Inserting checkbox values into multiple rows - php

I have an echo of a number of groups a user is a member of.
It will output multiple checkboxes. The value of the checkboxes (groupid)+ 2 hidden values need to be inserted into a row in a new table.
How do I insert the values per checkbox into separate rows?
<?php
$user=$_SESSION['SESS_USERID'];
if(isset($_REQUEST['user'])){
$user = preg_replace('#[^a-z0-9]#i', '', $_GET['user']);
}
if(isset($_REQUEST['id'])){
$id = preg_replace('#[^a-z0-9]#i', '', $_GET['id']);
}
$sql="SELECT * FROM groepsleden,groepen,werken WHERE groepsleden.groepid=groepen.groepid
AND groepsleden.userid='$user'AND werken.werkid='$id' ORDER BY groeplidid DESC ";
$result = $conn->query($sql) or die ("The query could not be completed. try again");
if ($result->num_rows > 0) {
echo" <table>";
while ($row = $result->fetch_assoc()) {
echo"<tr>
<td bgcolor='#1E1E1E'> </td>
<td bgcolor='#1E1E1E'><div align='right'>
<input name='groepid' type='checkbox' value=" . $row['groepid'] . ">
<input type='hidden' name='werkid' value=" . $row['werkid'] . ">
<input type='hidden' name='userid' value=" . $row['userid'] . ">
</div></td>
<td bgcolor='#1E1E1E'>Paats dit werk in <a href='groep.php?
id=" . $row['groepid'] . "'</a>".$row["groepsnaam"]."</span></td></tr></table>
}
}
?>
<?php
$sql=$dbo->prepare("INSERT INTO groepwerken(werkid,groepid,userid)
VALUES ('$werkid','$groepid','$user')");
$sql->bindParam(':werkid',$werkid,PDO::PARAM_INT);
$sql->bindParam(':groepid',$groepid,PDO::PARAM_INT);
$sql->bindParam(':userid',$userid,PDO::PARAM_INT);

Veldnamen als array ....
name="groepid[]"
name='werkid[]'
name='userid[]'
vervolgens kun je met foreach door deze waarden heen loopen
$werkid = $_POST['werkid'];
$userid = $_POST['userid'];
$x = 0;// teller zodat we juiste element kunnen selecteren
foreach($_POST['groepid'] as $groepen)
{
if($groepen!=="") // als niet leeg is
{
$juisterwerkid = $werkid[$x];
$juisteuserid= $userid[$x];
$degroep = $groepen;
$x++; // teller ophogen adhv element
echo "Groep: ".$groepen." juisterwerkid: ".$juisterwerkid." juisteuserid: ".$juisteuserid."";
}
}
zoiets zou moeten werken :)

Related

My SQL Update query won't update the specified record [duplicate]

