image->load($_FILES[$fieldname]['tmp_name']); if($image->getWidth() > 150) { //if the image is larger that 150. $image->resizeToWidth(500); //resize to 500. } $target_path = $image_folder.basename($filename); //image path. $image->save($target_path); //save image to a directory. //save array with filenames $images[$count] = $image_folder.$filename; $count = $count+1; }//end if }else{ $error = "Please use jpg, gif, png files"; }//end if }//end if }//end foreach if($error != ''){ echo $error; }else{ //error_reporting(E_ALL); //ini_set('display_errors','On'); $id = $_POST['id']; $id = substr($id, 0,5); if($id < 1 || $id > 99999) exit; $servername = "localhost"; $username = ""; $password = ""; if(!$_POST["title"] || !$_POST["rent"] || !$_POST["fees"]){ header('location: fields.php'); }else if (!(preg_match('#^\d+(\.(\d{2}))?$#',($_POST["rent"])))){ header('location: rent.php'); }else{ $conn = mysql_connect($servername,$username,$password)or die(mysql_error()); mysql_select_db("apts",$conn); // validate id belongs to user $sql_check = "SELECT * FROM apts WHERE id = '".$id."' AND username = '".$myusername."'"; $res = mysql_query($sql_check,$conn) or die(mysql_error()); $count = mysql_num_rows($res); if ($count > 0){ $sql_images = ""; foreach($images as $number => $imagedetail){ if($imagedetail != ''){ $sql_images .= "imageurl".$number." = '".mysql_real_escape_string($imagedetail)."',"; } } $sql = "UPDATE apartments SET title = '".mysql_real_escape_string($_POST['title'])."', description = '".mysql_real_escape_string($_POST['description'])."', cross_streets = '".mysql_real_escape_string($_POST['cross_streets'])."', county = '".mysql_real_escape_string($_POST['county'])."', town = '".mysql_real_escape_string($_POST['town'])."', service = '".mysql_real_escape_string($_POST['service'])."', phone = '".mysql_real_escape_string($_POST['phone'])."', contact = '".mysql_real_escape_string($_POST['contact'])."', office = '".mysql_real_escape_string($_POST['office'])."', pets = '".mysql_real_escape_string($_POST['pets'])."', email = '".mysql_real_escape_string($_POST['email'])."', rooms = '".mysql_real_escape_string($_POST['rooms'])."', bath = '".mysql_real_escape_string($_POST['bath'])."', square = '".mysql_real_escape_string($_POST['square'])."', rent = '".mysql_real_escape_string($_POST['rent'])."', fees = '".mysql_real_escape_string($_POST['fees'])."', service = '".mysql_real_escape_string($_POST['service'])."', feeornofee = '".mysql_real_escape_string($_POST['feeornofee'])."', lease = '".mysql_real_escape_string($_POST['lease'])."', video= '".mysql_real_escape_string($_POST['video'])."', zipcode = '".mysql_real_escape_string($_POST['zipcode'])."',".$sql_images." videotitle = '".mysql_real_escape_string($_POST['videotitle'])."' WHERE id = '".$id."'"; //replace info with the table name above $result = mysql_query($sql,$conn) or die(mysql_error()); header('location: apartments.php'); }else{ header('location: somethingwrong.php'); } } } ?> The option chosen in my form disappears -- both the initial default and the option chosen are no longer visible after the value is chosen. I use an ajax call to refresh the page, if that's any clue. Perhaps there's a way I can save the value and write to the drop-down field or something? Mark Hi- the code below lets me upload a CSV file to my database if I have 1 field in my database and 1 column in my CSV. I need to add to my db "player_id" from the CVS file and "event_name" and "event_type" from the form... any ideas??? here's the code: Code: [Select] <?php $hoststring =""; $database = ""; $username = ""; $password = ""; $makeconnection = mysql_pconnect($hoststring, $username, $password); ?> <?php ob_start(); mysql_select_db($database, $makeconnection); $sql_get_players=" SELECT * FROM tabel ORDER BY player_id ASC"; // $get_players = mysql_query($sql_get_players, $makeconnection) or die(mysql_error()); $row_get_players = mysql_fetch_assoc($get_players); // $message = null; $allowed_extensions = array('csv'); $upload_path = '.'; //same directory if (!empty($_FILES['file'])) { if ($_FILES['file']['error'] == 0) { // check extension $file = explode(".", $_FILES['file']['name']); $extension = array_pop($file); if (in_array($extension, $allowed_extensions)) { if (move_uploaded_file($_FILES['file']['tmp_name'], $upload_path.'