Can I please have some help with a problem I'm having updating a mysql database with PHP.
I'm sorry to ask a question that has been asked a lot of times before, it's just driving me a bit nuts, and I've looked through similar questions but the answers don't seem to help with my problem.
I'm using two files, an admin page (admin.php) to edit content with, and an update file that is meant to update the database when the submit button is pressed.
Everything seems to be working fine, the values are being posted to the update.php page (I can see them when I echo them out) but it wont update the database.
If anyone can please point me in the right direction or tell me what I'm doing wrong I'd be very grateful!
Thank you very much:)
This is my admin.php page;
<head>
<?php
/*
Check to see if the page id has been set in the url.
If it has, set it as the $pageid variable,
If it hasn't, set the $pageid variable to 1 (Home page)
*/
if (isset($_GET['pageid'])) {
$pageid = $_GET['pageid'];
}
else {
$pageid = '1';
}
//Database connection variables
$servername = "localhost";
$username = "root";
$password = "";
$database = "cms";
// Create connection
$conn = new mysqli($servername, $username, $password, $database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//Get information from the database
$sql = "SELECT title, sub_title, tab1, tab2, tab3, content FROM data WHERE id='$pageid'";
$result = $conn ->query($sql);
if ($result->num_rows > 0)
{
while($row = $result->fetch_assoc()) {
$conn->close();
//Store database information in variables to display in the form
$title = $row["title"];
$sub_title = $row["sub_title"];
$tab1 = $row["tab1"];
$tab2 = $row["tab2"];
$tab3 = $row["tab3"];
$content = $row["content"];
}
} else {
echo "0 results";
}
?>
</head>
<body>
//basic navigation
Page 1 | Page 2 | Page 3
<form action="update.php" method="post" name="adminform">
<input type="hidden" name="pageid" value="<?php echo "$pageid";?>">
NAME:<br>
<input type="text" name="title" value="<?php echo $title;?>"><br><br>
EMAIL:<br>
<input type="text" name="sub_title" value="<?php echo $sub_title;?>"><br><br>
CONTENT:<br>
<input type="text" name="tab1" value="<?php echo $tab1;?>"><br><br>
CONTENT:<br>
<input type="text" name="tab2" value="<?php echo $tab2;?>"><br><br>
CONTENT:<br>
<input type="text" name="tab3" value="<?php echo $tab3;?>"><br><br>
CONTENT:<br>
<textarea rows="4" cols="50" name="content">
<?php echo $content;?>
</textarea>
<br><br>
<input type="submit">
</form>
</body>
And this is the update.php page;
<?php
/*Values passed from the admin form, to be used as update variables*/
if (isset($_POST['adminform']))
{
$pageid = $_POST["pageid"];
$titleu = $_POST["title"];
$sub_titleu = $_POST["sub_title"];
$tab1u = $_POST["tab1"];
$tab2u = $_POST["tab2"];
$tab3u = $_POST["tab3"];
$contentu = $_POST["content"];
}
?>
<?php
if(isset($_POST['adminform']))
{
// Create connection
$conn = new mysqli($servername, $username, $password, $database);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//Update the database
$sql = "UPDATE data SET title='$titleu', sub_title='$sub_titleu', tab1='$tab1u', tab2='$tab2u', tab3='$tab3u', content='$contentu' WHERE id =='$pageid'";
$result = $conn ->query($sql);
$conn->close();
}
?>
You're using == instead of = on the where clause.
On the other hand, don't pass user values to the query without validation and sanitization if you don't want to be vulnerable to sql injection attacks.
$sql = "UPDATE data SET title='" . $conn->real_escape_string($titleu) . "', sub_title='" . $conn->real_escape_string($sub_titleu) . "', tab1='" . $conn->real_escape_string($tab1u) . "', tab2='" . $conn->real_escape_string($tab2u) . "', tab3='" . $conn->real_escape_string($tab3u) . "', content='" . $conn->real_escape_string($contentu) . "' WHERE id = " . (int)$pageid;
This will work, but is not very elegant solution. You may use prepared statements instead, to pass the correct types and prevent sql injection.