This question already has answers here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(2 answers)
Closed 3 years ago.
I'm making a system where you can assign guests to certain rooms. To unassign a guest, two UPDATE queries must be executed in two different tables, but only one of these is actually executing. The one that doesn't work, doesn't put out an error message, but doesn't seem to do anything.
I've already printed out all the variables I'm using in my query to check for mistakes, but I haven't found any data in those which was wrong. I also checked if there were errors in my query using phpMyAdmin, but they properly executed there. I used the same type of method on another page, updating something else, and that worked. Could someone take a look in my code and tell me what I've done wrong?
Thanks in advance!
if(isset($_GET['view'])) {
$viewid = $_GET['view'];
$sql = "SELECT * FROM renterinfo LEFT JOIN apartments ON apartments.renterid = renterinfo.renterid WHERE renterinfo.renterid = $viewid";
$xresult = mysqli_query($conn, $sql);
$xrow = mysqli_fetch_array($xresult);
}
$freeroomquery = "SELECT * FROM apartments WHERE status = 1 ORDER BY apartmentno ASC";
$freerooms = mysqli_query($conn, $freeroomquery);
$rentername = $xrow['name'];
$renteremail = $xrow['email'];
$roomno = $xrow['apartmentno'];
$enddate = $xrow['leaveroom'];
if(isset($_POST['submit'])){
//------------------------------------
$name = $_POST['newNaam'];
$email = $_POST['newEmail'];
$room = $_POST['newKamer'];
$leaveroom = $_POST['newEinde'];
$insertSql = "UPDATE renterinfo SET name = '$name', email = '$email', leaveroom = '$leaveroom'";
$updateRoom = "UPDATE apartments SET renterid = NULL WHERE apartmentno = '$roomno'";
The second UPDATE query doesn't update the renterid. The first UPDATE query does work.
if ($conn->query($insertSql) === TRUE) {
$success = "Huurder <strong>'" . $name . " - CobbenCampus'</strong> is succesvol toegevoegd.";
} else {
$error = "Er is een fout opgetreden bij het toevoegen van appartement <strong>'" . $apartmentno . " - CobbenCampus'</strong>. Zijn alle velden ingevuld?<br><strong>Foutmelding:</strong> " . $sql . "<br>" . $conn->error;
}
if ($conn->query($updateRoom) === TRUE) {
echo "Succeeded.";
} else {
echo "Query unsuccessfull.";
}
$conn->close();
}
HTML-code:
<form action="viewrenter.php" method="POST" enctype="multipart/form-data">
<label id="first">Voor- en achternaam huurder:</label><br>
<input type="text" name="newNaam" value="<?php echo $rentername ?>"><br><br>
<hr class="line-black">
<label id="first">E-mailadres van huurder:</label><br>
<input type="text" name="newEmail" value="<?php echo $renteremail ?>"><br><br>
<hr class="line-black">
<label id="first">Kamernummer:</label><br/>
<p>Een kamer toewijzen aan een huurder, zal de kamer automatisch op bezet zetten.</p>