/'.$_FILES['file']['name'])) { if (($handle = fopen($upload_path.'/'.$_FILES['file']['name'], "r")) !== false) { $keys = array(); $out = array(); $insert = array(); $line = 1; while (($row = fgetcsv($handle, 0, ',', '"')) !== FALSE) { foreach($row as $key => $value) { if ($line === 1) { $keys[$key] = $value; } else { $out[$line][$key] = $value; } } $line++; } fclose($handle); if (!empty($keys) && !empty($out)) { $db = new PDO( 'mysql:host=host;dbname=db', 'user', 'pw'); $db->exec("SET CHARACTER SET utf8"); foreach($out as $key => $value) { $sql = "INSERT INTO `table` (`"; $sql .= implode("`player_id`", $keys); $sql .= "`) VALUES ("; $sql .= implode(", ", array_fill(0, count($keys), "?")); $sql .= ")"; $statement = $db->prepare($sql); $statement->execute($value); } $message = '<span>File has been uploaded successfully</span>'; } } } } else { $message = '<span>Only .csv file format is allowed</span>'; } } else { $message = '<span>There was a problem with your file</span>'; } } ob_flush();?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>CSV File Upload</title> </head> <body> <form class="form" action="" method="post" enctype="multipart/form-data"> <h3>Select Your File</h3> <p><?php echo $message; ?></p> <input type="file" name="file" id="file" size="30" /> <br/> <label>Event Name:</label><input name="event_name" type="text" value="" /> <br/> <label>Event Type:</label><input name="event_type" type="text" value="" /> <br/> <input type="submit" id="btn" class="button" value="Submit" /> </form> <br/> <h3>Results:</h3> <?php do { ?> <p><?php echo $row_get_players['player_id'];?></p> <?php } while ($row_get_players = mysql_fetch_assoc($get_players)); ?> </body> </html> Hi all, What I am trying to achieve is, I thought quite simple! Basically, a user signs up and chooses a package, form is submitted, details added to the database, email sent to customer, then I want to direct them to a paypal payment screen, this is where I am having issues! Is their any way in php to submit a form without user interaction? Here is my code for the form process page Code: [Select] <?php include('config.php'); require('scripts/class.phpmailer.php'); $package = $_POST['select1']; $name = $_POST['name']; $email = $_POST['email']; $password = md5($_POST['password']); $domain = $_POST['domain']; $a_username = $_POST['a_username']; $a_password = $_POST['a_password']; $query=mysql_query("INSERT INTO orders (package, name, email, password, domain, a_username, a_password) VALUES ('$package', '$name', '$email', '$password', '$domain', '$a_username', '$a_password')"); if (!$query) { echo "fail<br>"; echo mysql_error(); } else { $id = mysql_insert_id(); $query1=mysql_query("INSERT INTO customers (id, name, email, password) values ('$id', '$name', '$email', '$password')"); if (!$query1) { echo "fail<br>"; echo mysql_error(); } if($package=="Reseller Hosting") { //email stuff here - all works - just cutting it to keep the code short if(!$mail->Send()) { echo "Message could not be sent. <p>"; echo "Mailer Error: " . $mail->ErrorInfo; exit; } ?> <form name="_xclick" action="https://www.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_xclick-subscriptions"> <input type="hidden" name="business" value="subscription@jollyhosting.com"> <input type="hidden" name="currency_code" value="USD"> <input type="hidden" name="item_name" value="Jolly Hosting Reseller Packages"> <input type="hidden" name="no_shipping" value="1"> <!--1st month --> <input type="hidden" name="currency_code" value="USD"> <input type="hidden" name="a3" value="3.00"> <input type="hidden" name="p3" value="1"> <input type="hidden" name="t3" value="M"> <input type="hidden" name="src" value="1"> <input type="hidden" name="sra" value="1"> </form>'; <?php } //last } //end ?