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PHP - Insert Value Into Mysql After Confirmation
Hi all , sorry for posting question again , I really tried this for a whole day already but still can't get any solution...
Now I completed my submit form with validation , when I click submit , above the form will echo the values entered by user and a button for confirmation , if everything okay then just click the confirm button , it will update into mysql . (I do this using php function ) But even if I click the confirm button , it will just refresh my page and won't update into mysql . I know that it is because this button has no related to the submit form so there is nothing for it to update , but I still don't know how to find other ways to do it . I even tried to put a disabled button first , when I click submit button then it will allow to click . Hmm...not a good way I know...(and won't works with my limited php knowledge) Can I get a little hints for this problem ? In my submit form Code: [Select] if(isset($_POST['Submit'])) { if($validator->ValidateForm()) { $myfunction->show_confirmation(); } For the function part: ( another php file ) Code: [Select] function show_confirmation(){ echo "Total recipient(s):".count($total)."<br>"; echo "<br>" ; echo "Recipient Number(s):<br>".$_POST['cellphonenumber']."<br>" ; echo "<br>" ; echo "Message:<br>".$_POST['inputtext']."<br>" ; echo "<br>" ; echo "Date:".$_POST['datetime']."<br>" ; echo "<br>"; echo "<br>"; echo " Proceed ?"; echo "<br>"; echo "<form>"; echo "<input type=\"submit\" name=\"Submit2\"> echo "</form>"; if(isset($_POST['Submit2'])) { $this->submit(); } } submit() is just a function that will insert values into mysql table . This is my normal form: After click submit button: So...what should I do to make the Submit2 button works ? Any hints or advices will be greatly appreciate . I'm not asking for a whole complete solution but just some hints...thanks for every reply . Similar TutorialsI want to do an sql query, but I want it to happen only after my user has clicked on a confirmation link. The link will be sent to their email address. What is the usual procedure for this? For the time being Ive chucked what ive got into 1 php file, but I pressume I will need a couple of php files, and some how pass some variables from 1 page to the next??? $first = $_GET['first']; $last = $_GET['last']; $age = $_GET['age']; $$emailad = $_GET ['emailad'] $query =mysql_query ("INSERT INTO treesurgeons (FirstName, LastName, Age) VALUES ('$first', '$last', '$age')"); echo mysql_error(); echo "<p>Thankyou for submiting your details.\r\n Please check your inbox and click on your confirmation link.</p>"; $message = "The following details have been submited to the tree directory and are awaiting confirmation; $first $last, $age. \r\n Please click on the link below to confirm these details are correct"; //Do I include a link to an html file or php file? $subject = "Your tree directory details"; $to = "$email"; $email = "$emailad"; mail( "$to", "$subject","$message", "$headers"); //How do I trigger this email upon confirmation ?></body></html> Can anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh! <?php $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'"; $results = mysqli_query($cxn, $query); $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($row = mysqli_fetch_array($results)) { $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error(); $insert = mysqli_query($cxn, $insert_city_query); if (!$insert) { echo "INSERT is NOT working!"; exit(); } echo $row['City'] . "<br />"; echo "<pre>"; echo print_r($row); echo "</pre>"; } //while ($rows = mysqli_fetch_array($results)) } //if (mysqli_num_rows($results)) else { echo "No results to get!"; } ?> Here is my all_illinois INSERT table structu CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Here is my source table geo_cities structu CREATE TABLE IF NOT EXISTS `1` ( `CityId` varchar(255) NOT NULL, `CountryID` varchar(255) NOT NULL, `RegionID` varchar(255) NOT NULL, `City` varchar(255) NOT NULL, `Latitude` varchar(255) NOT NULL, `Longitude` varchar(255) NOT NULL, `TimeZone` varchar(255) NOT NULL, `DmaId` varchar(255) NOT NULL, `Code` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated I am trying to use text boxes to insert numbers into the database based on what is inputed. If I have a string, like this for example: $variable = 09385493; And I want to insert it into the database like this: mysql_query("INSERT INTO integers(number) VALUES ('$variable')"); When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as 9385493 Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too. Thanks well this is truely embarrising...i have a insert statement which works within phpmyadmin but when using mysqli_query it returns a error.