<select name="newKamer">
<option value="<?php echo $roomno ?>"><?php echo $roomno ?></option>
<option value="0">Geen kamer</option>
<?php
while($rooms = mysqli_fetch_array($freerooms)) {
echo "<option value='" . $rooms['apartmentno'] . "'>" . $rooms['apartmentno'] . "</option>";
}
?>
</select><br><br>
<hr class="line-black">
<label id="first">Afloopdatum huurdersovereenkomst:</label><br>
<input type="text" name="newEinde" placeholder="Als: 25-04-2019" value="<?php echo $enddate ?>"><br><br>
<button class="spaced btn btn-primary btn-wide" type="submit" name="submit">Voeg huurder toe</button>
<hr class="bigspacer">
</form>
UPDATE apartments SET `renterid` = NULL WHERE `apartmentno` = `$roomno`
Write query like this. I think it will work.

how can i insert records of the option that was selected from database?

so as you all can see this is a little piece of the form that is causing a problem.
i am getting all the records in the form with a dropdown and want to send the selected result to the database.
after submitting records are inserted correct except this two: $locatie =$_POST["locatie"]; and $vestiging = $_POST["vestiging"];
are inserted as NULL.
i would like it to send the records that was selected in the form.
can someone help me?
<form name="nieuwTicket" action="nieuwTicketNieuwKlant.php" method="POST">
<label class="hidden02">locatie:</label>
<select name="locatie" class="hidden2">
<option value = "">---Select---</option>
<?php
$ophaall = "SELECT * FROM locatie ";
$resultl = mysqli_query($connectie, $ophaall);
while ($l = mysqli_fetch_assoc($resultl)) {
echo "<option value='{" . $l['locatieId'] . "}'>" . $l['locatieId'] . " " . $l['locOmschrijving'] . "</option>";
}
?>
</select><br>
<label class="hidden02">vestiging:</label>
<select name="vestiging" class="hidden2"> <!-- Disabled, gaan we nog niets mee doen-->
<option value = "">---Select---</option>
<?php
$ophaalv = "SELECT * FROM vestigingen ";
$resultv = mysqli_query($connectie, $ophaalv);
while ($v = mysqli_fetch_assoc($resultv)) {
echo "<option value='{" . $v['vestigingId'] . "}'>" . $v['vestigingId'] . " " . $v['vesOmschrijving'] . "</option>";
}
?>
</select><br>
<input type="submit" name="submit1" value="invoeren" class="hidden" />
</form>
<!--submit insert to database-->
<?php
$locatie = $_POST["locatie"];
$vestiging = $_POST["vestiging"];
if (isset($_POST['submit1'])) {
// nieuwe klant
$insertklant = $connectie->prepare("INSERT INTO klant (klantId, klantAchternaam, klantNaam, klantTel,
klantAdres, klantPostc, klantStad, klantEmail, instantieId, locatieId)
VALUES ('',?,?,?,?,?,?,?,?,?)");
if ($insertklant) {
$insertklant->bind_param('sssssssii', $achternaam, $naam, $tel, $adres, $postcode, $stad, $email, $locatie, $vestiging);
if ($insertklant->execute()) {
echo ' klant gemaakt!';
}
}
?>