> I am working on a project that uses a drop down list to chose the category when inserting new data into the database. What I want to do now is make the drop down list default to the chosen category on the list records page and the update page. I have read several tutorials, but they all say that I have to list the options and then select the default. But since it is possible to add and remove categories, this approch won't work. I need the code to chose the correct category on the fly. There are two tables, one that has the category ID and category name. The second table has the data and the catid which is referenced to the category id in the first table. Code: [Select] -- -- Table structure for table `categories` -- DROP TABLE IF EXISTS `categories`; CREATE TABLE IF NOT EXISTS `categories` ( `id` int(11) NOT NULL AUTO_INCREMENT, `categories` varchar(37) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ; -- -------------------------------------------------------- -- -- Table structure for table `links` -- DROP TABLE IF EXISTS `links`; CREATE TABLE IF NOT EXISTS `links` ( `id` int(4) NOT NULL AUTO_INCREMENT, `catid` int(11) DEFAULT NULL, `name` varchar(255) NOT NULL DEFAULT '', `url` varchar(255) NOT NULL DEFAULT '', `content` varchar(255) NOT NULL DEFAULT '', PRIMARY KEY (`id`), KEY `catid` (`catid`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ; Then is the list records file, I have Code: [Select] <?php include ("db.php"); include ("menu.php"); $result = mysql_query("SELECT categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $categories=$row["categories"]; $options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>'; }; $id = $_GET['id']; $query="SELECT * FROM links ORDER BY catid ASC"; $result=mysql_query($query); ?> <table width="65%" align="center" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <table width="100%" border="1" cellspacing="0" cellpadding="3"> <tr> <td colspan="7"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>Category ID</strong></td> <td align="center"><strong>Category ID</strong></td> <td align="center"><strong>Name</strong></td> <td align="center"><strong>URL</strong></td> <td align="center"><strong>Content</strong></td> <td align="center"><strong>Update</strong></td> <td align="center"><strong>Delete</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td> <SELECT NAME=catid> <OPTION>Categories</OPTION> <?php echo $options; ?> </SELECT> </td> <td><? echo $rows['catid']; ?></td> <td><? echo $rows['name']; ?></td> <td><a href="<? echo $rows['url']; ?>"><? echo $rows['url']; ?></a></td> <td><? echo $rows['content']; ?></td> <td align="center"><a href="update.php?id=<? echo $rows['id']; ?>">update</a></td> <td align="center"><a href="delete.php?id=<? echo $rows['id']; ?>">delete</a></td> </tr> <?php } ?> </table> </td> </tr> </table> <?php mysql_close(); ?> So, how do I get this code Code: [Select] $result = mysql_query("SELECT categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $categories=$row["categories"]; $options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>'; }; <SELECT NAME=catid> <OPTION>Categories</OPTION> <?php echo $options; ?> </SELECT> to give me an output that will be something like if catid exactly matches categories.id echo categories.categorie ??? so far everything I have done produces either a default category of the last category, the catid (which is a number), all of the categories (logical since catid will always be = id, or nothing. How do I get just the category name? I will keep reading and try to figure this out, but any help would be greatly appreciated. Thanks in advance This is the code that I'm using for a shopping cart in order to update the quantity if the same product id is chosen. $cart_items is a two-dimensional session array composed of the id session array and the quantity session array. While looping through $cart_items, I create a temporary array $temp_array so that I can execute some logic on the array without changing the normal output. $temp_array is the $cart_items array without the last items which would be the $_REQUEST id and the $_REQUEST quantity. So the two $_REQUESTs are popped off with array_pop and then the remaining $_SESSIONs are compared to the $_REQUEST id with
$j = array_search($_REQUEST["element_id_$i"], $temp_array); if( $j==$temp_array[$i]) If $j is the key of the item in $temp_array, then the $_REQUEST is unset and the quantities for the item are added together.