INSERT INTO users (username, timestamp) VALUES ('test', UTC_TIMESTAMP())
Unknown column 'timestamp' in 'field list'
i've been playing about with this for a few hours now ...tried changing the column name (timestamp), adding ` around column names as well as table name.
the column exists which is the strangest part, and ive even checked there is no space after the column name in the db.
whats going on please?
Hi guys I have a registration form working fine, my database is as below: userid username password repeatpassword I have added another column which is "name", users can update their profile once they have logged in so I have created updateprofile.php and when I login-->go to update profile and insert my name nothing adds to mysql name column this is my code below: <?php include ("global.php"); //username session $_SESSION['username']=='$username'; $username=$_SESSION['username']; //welcome messaage echo "Welcome, " .$_SESSION['username']."!<p>"; if ($_POST['register']) { //get form data $name = addslashes(strip_tags($_POST['name'])); $update = mysql_query("INSERT INTO users (name) VALUES ('$_POST[name]') WHERE username='$username'"); } ?> <form action='updateprofile.php' method='POST'> Company Name:<br /> <input type='text' name='name'><p /> <input type='submit' name='register' value='Register'> </form> can you please tell me where in this code is wrong? Im new in php so please excuse me if I have silly mistakes. thanks in advance I have this code: <?php $con = mysql_connect("localhost","hhh","hhh"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("hhh", $con); // -------------------- // Avatar insert check // -------------------- session_start(); $name = $_POST[name]; $group = $_POST[group]; $age = $_POST[age]; $usernameid = $_SESSION[id]; $result = mysql_query("SELECT * FROM avatars WHERE name='$_POST[name]'"); $num = mysql_numrows($result); if ($num == 0) { mysql_query("INSERT INTO avatars (id, usernameid, name, group, age, xp) VALUES ('', '$usernameid', '$name', '$group', '$age', '0')"); header( 'Location: me/' ) ; } else echo 'Sorry, please pick a new name'; ?> And it does everything but put the data into the datebase. If I add a session befor and after '$request' they both run, but the sql doesn't. No error returns, if just redirects to the other page. Any help? I don't understand where the empty value is. I've substituted the variables for text and still have the same problem. Code: Code: [Select] $sql = "INSERT INTO courses (course#, name, subject, semester, ap)VALUES('$courseNum', '$courseName', '$subject', '$semester', '$ap')"; Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 I need help badly! What I want to do is insert into database the value from the selected radio group buttons.. All of them. There are 10 radio groups total (they can be less, but not more). Thanks! Code: [Select] <?php require_once('Connections/strana.php'); mysql_select_db($database_strana, $strana); ?> <link href="css/styles.css" rel="stylesheet" type="text/css" /> <table width="100%" height="100%" style="margin-left:auto;margin-right:auto;" border="0"> <tr> <td align="center"> <form action="" method="post" enctype="multipart/form-data" name="form1"> <table> <?php $tema = mysql_query("SELECT * from prasanja where tip=2")or die(mysql_error()); function odgovor1($string) { $string1 = explode("/", $string); echo $string1[0]; } function odgovor2($string) { $string1 = explode("/", $string); echo $string1[1]; } while ($row=mysql_fetch_array($tema)) { $id=$row['prasanje_id']; $prasanje=$row['prasanje_tekst']; $tekst=$row['odgovor']; ?> <tr> <td> </td> </tr> <tr> <td class="formaP"> <?php echo $prasanje?> </td> </tr> <tr> <td class="formaO"> <p> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor1($tekst) ?>" /> <?php odgovor1($tekst) ?></label> <br /> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor2($tekst) ?>" /> <?php odgovor2($tekst) ?></label> <br /> </p></td> </tr> <tr> <td> <br /> </td> </tr> <?php } ?> </table> <input align="left"type="submit" name="submit" value="Внеси" > </form> </td> </tr> </table> prasanje = question tekst/odgovor = answer The answer table: id - primary question_id - the questions ID whose answer is selected in the radio group user_id - cookie takes care of this answer - the value from radio group date - automatic Hi,
The following code was written by someone else. It allows me to upload images to a directory while saving image name in the mysql table.