How to keep file from deleting

I am creating a website where I can edit texts, and upload a picture for my texts. The picture get uploaded to a server and the file path is saved in the MySQL database. I made a function to update content, including the picture. If I upload a new picture, the old one gets deleted from the server.
The problem is if I DON'T upload the picture, the old picture still gets deleted!
I just started programming and I need your help with this guys!
My code
<?php
include_once('../include/config.php');
include_once('../include/functions.php');
getHeader('Boek updaten', 'Schrijf een boek!');
if (isset($_GET['action']) == 'save')
{
$id = $_GET["id"];
$boek_naam = $_POST['boek_naam'];
$boek_text = $_POST['boek_text'];
$boek_genre = $_POST['boek_genre'];
$allow = array("jpg", "jpeg", "gif", "png");
if(isset($_FILES['cover']['name'])) {
$query = $db->prepare("
SELECT `boeken_cover` FROM `boeken` WHERE boeken_id = '$id'");
$query->execute();
foreach($query->fetchAll() as $row){
$old_cover = $row['boeken_cover'];
}
chmod($old_cover, 0600);
unlink($old_cover);
$imgdir = "../images/";
$fulldir = $imgdir.$_FILES['cover']['name'];
if (!!$_FILES['cover']['tmp_name']) {
$info = explode('.', strtolower($_FILES['cover']['name']));
if (in_array(end($info), $allow)) {
if (move_uploaded_file($_FILES['cover']['tmp_name'], $imgdir . basename($_FILES['cover']['name']))) {
echo "Foto uploaden gelukt";
}
}
}
try
{
$query=$db->prepare("
UPDATE `boeken` SET `boeken_naam` = :naam, `boeken_text` = :text, `boeken_genre` = :genre, `boeken_cover` = :fulldir WHERE `boeken_id` ='$id'");
$query->bindParam(':naam', $boek_naam);
$query->bindParam(':text', $boek_text);
$query->bindParam(':genre', $boek_genre);
$query->bindParam(':fulldir', $fulldir);
$query->execute();
echo "Het boek is geupdated. Klik <a href='index.php'>hier</a> om naar het overzicht te gaan.";
}
catch(PDOException $e)
{
$sMsg = '<p>
Regelnummer: '.$e->getLine().'<br/>
Bestand: '.$e->getFile(). '<br/>
Foutmelding: '.$e->getMessage().'
</p>';
trigger_error($sMsg);
}
}
else{
try
{
$query=$db->prepare("
UPDATE `boeken` SET `boeken_naam` = '$boek_naam', `boeken_text` = '$boek_text', `boeken_genre` = '$boek_genre' WHERE `boeken_id` ='$id'");
$query->execute();
echo "Het boek is geupdated. Klik <a href='index.php'>hier</a> om naar het overzicht te gaan.";
}
catch(PDOException $e)
{
$sMsg = '<p>
Regelnummer: '.$e->getLine().'<br/>
Bestand: '.$e->getFile(). '<br/>
Foutmelding: '.$e->getMessage().'
</p>';
trigger_error($sMsg);
}
}
}
else {
$id = $_GET["id"];
try {
$query = $db->prepare("
SELECT * FROM `boeken` WHERE boeken_id = '$id'");
$query->execute();
} catch (PDOException $e) {
$sMsg = "<p>
Regelnummer: " . $e->getLine() . "<br />
Bestand: " . $e->getFile() . "<br />
Foutmelding: " . $e->getMessage() . "
</p>";
trigger_error($sMsg);
}
while ($rij = $query->fetch(PDO::FETCH_ASSOC)) {
$boeken_id = $rij["boeken_id"];
$boeken_naam = $rij["boeken_naam"];
$boeken_genre = $rij["boeken_genre"];
$boeken_text = $rij["boeken_text"];
}
echo "
<form name = \"boeken updaten\" action=\"?action=save&id=$id\" method=\"post\" enctype=\"multipart/form-data\">
<table>
<tr>
<td>Boek naam</td>
<td><input type=\"text\" name=\"boek_naam\" value=\"$boeken_naam\" required> </td>
</tr>
<tr>
<td>Boek genre</td>
<td><input type=\"text\" name=\"boek_genre\" value=\"$boeken_genre\" required> </td>
</tr>
<tr>
<td>Cover</td>
<td><input type='file' name='cover' id='cover'></td>
</tr>
<tr>
<td>Inhoud</td>
<td><textarea cols='50' rows='20' name='boek_text'>$boeken_text</textarea></td>
</tr>
<tr>
<td colspan=\"2\" ><input type=\"reset\" name=\"reset\" value=\"Leeg maken\">
<input type=\"submit\" name=\"submit\" value=\"Updaten\"</td>
</tr>
</table>
</form>";
}
getFooter();
?>
Thank you guys!
Ali
You have to check in posted form if it having picture or not
if(!empty($_FILES['NAME_OF_YOUR_INPUT_TYPE_FILE']['tmp_name'])){
unlink($oldFile);
}
This code with the extra ! will have the effect of reversing the state
if (!!$_FILES['cover']['tmp_name']) {
says if ( not not $_FILES['cover']['tmp_name']) {
use only one !
if (!$_FILES['cover']['tmp_name']) {
Try to change the following line
if(isset($_FILES['cover']['name'])) {}
to
if(isset($_FILES['cover']['name']) && !empty($_FILES['cover']['name'])) {}