Check your DB Connections and test whether you are connected to DB or not.
Change your query as below
$sql = "UPDATE data SET title='".$titleu."', sub_title='".$sub_titleu."', tab1='".$tab1u."', tab2='".$tab2u."', tab3='".$tab3u."', content='".$contentu."' WHERE id ='$pageid'";
Related
I'm making a page where you have to enter a text in a textbox and the click send, another page will save it.
Also, on the first page, the text that was stored previously in the database, has to load. This is the code that i've got:
<?php
$databaseid = 3;
$servername = "jog4fun.be.mysql";
$username = "jog4fun_be";
$password = "****";
$dbname = "jog4fun_be";
$gettitel1 = null;
$gettext1 = null;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Id,Titel,Tekst FROM Teksten";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
if ($row["Id"]== $databaseid ){
$gettitel1 = $row["Titel"];
$gettext1 = $row["Tekst"];
}
}
} else {
echo "0 results";
}
$conn->close();
$gettitel1 = strip_tags($gettitel1, '<br>');
$link1 = '<textarea id = "klein" rows="4" cols="50" name="titel3" form="usrform">' . $gettitel1 . '</textarea>';
$link2 = '<textarea id = "groot" rows="4" cols="50" name="text3" form="usrform">' . $gettext1 . '</textarea>';
echo $link1;
echo $link2;
?>
The problem is that it sends the text from textbox with name text3 as text1 with the post function. Can someone figure out what's wrong with it? I've been tying for an hour and i did not find it.
Thanks for your time and help,
Jonas
Simply here:
while($row = $result->fetch_assoc()) {
if ($row["Id"]== $databaseid ){
$gettitel1 = $row["Titel"];
$gettext1 = $row["Tekst"];
}
you overwrite the value of the two variables each time you iterate. So after this block of code you will keep stored the last values returned from the db.
YOu should add to your query a WHERE clause to identify the one and only record you need so you will fetch only the relevant data. Example:
$sql = "SELECT Id,Titel,Tekst FROM Teksten WHERE Id='1'";
i have a simple form in html where user selects and ID from the table on the website (required data) and then he can / or not change 2 fields. First is a dropdown list of 2 values (strings), and the second is number of open spots!
So if a user leaves both fields empty and click send by mistake nothing should happen. If a user only changes one of the fields only that one should change!
I have checked every forum and almost all posts in here and i still cannot get it to work.
<form action="viv_settings_tecaji.php" method="post">
Datum termina (izberi ID):
<input type="number" name="ID" required><br><br>
<!--Sprememba tega datuma (če ne želiš spremenit pusti prazno):
<input type="date" name="nov_datum"><br><br>-->
Sprememba statusa (če želiš da ostane isto vpiši trenutni status!:
<select name="STATUS">
<option></option>
<option value="zaprt">Zaprt</option>
<option value="odprt">Odprt</option>
</select><br><br>
Sprememba števila odprtih mest
<input type="number" name="st_odprtih_mest"><br><br>
<input type="submit">
</form><br>
php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "viverius_education";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$update_status = $_POST['STATUS'];
$update_st_odprtih_mest = $_POST['st_odprtih_mest'];
$update_ID = $_POST['ID'];
if (empty($update_status) AND empty($update_status)){
header('Location: viv_settings_tecaji_main.php'); exit;
}
else{
$sql = "UPDATE razpisani_tecaji
SET
STATUS = IF('$update_status'='',STATUS,'$update_status'),
ST_ODPRTIH_MEST = IF('$update_st_odprtih_mest'='',STATUS,'$update_st_odprtih_mest'),
WHERE ID_TECAJA = $update_ID";
}
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
header('Location: viv_settings_tecaji_main.php'); exit;
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}$conn->close();
?>
A decent form example would be like the following
<form action='' method='post'>
<input type='text' name='field1' required>
<select name='fieldSelect'>
<option value='value1'>Value1</option>
<option value='value2'>Value2</option>
</select>
<input type='submit' name='send'>
</form>
then the PHP would be like
<?php
if(isset($_POST['send'] && (!empty($_POST['field1']) || !empty($_POST['fieldSelect']))){
$field1 = $_POST['field1'];
$fieldSelect = $_POST['fieldSelect'];
//YOUR SQL CODE
} else {
echo "Please Insert Some Data";
}
?>
In brief:
Give your Submit button a name
Check if the submit button is clicked or not by if(isset($_POST['submit-button-name'])){}
define your form's $_POST variables with names.