I also want the code to allow me save other data (surname, first name) along with the image name into the table, but my try is not working, only the images get uploaded.
What am I missing here?
if(isset($_POST['upload']))
{
$path=$path.$_FILES['file_upload']['name'];
if(move_uploaded_file($_FILES['file_upload']['tmp_name'],$path))
{
echo " ".basename($_FILES['file_upload']['name'])." has been uploaded<br/>";
echo '<img src="gallery/'.$_FILES['file_upload']['name'].'" width="48" height="48"/>';
$img=$_FILES['file_upload']['name'];
$query="insert into imgtables (fname,imgurl,date) values('$fname',STR_TO_DATE('$dateofbirth','%d-%m-%y'),'$img',now())";
if($sp->query($query)){
echo "<br/>Inserted to DB also";
}else{
echo "Error <br/>".$sp->error;
}
}
else
{
echo "There is an error,please retry or check path";
}
}
?>
joseph
Hi anyone here know how to insert $GET variable into mysql, i don't know how to put this variable between curly bracket, when i put on top insert query, i got error 'Could not insert admin'...please help Code: [Select] <?php mysql_connect("localhost","root","") or die(mysql_error()); mysql_select_db("healthsystem") or die(msql_error()); [color=red]//GET varibable $id = $_GET['id'];[/color] // file properties $file = $_FILES['image']['tmp_name']; if (!isset($file)) echo "Please select an image"; else { $image = addslashes(file_get_contents($_FILES['image']['tmp_name'])); $image_name = addslashes($_FILES['image']['name']); $image_size = getimagesize($_FILES['image']['tmp_name']); if ($image_size==FALSE) echo "That's not an image."; else { $insert = "INSERT INTO image_tbl(m_id,name,image) VALUES ('$id','$image_name','$image')"; $insert2=mysql_query($insert) or die("Could not insert admin"); print "Personal Wellness Successfully Submitted"; } } ?> Hello every body,, i want to create a new database (auto generate duty assigned using form) in PHP and mysql..
i have four input fields:-
Name, Subject, Class & weekly lectures
now i want to insert name,subject & class into database, when i insert number of weekly lecture in field four..
insert automaticlly in database multiple time which i write in field four (weekly lecture)
database: table structure i have already is:::
id-name-subject-class-period-monday-tuesday-wednesday-thursday-friday-saturday
anybody please help me to create this database or just insert query ....
guys, im having a problem here thats making me crazy im making a system and i want to insert the current date in the mysql. I have the field called 'data' in the DB, but when i make the code to insert all the other fields, including the 'data', its work perfectly... unless that damn date! $query = "INSERT INTO news (id, titulo, mensagem, data) VALUES (NULL, '$titulo', '$mensagem', 'date(\"d/m/Y\")')"; whats wrong with that? its stores in DB as '0000-00-00'. I am trying to add a value, input into a form, to a MySQL database. However, something must be wrong with the casting, because if there is a space in the form value, then I get an error, as in: //$_POST['string'] == '1blah 2blah'; sql = "INSERT INTO table (some_string) VALUES ($_POST[string])"; $sql_result = mysql_query($sql) or die ('The error is as follows: ' . mysql_error() . '<br /><br />Value could not be added.'); Then I get the following error: The error is as follows: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '2blah' at line 1 I've entered paragraphs into a database before, so this error is now to me. The column 'some_value' is a type: varchar(50).