PHP - what is wrong here?

I'm making a Car rental system in school and i have run into an issue I cannot fix.
When I have completed my Query to ask for all the cars matching the users search I have a buttun they can click to rent that car and it redirects them to another page. Lets say you want to rent a Ferrari F12 and click on rent. When you get to the other page it says "Rent Opel Corsa" no matter what car i choose. Probably just a silly oversight on my side.
Here is the Search Function
function søk(){
define('DB_HOST', 'db-kurs.hit.no');
define('DB_NAME', 'v15gdb1');
define('DB_USER','v15g1');
define('DB_PASSWORD','pw1');
$con=mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Kunne ikke koble til MySQL: " . mysql_error());
$db=mysql_select_db(DB_NAME,$con) or die("Kunne ikke koble til MySQL: " . mysql_error());
$sql = mysql_query("SELECT * FROM Biler WHERE Sted= '$_POST[by]' AND Merke = '$_POST[biler]'");
$resultat = $sql;
$mld = "Du har valgt å søke etter en <b>$_POST[biler]</b> i <b>$_POST[by]</b>.";
if ($_POST['by'] == 'Alle'){
$sql = mysql_query("SELECT * FROM Biler WHERE Merke = '$_POST[biler]'");
$resultat = $sql;
$mld = "Du har valgt å søke etter en <b>$_POST[biler]</b> i <b>alle</b> byer.";
}
if ($_POST['biler'] == 'Alle'){
$sql = mysql_query("SELECT * FROM Biler WHERE Sted= '$_POST[by]' ");
$resultat = $sql;
$mld = "Du har valgt å søke etter <b>alle</b> bilene i <b>$_POST[by]</b>.";
}
if ($_POST['by'] == 'Alle' and $_POST['biler'] == 'Alle'){
$sql = mysql_query("SELECT * FROM Biler");
$resultat = $sql;
$mld = "Du har valgt å søke etter <b>alle</b> bilene i <b>alle</b> byene.";
}
echo "<div id='søkmld'>$mld Husk at du må logge inn for å leie en bil.</div>";
while($row = mysql_fetch_array($resultat, MYSQL_ASSOC))
{
$merke = $_SESSION['merke'] = $row['Merke'];
$modell = $_SESSION['modell'] = $row['Modell'];
$reg = $row['RegNr'];
$pris = $row['Dagspris'];
$sted = $row['Sted'];
$status = $row['erLedig'];
$bilde = $row['Bilde'];
$_SESSION['name'] = $merke. " " .$modell;
echo" <div id='bilsøk'>
<div id='biltype'><h3> $merke $modell</div>
<div id='lei'>
<form method='POST' action='leibil.php' name='leibil'>";
if(isset($_SESSION['brukerID'])){
echo"<input type='submit' value='Lei Nå' name='lei $merke $modell'>
</div>
<div id='bilsøktxt'>
<table>
<tr>
<td>RegNr:</td>
<td>$reg</td>
</tr>
<tr>
<td>Pris:</td>
<td>$pris</td>
</tr>
<tr>
<td>Sted:</td>
<td>$sted</td>
</tr>
<tr>
<td>Status:</td>
<td>";
if ($status == 1){
echo'Ledig';}
if ($status == 0){
echo'Utleid';}
echo "</td>
</tr>
</table>
</div>
<div id='søkimg'><center><img src='images/cars/$bilde.jpg'></center></div>
</div>
";
}
}}
And here is the div on the page you are redirected to:
<div id="hoved">
<?php
echo"<h1>Lei $_SESSION[name]</h1>"
?>
<?php
status();
?> <hr/>
</div>
The Status() function is set just so you know.
#dbinns66 is correct, you are inconsistent with your session and post variables and this could cause problems.
Also, a stab in the dark here...
When you do:
while($row = mysql_fetch_array($resultat, MYSQL_ASSOC))
What happens if you have more than 1 row?
Lets say your query returns 2 rows. When you do:
$merke = $_SESSION['merke'] = $row['Merke'];
$modell = $_SESSION['modell'] = $row['Modell'];
...
$_SESSION['name'] = $merke. " " .$modell;
$_SESSION['name'] will always contain the make and model of the car that is in the last row of your query (in this case, row 2).
Anyway...
To fix this, try adding:
...
echo"<input type='submit' value='Lei Nå' name='lei $merke $modell'>
<input type='hidden' value='$merke $modell' name='makeAndModel'>
...
In your redirected page (leibil.php):
<div id="hoved">
<?php
echo"<h1>Lei " . $_POST['makeAndModel'] . "</h1>"
?>
...
Close your form in your first echo statement:
...
<div id='søkimg'><center><img src='images/cars/$bilde.jpg'></center
</div>
</form>
...
For starters, could you try changing
$_SESSION[name]
to
$_SESSION["name"]
Also, I noticed in building your SQL you have clauses like
... Sted= '$_POST[by]' AND ...
Since "by" is the hash key of the value you're looking for, I'd try changing those to be more like
... Sted= '" . $_POST["by"] . "' AND