Continue with SQL.
There are a couple of ways you can avoid inserting "empty" records in your database. It depends if you want to do it in the client-side (before the form submits) or server-side (when the form is submitted to the server).
I'll show you how to do it server-side, since you posted your php script.
In viv_settings_tecaji.php
...
$update_status = $_POST['STATUS'];
$update_st_odprtih_mest = $_POST['st_odprtih_mest'];
$update_ID = $_POST['ID'];
if($update_status == "" || $update_st_odprtih_mest == "" || $update_ID == ""){
die("One of the form fields was empty");
}
...
That would kill your script if any of the form fields was empty. Potentially, you could check for null or use PHP's empty() function.
I hope this helps!
So i manege to get it to work using the following code. So for the part if user leaves both field empty i will put together a script for at least 1 to be required.
$sql = "UPDATE razpisani_tecaji
SET
STATUS = IF(LENGTH('$update_status')=0, STATUS, '$update_status'),
ST_ODPRTIH_MEST = IF(LENGTH('$update_st_odprtih_mest')=0, ST_ODPRTIH_MEST, '$update_st_odprtih_mest')
WHERE ID_TECAJA = $update_ID";
PROBLEM: I got a problem updating my input into sql using PHP, the PHP updates all empty values into sql which I don't want to.
ACHIEVEMENT: So I hope to achieve when user submit their data either empty or filled then PHP might be able to pickup and update only filled data into my sql. I tried using input with value=">php echo here<" but it won't work with textarea, so I couldn't find any solution since I'm new to PHP and SQL. Tried to find similar posts but I couldn't make them work like I wanted to :(
<?php include 'config/sqlconnect.php'; ?>
<form method="post" action"config/sqlconnect.php">
</p>MainPage info</p>
<input type="text" name="mainPageInfo"/>
<br>
</p>MiddlePage info</p>
<textarea name="middlePageInfo"></textarea>
<br>
</p>Container info</p>
<input type="text" name="containerInfo"/>
<br>
</p>Content</p>
<input type="text" name="content"/>
<br>
</p>Second content</p>
<input type="text" name="secondContent"/>
<input type="submit" name="submit" class="btn-block"/>
<br>
</form>
in PHP script
<?php
$servername = "localhost";
$username = "root";
$password = "pass";
$dbname = "pagesDb";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
mysqli_set_charset($conn,"utf8");
$sql = "SELECT * FROM myPages";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$mainPageInfo = $row['mainPageInfo'];
$middlePageInfo = $row['middlePageInfo'];
$containerInfo = $row['containerInfo'];
$content = $row['content'];
$secondContent = $row['secondContent'];
}
} else {
echo "0 results";
}
if (isset($_POST['submit'])) {
$mainPageInfo = $_POST['mainPageInfo'];
$middlePageInfo = $_POST['middlePageInfo'];
$containerInfo = $_POST['containerInfo'];
$content = $_POST['content'];
$secondContent = $_POST['secondContent'];
$sql = "UPDATE myPages SET mainPageInfo='$mainPageInfo',
middlePageInfo='$middlePageInfo',
containerInfo='$containerInfo',
content='$content',
secondContent='$secondContent'
WHERE id=0";
if ($conn->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $conn->error;
}
}
$conn->close();
?>
Second Attempts: It doesn't update my data somehow... please help I tried more than 8 hours with no results :(
if (isset($_POST['submit'])) {
foreach($_POST as $name => $value) {
$sql = "UPDATE myPages SET $name = '$value' WHERE id=1";
}
if ($conn->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $conn->error;
}
}
Help would be appreciated, thanks everyone!