Array ( [data] => Array ( [0] => Array ( [latitude] => 22.934566 [longitude] => 79.08728 [type] => county [distance] => 44.328 [name] => Narsinghpur [number] => [postal_code] => [street] => [confidence] => 0.5 [region] => Madhya Pradesh [region_code] => MP [county] => Narsinghpur [locality] => [administrative_area] => [neighbourhood] => [country] => India [country_code] => IND [continent] => Asia [label] => Narsinghpur, India ) ) ) how can i make this code to insert a single, multiple and update rows in the database. The code only insert new rows in the database. Code: [Select] if (($handle = fopen('inventorylist.csv', "r")) !== FALSE) { while (($data = fgetcsv($handle, 100000, ",")) !== FALSE) { $num = count($data); $sql="INSERT into inventory(itemNumber,itemDesc,quantityHand,category,Whse) values('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]')"; mysql_query($sql) or die(mysql_error()); } fclose($handle); } I have problem with this code. It does absolutely nothing. When INSERT is over it should redirect to index.php but it does nothing. There is no error, when the submit is clicked the page just refresh itself and because of echo function it write all the values. What seems to be the problem (I going slightly mad ) Code: [Select] <?php require_once("public/includes/session.php"); ?> <?php require_once("public/includes/connection.php"); ?> <?php require_once("public/includes/functions.php"); ?> <?php include_once("public/includes/form_functions.php"); include_once("public/includes/header.php"); if (isset($_POST['submit'])) { // Form has been submitted. $errors = array(); $required_fields = array('nik', 'lozinka', 'ime', 'prezime', 'adresa', 'grad', 'postanskiBroj', 'fiskni', 'moblini', 'email'); $errors = array_merge($errors, check_required_fields($required_fields, $_POST)); $username = trim(mysql_prep($_POST['nik'])); $password = trim(mysql_prep($_POST['lozinka'])); $hashed_password = sha1($password); $ime = trim(mysql_prep($_POST['ime'])); $prezime = trim(mysql_prep($_POST['prezime'])); $adresa = trim(mysql_prep($_POST['adresa'])); $grad = trim(mysql_prep($_POST['grad'])); $postanskiBroj = trim(mysql_prep($_POST['postanskiBroj'])); $fiskni = trim(mysql_prep($_POST['fiksni'])); $moblini = trim(mysql_prep($_POST['mobilni'])); $email = trim(mysql_prep($_POST['email'])); echo $username . $hashed_password . $ime . $prezime . $adresa . $grad . $postanskiBroj . $fiskni . $moblini . $email; if ( empty($errors) ) { $query = " INSERT INTO `gume`.`korisnik` (`id`, `korisnicko_ime`, `lozinka`, `ime`, `prezime`, `adresa`, `grad`, `postanskiBroj`, `fiksni_telefon`, `mobilni_telefon`, `email`) VALUES (NULL, '$username', '$hashed_password', '$ime', '$prezime', '$adresa', '$grad', '$postanskiBroj', '$fiskni, $moblini', '$email' )"; $result = mysql_query($query, $connection) or die(mysql_error); if ($result) { redirect_to("index.php"); } else { $message = "The user could not be created."; $message .= "<br />" . mysql_error(); } } else { if (count($errors) == 1) { $message = "There was 1 error in the form."; } else { $message = "There were " . count($errors) . " errors in the form."; } } } else { // Form has not been submitted. $username = ""; $password = ""; } ?> <div id="telo"> <div id="kreiranjeNaloga"> <script src="SpryAssets/SpryValidationTextField.js" type="text/javascript"></script> <link href="SpryAssets/SpryValidationTextField.css" rel="stylesheet" type="text/css" /> <script src="SpryAssets/SpryValidationPassword.js" type="text/javascript"></script> <script src="SpryAssets/SpryValidationConfirm.js" type="text/javascript"></script> <link href="SpryAssets/SpryValidationPassword.css" rel="stylesheet" type="text/css" /> <link href="SpryAssets/SpryValidationConfirm.css" rel="stylesheet" type="text/css" /> <p>Polja sa * su obavezna</p> <form action="new_user.php" method="post"> <span id="sprytextfield1"> <label>Korisnicko ime: </label> <input type="text" name="nik" id="nik" size="40" value=""/> *<span class="textfieldMinCharsMsg">Korisnicko ime ne moze imati manje od 5 karaktera</span><span class="textfieldMaxCharsMsg">Korisnicko ime moze imati najvise 30 karaktera.