mysqli_real_escape_query, query seems ok, but nothing written in database

I have a form, method="post" where users can input info like their name and email, that then get's inserted in a database. For safety I tried to use mysqli_real_escape_string.
Now, the query says it worked but no data get's inserted in my database. Without the escape everything worked allright too (except for not being escaped)
CODE:
(Updated missing quote, it's there in my original code, so that's not the problem. Sry for that)
if(isset($_POST['submit'])) {
$email = explode('#',$_POST['mail']); //explode because I only need the prefix
$maila = mysqli_real_escape_string($link,$email[0]);
$name = mysqli_real_escape_string($link,$_POST['name']);
$query = "INSERT INTO base(mail,name) VALUES ('$maila','$name')";
if(mysqli_query($link,$query)) {
echo "SUCCES";
}
else {
echo "FAIL";}
}
So when I process the query, SUCCES comes up but the mail and name don't arrive in my table.
I googled and searched here, but couldn't find a solution (excuse me if I overlooked it). I also hope I posted enough of my code.
Extra info:
Before the SQL-query goes into action the form is checked in a way like
if($_POST['name'] == null){echo "an error message";}
EDIT; FULL CODE (I am aware that there are mistakes/stupid things in my if-statements, but these work fine without escaping so I will check these later)
<?php
if(isset($_POST['submit'])) {
if($_POST['ios'] == null ) {$resios = 0;} else {$resios = $_POST['ios'];}
if($_POST['android'] == null) {$resand = 0;} else {$resand = $_POST['android'];}
if($_POST['windows'] == null) {$reswin = 0;} else {$reswin = $_POST['windows'];}
//Check for errors
if($_POST['naam'] == null) {echo "<span class=\"error\">Gelieve een naam in te vullen</span><br />";}
if($_POST['opleiding'] == 0) {echo "<span class=\"error\">Selecteer een opleiding</span><br />";}
if($resios > $ios) {$resios = $ios; echo "<span class=\"error\">Aantal iOS tablets overschreden. Maximum " . $ios . " tablets beschikbaar.</span><br />";}
if($resand > $android) {$resand = $android; echo "<span class=\"error\">Aantal Android tablets overschreden. Maximum " . $android . " tablets beschikbaar.</span><br />";}
if($reswin > $windows) {$reswin = $reswin; echo "<span class=\"error\">Aantal Windows tablets overschreden. Maximum " . $windows . " tablets beschikbaar.</span><br />";}
if($resios < 0) {$resios = 0; echo "<span class=\"error\">Aantal tablets kan niet lager zijn dan 0!</span><br />";}
if($resand < 0) {$resand = 0; echo "<span class=\"error\">Aantal tablets kan niet lager zijn dan 0!</span><br />";}
if($reswin < 0) {$reswin = 0; echo "<span class=\"error\">Aantal tablets kan niet lager zijn dan 0!</span><br />";}
if($_POST['terms'] != 'on') {echo "<span class=\"error\">Reglement moet aanvaard worden.</span><br />";}
if($resios == 0 && $resand == 0 && $reswin == 0) {echo "<span class=\"error\">Er moet minstens 1 tablet gereserveerd worden</span>";}
else {
//ESCAPE + INSERT
$email = explode('#',$_POST['mail']);
$maila = mysqli_real_escape_string($link,$email[0]);
$opleiding = mysqli_real_escape_string($link,$_POST['opleiding']);
$naam = mysqli_real_escape_string($link,$_POST['naam']);
$datum = mysqli_real_escape_string($link,$datum);
$resios = mysqli_real_escape_string($link,$resios);
$resand = mysqli_real_escape_string($link,$resand);
$reswin = mysqli_real_escape_string($link,$reswin);
$opmerking = mysqli_real_escape_string($link,$_POST['opmerking']);
$query = "INSERT INTO reservaties(oplid,naam,datum,ios,android,windows,emailname,opmerking) VALUES ('$opleiding','$naam','$datum','$resios','$resand', '$reswin','$maila', '$opmerking')";
if(mysqli_query($link,$query)) {
echo "<p class=\"succes\">U hebt succesvol " . $resios . " iOS-tablets, " . $resand . " Android-tablets en " . $reswin . " Windows-tablets gereserveerd op " . $disdate . "</p>";