Using your Second Attempt as a starting point.
The problem with just using the POST array without being specific is that, in this example you are going to try an update a column on the database called submit i.e. your submit button. Later there may be data on the page that belongs in 2 or more tables.
So create an controlling array containing all the field names from the form that you want to process onto your table.
$db_fields = array('mainPageInfo', 'middlePageInfo', 'containerInfo',
'content', 'secondContent');
$sql = ''; // will hold the query we build dynamically
// was this a user sending data
if ( $_SERVER['REQUEST_METHOD' == 'POST' ) {
foreach($db_fields as $fieldname) {
if ( ! empty($_POST[$fieldname] ) {
$sql .= "$fieldname = '{$_POST[$fieldname]}', ";
}
}
}
$sql = rtrim($sql, ','); // remove the trailing comma
$sql = "UPDATE myPages SET $sql WHERE id=1";
if ($conn->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $conn->error;
}
I am working on a test login-check with PHP/HTML and MySQL. I got it working great; it successfully connects to the database, it can grab my database values and save them in a variable, etc., but I ran into one slight problem.
I'm using two PHP pages to do the check. The login.php page, which only contains the forum, and the welcome.php page, which does the database connecting. When I ran a test page to just have it echo the database info, it printed out right (testUser, testEmail#email.com, testPassword, 1/1/1900). So when I tried to run my login-authentication check, it just says 'Unknown user!' twice, even when I try the usernames 'usr', 'testUser', and 'testUser2' (I made two tables, and the second one is the same with 2 added to the end). Here's my code.
<html>
<head>
<?php
$title = ucfirst(basename($_SERVER['PHP_SELF'], ".php"));
echo "<title>$title</title>";
?>
</head>
<body>
<form name="form" accept-charset="utf-8" action="welcome.php" method="post">
<span class="header">Username</span><input type="text" name="usr" value="usr"></input><br>
<span class="header">Password</span><input type="text" name="pass" value="pass"></input>
<input type="submit">
</form>
</body>
</html>
<?php
$servername = removed;
$username = removed;
$password = removed;
$dbname = removed;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT ID, USER, PASSWORD FROM usrdatabase";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
// the given info from the form
$usrUser = $_POST["usr"];
$usrPass = $_POST["pass"];
// convert the findings to uppercase to get rid of sensitivity
if (strtoupper($usrUsr) == strtoupper($row["USER"]) && strtoupper($usrPass) == strtoupper($row["PASSWORD"])) {
echo "Welcome $usrUser!<br>Your login was successful! ?>";
}
elseif (strtoupper($usrUsr) == strtoupper($row["USER"]) && strtoupper($usrPass) != strtoupper($row["PASSWORD"])) {
echo "Login failed as $usrUser!";
}
else {
echo "Unknown user!";
}
}
} else {
echo "0 results";
}
$conn->close();
?>
This always produces a 'Unknown user!' Is there something wrong with my check? I want it to go through each user in the database to check the info with each existing user.
Change
strtoupper($usrUsr) == strtoupper($row["USER"])
To
strtoupper($usrUser) == strtoupper($row["USER"])
Fetch single user from the database by using the username since they are unique for each user.
$sql = "SELECT ID, USER, PASSWORD FROM usrdatabase WHERE USER = '" . mysqli_real_escape_string($_POST['usr']) . "' AND PASSWORD = '" . mysqli_real_escape_string($_POST['pass']) . "'";
hey i see your if else contains $usrUsr shoudn't it be $usrUser ? (forgot the e)
I have a table displaying the content of a MySQL table. For every row I added an 'edit button' so our users can update the content.
The 'edit button' goes to a link ?edit_entry.php?sid=4 with 4 the sid of the entry.
This works but I get a blank form.
Question 1: Is there any way to already display the content of the specific MySQL row in the text fields of the form?