</span></span><br /> <span id="sprypassword1"> <label>Lozinka:</label> <input type="password" name="lozinka" id="lozinka" size="40" value=""/> *<span class="passwordMinCharsMsg">Sifra mora sadrzati najmanje 5 karaktera.</span><span class="passwordMaxCharsMsg">Sifra moze imati najvise 30 karaktera.</span></span> <br /> <span id="spryconfirm1"> <label>Potvrdite lozinku:</label> <input type="password" name="password1" id="password1" size="40" value=""/> <span class="confirmRequiredMsg">*</span>Obe lozinke moraju da budu iste.</span> <br /> <span id="sprytextfield2"> <label>Ime:</label> <input type="text" name="ime" id="ime" size="40" value=""/> * </span> <br /> <span id="sprytextfield3"> <label>Prezima</label> <input type="text" name="prezime" id="prezime"size="40" value="" /> *</span> <br /> <span id="sprytextfield4"> <label>Adresa:</label> <input type="text" name="adresa" id="adresa" size="40" value=""/> * </span> <br /> <span id="sprytextfield7"> <label>Grad:</label> <input type="text" name="grad" id="grad" size="40" value="" /> * </span> <br /> </span><span id="sprytextfield9"> <label>Postanski Broj: </label> <input type="text" name="postanskiBroj" id="postanskiBroj" size="10" value=""/> * <span class="textfieldInvalidFormatMsg">Postanski broj nije pravilno upisan.</span></span><br /> <span id="sprytextfield5"> <label>Broj fiksnog telefona: </label> <input type="text" name="fiksni" id="Broj fiksnog telefona" size="40" value="" /> *<span class="textfieldInvalidFormatMsg">Broj telefona nije pravilno upisan</span></span> <br /> <span id="sprytextfield6"> <label>Broj mobilnog telefona: </label> <input type="text" name="mobilni" id="mobilni" size="40" value="" /> *<span class="textfieldInvalidFormatMsg">Broj telefona nije pravilno upisan</span></span><br /> <span id="sprytextfield10"> <label>Email:</label> <input type="text" name="email" id="email" size="40" value="" /> *<span class="textfieldInvalidFormatMsg">Email adresa nija pravilno upisana.</span></span><br /> <input name="submit" type="submit" id="submit" value="Kreiraj korisnika" /> </form> </div> </div> <?php include("public/includes/footer.php"); ?> Im trying to insert some values automatically into a table once the form loads, but Im getting an error. Here is the code Code: [Select] <?php $aid = $_GET['aid']; $sd = $_GET['sd']; ?> <style> #message {margin:20px; padding:20px; display:block; background:#cccccc; color:#cc0000;} </style> <div id="message">Your notification has been submitted.</div> <div style="text-align:center "> <?php $connection = mysql_connect("localhost", "username", "password"); mysql_select_db("articles", $connection); $query="INSERT INTO broken_links (articleid, article) VALUES ('$aid', '$sp')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "Submitted"; mysql_close($con) ?> <table border="0" cellpadding="3" cellspacing="3" style="margin:0 auto;" > <input type="submit" id="Login" value=" Thank you. Please press to close " onclick="tb_remove()"></td> </tr> </table> </div> Any help will be appreciated Hi guys, new to the forum. Im in need of some advice. I am setting up a shopping cart using opencart. I have an xml product feed, and am writing a parser script to import/update products. The xml parsing via SimpleXML, image downloads, and category inserts work without a hitch. The product Inserts, not so much. there's 4 tables involved: product, product_description, product_to_category, product_to_store all INSERTs are successful for 3 out of 4 tables. 