echo "<p>Een bevesting van uw reservatie via mail? <form style=\"display:inline;\" target=\"_blank\" action=\"print.php\" method=\"post\"><input type=\"text\" name=\"mail\" value=\"".$maila."\" />#arteveldehs.be <input type=\"hidden\" name=\"naam\" value=\"".$_POST['naam']."\"/><input type=\"hidden\" name=\"datum\" value=\"". $datum . "\"/><input type=\"submit\" name=\"print\" value=\"mail\"></form></p>";
}
else {
echo "<p class=\"error\">Er is een fout opgetreden. Probeer opnieuw, of neem contact op met de Mediatheek.</p>";}
}
}
?>
<!-- my form-->
<form action="#" method="post">
<table>
<tr><td colspan="3"><span class="required">*</span> = verplicht veld</td></tr>
<tr><td>Naam:<span class="required">*</span></td><td><input type="text" name="naam" placeholder="Naam" /></td></tr>
<tr><td>Email:</td><td><input type="text" name="mail" placeholder="voornaam.naam" />#arteveldehs.be</td></tr>
<tr><td>Opleiding:<span class="required">*</span></td><td colspan="2">
<select name="opleiding">
<option value="0">Selecteer een opleiding</option>
<?php
$sql2 = "SELECT SUM(ios) as iostotal,SUM(android) as androidtotal,SUM(windows) as windowstotal FROM reservaties WHERE '$datum' = datum";
$check2 = mysqli_query($link,$sql2) or die(mysql_error());
while ($free2 = mysqli_fetch_array($check2)) {
$iosall = 16;
$andall = 18;
$winall = 20;
$ios2 = $iosall - $free2['iostotal'];
$android2 = $andall - $free2['androidtotal'];
$windows2 = $winall - $free2['windowstotal'];
}
$opleidingen = "SELECT * FROM opleidingen";
$values = mysqli_query($link,$opleidingen) or die(mysql_error());
while ($row = mysqli_fetch_array($values)) {
$oplid = $row['oplid'];$opleiding = $row['opleiding'];
echo "<option value=\"".$oplid."\">".$opleiding."</option>";
}
?>
</select>
</td></tr>
<tr><td>Aantal iOS</td><td><input type="text" name="ios" placeholder="<?= $ios2;?>" ><span class="max">(maximum <?= $ios2;?> beschikbaar)</span></td></tr>
<tr><td>Aantal Android</td><td><input type="text" name="android" placeholder="<?= $android2;?>" ><span class="max">(maximum <?= $android2;?> beschikbaar)</span></td></tr>
<tr><td>Aantal Windows</td><td><input type="text" name="windows" placeholder="<?= $windows2;?>" ><span class="max">(maximum <?= $windows2;?> beschikbaar)</span></td></tr>
<tr><td>Opmerking:</td><td colspan="2"><textarea maxlength="512" rows="5" cols="50" name="opmerking" placeholder="Bv. Tijdstip van oppikken/terugbrengen - vraag aan de mediatheek - ..." ></textarea></td></tr></table>
<input type="checkbox" name="terms" value="on" /> Hiermee verklaar ik me akkoord met het <a target="_blank" href="reglement.php">reglement</a> dat geldt voor het gebruik van deze tablets.<span class="required">*</span>
<p><input type="submit" name="submit" value="Reserveer"/></p>
</form>
<?php }
?>
DATABASE reservaties
resid int(9) PRIMARY KEY
oplid int(9)
naam varchar(55) latin1_swedish_ci
datum date
ios varchar(3)
android varchar(3)
windows varchar(3)
emailname
opmerking varchar(512) latin1_swedish_ci
As per OP's original posted question
You have a missing quote for your query:
$query = "INSERT INTO base(mail,name) VALUES ('$maila','$name') ;
// right there --^
do:
$query = "INSERT INTO base(mail,name) VALUES ('$maila','$name')";
Using error reporting would have helped shown the error http://www.php.net/mysqli_error
Since no error was thrown (at you), this tells me that you are not using error reporting.
if (!mysqli_query($link,$query))
{
die('Error: ' . mysqli_error($link));
}
These or die(mysql_error()) need to be changed to or die(mysqli_error()) since mysql_* and mysqli_* based functions do not mix with each other.
Try this, You have missed to close "
$query = "INSERT INTO `base` (`mail`,`name`) VALUES ('$maila','$name')";
you have missed end " in your query string.
replace this line:
$query = "INSERT INTO base(mail,name) VALUES ('$maila','$name');
by this:
$query = "INSERT INTO base(mail,name) VALUES ('$maila','$name')";

Categories