Here is the edit_entry.php code:
<?php require('includes/config.php');
//if not logged in redirect to login page
if(!$user->is_logged_in()){ header('Location: login.php'); }
// Create connection
$conn = new mysqli(DBHOST, DBUSER, DBPASS, DBNAME);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sid = $_GET['sid'];
$sql = "SELECT * FROM orders WHERE sid = '$sid'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = mysqli_fetch_array($sql)) {
$sid = $row['sid'];
$q1_requested_by = $row['q1_requested_by'];
$q2_productname = $row['q2_productname'];
$q3_supplier = $row['q3_supplier'];
$q4_productnumber = $row['q4_productnumber'];
$q5_quantity = $row['q5_quantity'];
$q6_price = $row['q6_price'];
$q7_budget = $row['q7_budget'];
$q8_link = $row['q8_link'];
}
?>
<form action="update_script.php" method="post">
<input type="hidden" name="sid" value="<?=$sid;?>">
Requested by: <input id="q1" type="text" style="width:400px" name="ud_q1_requested_by" value="<?=$q1_requested_by?>" required="true" tabindex="1"><br>
Product name: <input id="q2" type="text" style="width:400px" name="ud_q2_productname" value="<?=$q2_productname?>" required="true" tabindex="2"><br>
Supplier: <input id="q3" type="text" style="width:400px" name="ud_q3_supplier" value="<?=$q3_supplier?>" required="true" tabindex="3"><br>
Product number: <input id="q4" type="text" style="width:400px" name="ud_q4_productnumber" value="<?=$q4_productnumber?>" required="true" tabindex="4"><br>
Quantity: <input id="q5" type="text" style="width:400px" name="ud_q5_quantity" value="<?=$q5_quantity?>" required="true" tabindex="5"><br>
Price: <input id="q6" type="text" style="width:400px" name="ud_q6_price" value="<?=$q6_price?>" tabindex="6"><br>
Budget: <input id="q7" type="text" style="width:400px" name="ud_q7_budget" value="<?=$q7_budget?>" tabindex="7"><br>
Link: <input id="q8" type="text" style="width:400px" name="ud_q8_link" value="<?=$q8_link?>" tabindex="8"><br>
<input type="submit" name="submit" id="submit" value="Update your input!" tabindex="9" />
</form>
<?php
}else{
echo 'No entry found. Go back';
}
?>
And here is update_script.php:
<?php require('includes/config.php');
//if not logged in redirect to login page
if(!$user->is_logged_in()){ header('Location: login.php'); }
// Create connection
$conn = new mysqli(DBHOST, DBUSER, DBPASS, DBNAME);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sid = $_POST["sid"];
$ud_q1_requested_by = mysqli_real_escape_string($_POST["ud_q1_requested_by"]);
$ud_q2_productname = mysqli_real_escape_string($_POST["ud_q2_productname"]);
$ud_q3_supplier = mysqli_real_escape_string($_POST["ud_q3_supplier"]);
$ud_q4_productnumber = mysqli_real_escape_string($_POST["ud_q4_productnumber"]);
$ud_q5_quantity = mysqli_real_escape_string($_POST["ud_q5_quantity"]);
$ud_q6_price = mysqli_real_escape_string($_POST["ud_q6_price"]);
$ud_q7_budget = mysqli_real_escape_string($_POST["ud_q7_budget"]);
$ud_q8_link = mysqli_real_escape_string($_POST["ud_q8_link"]);
$sql= "UPDATE orders
SET q1_requested_by = '$ud_q1_requested_by', q2_productname = '$ud_q2_productname', ud_q3_supplier = '$ud_q3_supplier', ud_q4_productnumber = '$ud_q4_productnumber', ud_q5_quantity = '$ud_q5_quantity', ud_q6_price = '$ud_q6_price', ud_q7_budget = '$ud_q7_budget', ud_q8_link = '$ud_q8_link'
WHERE sid='$sid'";
$result = $conn->query($sql);
if(mysqli_affected_rows()>=1){
echo "<p>($sid) Record Updated<p>";
}else{
echo "<p>($sid) Not Updated<p>";
}
?>
There must be a problem in this last part because I get the (4) Not updated message.