There are two problems with this code. First of all the var_dump of the $temp_array is always an empty array but I am expecting to get an array with just the last element popped off. Secondly,
foreach($cart_items as $item) {
foreach($item["id"] as $key => $val) {
foreach($item["quantity"] as $i => $value) {
if ($key == $i){
$temp_array = $cart_items;
$array = array_pop($temp_array);
var_dump($temp_array);
if (isset($_REQUEST["element_id_$i"]) && isset($_REQUEST["quantity_$i"])&& isset($value)){
$j = array_search($_REQUEST["element_id_$i"], $temp_array);
if( $j==$temp_array[$i]) {
echo "<b>SAME PRODUCT ADDED</b>";
$value += $_REQUEST["quantity_$i"];
unset($_REQUEST["element_id_$i"]);
}}}}}}
Can anybody see as to why, if any, this form takes forever to execute? Code: [Select] <?php if(isset($_POST['submitted'])){ $z = $_POST['zipcode']; $r = $_POST['radius']; $sql = mysql_query("SELECT DISTINCT m.LocAddZip, m.MktName,m.LocAddSt,m.LocAddCity,m.LocAddState,m.x1,m.y1,z1.lat,z2.long FROM mrk m, zip z1, zip z2 WHERE m.LocAddZip = z1.zipcode AND z2.zipcode = $z AND ( 3963 * acos( truncate( sin( z2.lat / 57.2958 ) * sin( z1.lat / 57.2958 ) + cos( z2.lat / 57.2958 ) * cos( z1.lat / 57.2958 ) * cos( z1.long / 57.2958 - z2.long / 57.2958 ) , 8 ) ) ) <= $r ") or die(mysql_error()); while($row = mysql_fetch_array( $sql )) { $store = $row['MktName']."<br />"; $store .= $row['LocAddSt']."<br />"; $store .= $row['LocAddCity'].", ".$row['LocAddState']." ".$row['LocAddZip']; $lat1 = $row['lat']; $lon1 = $row['long']; $lat2 = $row['y1']; $lon2 = $row['x1']; $dis = distance($lat1, $lon1, $lat2, $lon2); echo "<p>".$store."</p>"; echo ceil($dis) . " mile(s) away"; echo "<hr/>"; } } ?> I get a timeout error sometimes, and sometimes I don't. Is it the form, or would it be on the server? Thanks in advance Hi, I want to submit a form using ajax ans jquery with two fields input text and file. Code: [Select] <script> $(document).ready( function() { $('#basicinfofrm').ajaxForm(function() { url: ''+PN+'.php', type:'POST', data:'action=yes&'+frmFeilds, success: function(html) { alert(html); } }).submit(); </script> <form action='' method='post'> <label>name</label> <input type='text' name='txtname' /> <input type='file' name = 'txtfile' /> <input type='submit' value='save' /> </form> but is is not working... can anyone tell me how to? i don't want to refresh my page Thanks Code: [Select] <?php if (isset($_POST['checking'])) { echo $_POST['checking']; } if (isset($_POST['test'])) { echo $_POST['test']; } ?> <form id="testform2" action="testform.php" method="post"> <select name="checking" onchange="this.form.submit()"><option value="5">5</option><option value="6">6</option></select> <input type="text" name"test" id="test"> <a href="#" onclick="this.form.submit()">submit it!</a> </form> It works when I try to submit by selecting a new value from the dropdown box but when I try to click the link it won't display the text field value. That is a test case for a problem I m having in one of my codes. Hi, I have setup a basic enquiry form with a Captcha - once the code has been inserted and is correct I want to action the form (submit) in PHP. <?php session_start(); if($_SERVER['REQUEST_METHOD'] == 'POST'){ $vResult = ''; if(strtolower($_SESSION['security_code']) != strtolower($_POST['security_code'])){ $vResult = 'Invalid code!'; } else{ "/enquiry.php" } } ?> I want to submit the form to /enquiry.php - I tried header(Location.. and realised that just redirects, and doesn't submit the form. Any suggestions or tips would be great. Cheers, Paul |
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