2,540 products get inserted. However, the product_description table shows only 2,143 rows. For some reason, not all INSERTs to this table produce a valid row. I cleared the tables and run the script many times, and each time produces the same results which leads me to believe the problem has something to do with the data instead of the script, but i have no idea where to start troubleshooting. ive included the applicable code class SpicyDB extends mysqli{ function doesExist ($name, $type) { switch ($type) { case "product": $query = "SELECT product_id from spicyvib_product WHERE model='$name'"; break; case "category": $query = "SELECT category_id from spicyvib_category_description WHERE name='$name'"; break; } $result = $this->query($query); if ($result->num_rows >= 1) $status = $result->fetch_row(); else $status = false; return $status; } function addCategory ($name) { $query = "INSERT INTO spicyvib_category (parent_id) VALUES (0)"; $this->query($query); $last_id = $this->insert_id; $query2 = "INSERT INTO spicyvib_category_description (category_id, name) VALUES ($last_id, '$name')"; $this->query($query2); $query3 = "INSERT INTO spicyvib_category_to_store (category_id, store_id) VALUES ($last_id, 0)"; $this->query($query3); } function getCatIDS () { $catIDS = array(); $result = $this->query("SELECT category_id, name FROM spicyvib_category_description"); while ($row = $result->fetch_assoc()) { $name = $row['name']; $id = $row['category_id']; $catIDS[$name] = $id; } return $catIDS; } function addProduct ($data, $catIDS) { $this->query("INSERT INTO spicyvib_product SET model = '" . $data['model'] . "', quantity = '1', minimum = '1', subtract = '0', stock_status_id = '7', image = '" . $data['image'] . "', date_available = NOW(), manufacturer_id = '0', price = '" . (float)$data['price'] . "', cost = '" . (float)$data['cost'] . "', weight_class_id = '5', length_class_id = '3', status = '1', tax_class_id = '0', date_added = NOW()"); $last_id = $this->insert_id; $this->query("INSERT INTO spicyvib_product_to_store SET product_id = '" . (int)$last_id . "', store_id = '0'"); $this->query("INSERT INTO spicyvib_product_description SET product_id = '" . (int)$last_id . "', language_id = '1', name = '" . $data['name'] . "', description = '" . $data['description'] . "'"); $categoryList = $data['category']; $categoryArr = explode(';', $categoryList); foreach ($categoryArr as $currCategory) { if (!empty($currCategory)) { if (!isset($catIDS[$currCategory])) { $currCategory = 'Miscellaneous'; } $this->query("INSERT INTO spicyvib_product_to_category SET product_id = '" . (int)$last_id . "', category_id = '" . (int)$catIDS[$currCategory] . "'"); } } } } $db = new SpicyDB(DB_HOST, DB_USER, DB_PW, DB_NAME); $categoryXml = new SimpleXMLElement(CAT_FNAME, NULL, TRUE); foreach ($categoryXml->category as $cat) { $name = (string)$cat->name; $exists = $db->doesExist($name, 'category'); if (!$exists) { $db->addCategory($name); } } $category_ids = $db->getCatIDS(); $finalXML = new SimpleXMLElement(OLD_FNAME, NULL, TRUE); foreach ($finalXML->items->item as $item) { $data = array(); $updated = (string)$item->lastupdated; $data['model'] = (string)$item->model; $data['name'] = (string)$item->title; $image_str = (string)$item->image; $image = basename($image_str); $data['image'] = 'data/' . $image; $data['price'] = (float)$item->suggested_retail; $data['cost'] = (float)$item->price; if (empty($data['price'])) { $data['price'] = ($data['cost'] * 2); } $data['description'] = (string)$item->description; $data['category'] = (string)$item->category; switch ($updated) { case "UPDATE": $model = $data['model']; if (!$db->doesExist($model, 'product')) { $db->addProduct($data, $category_ids); } break; } } Hi. I think you all know me by now so I'll cut to the chase. Code: [Select] <?php $host="edited"; $username="edited"; $password="edited"; $db_name="edited"; $tbl_name="topic"; // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $name=$_POST['name']; $detail=$_POST['details']; $sql="INSERT INTO $tbl_name(topic, detail, datetime)VALUES('$name', '$detail', NOW())"; $result=mysql_query($sql); if($result){ header("location:site.html");} else{ echo("I have failed you master.");} ?> Displayed error: "I have failed you master." Anyone know a possible cause? Thanks. Bye. |
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