Question 2: Does anyone see the problem here?
I've been trying a few things to tackle the problem but neither are working.
Thank you
mysqli_real_escape method requires the connection to be provided; this was not the case in deprecated mysqli_* methods..
see documentation at http://php.net/manual/en/mysqli.real-escape-string.php
In your case, since you are using object of mysqli:
$conn->real_escape_string($string)
Also for the record, you have a possible inject despite your attempts not to.
You should update $sid = $_POST["sid"]; to $sid = (int) $_POST["sid"]; if it is supposed to be an integer or escape it as well.
With this many variables needing to be escaped though, you should probably look at how to conduct a prepared statement. http://php.net/manual/en/mysqli.quickstart.prepared-statements.php
You use mysqli, not mysql that is good news.
But you continue to use old techniques to pass parameter to query.
So let just try to bind parameters as it should be done with mysqli:
$sql= "UPDATE orders
SET
q1_requested_by = ? ,
q2_productname = ?,
ud_q3_supplier = ?,
ud_q4_productnumber = ?,
ud_q5_quantity = ?,
ud_q6_price = ?,
ud_q7_budget = ?,
ud_q8_link = ?
WHERE sid=? ";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param('ssssiddsi', $ud_q1_requested_by, $ud_q2_productname, $ud_q3_supplier, $ud_q4_productnumber, $ud_q5_quantity, $ud_q6_price, $ud_q7_budget, $ud_q8_link, $sid);
$result = $stmt->execute();
if($result && $stmt->affected_rows>0){
echo "<p>($sid) Record Updated<p>";
}else{
echo "Error:\n";
print_r($stmt->error_list);
echo "<p>($sid) Not Updated<p>";
}
I got it to work using the following code:
<?php require('includes/config.php');
//if not logged in redirect to login page
if(!$user->is_logged_in()){ header('Location: login.php'); }
$conn = new mysqli(DBHOST, DBUSER, DBPASS, DBNAME);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sid = (int)$_POST["sid"];
$ud_q1_requested_by = $conn->real_escape_string($_POST["ud_q1_requested_by"]);
$ud_q2_productname = $conn->real_escape_string($_POST["ud_q2_productname"]);
$ud_q3_supplier = $conn->real_escape_string($_POST["ud_q3_supplier"]);
$ud_q4_productnumber = $conn->real_escape_string($_POST["ud_q4_productnumber"]);
$ud_q5_quantity = $conn->real_escape_string($_POST["ud_q5_quantity"]);
$ud_q6_price = $conn->real_escape_string($_POST["ud_q6_price"]);
$ud_q7_budget = $conn->real_escape_string($_POST["ud_q7_budget"]);
$ud_q8_link = $conn->real_escape_string($_POST["ud_q8_link"]);
$sql= "UPDATE orders
SET
q1_requested_by = '$ud_q1_requested_by',
q2_productname = '$ud_q2_productname',
q3_supplier = '$ud_q3_supplier',
q4_productnumber = '$ud_q4_productnumber',
q5_quantity = '$ud_q5_quantity',
q6_price = '$ud_q6_price',
q7_budget = '$ud_q7_budget',
q8_link = '$ud_q8_link'
WHERE sid='$sid'";
$result = $conn->query($sql);
header("Location: edit_orders.php");
?>
which is just simple query, as the original form to enter a new row of data also does. I also decided to remove the error handling at the end since it didn't seem to work with mysqli_affected_rows()>0...
It's probably not a very elegant solution, but it works. Still I'd like to learn more so if anybody would have a useful link explaining php+mysqli basics that would help me much. The links to php.net or mysql.com are for me too brief at this moment though, for me they don't explain what is going on. I'm a total novice to php and mysql and could use some more explanatory/introductary text, maybe with examples, but mostly providing me with an overview of what is going on... Thanks anyway for all the help!