Question

所以你们都可以看到这是造成问题的一小部分形式。

我通过下拉列表获取表单中的所有记录，并希望将所选结果发送到数据库。提交记录后插入正确的除了这两个：$ locatie = $ _ POST [＆＃34; locatie＆＃34;];和$ vestiging = $ _POST [＆＃34; vestiging＆＃34;]; 插入为NULL。

我希望它能够发送在表单中选择的记录。有人能帮助我吗？

<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!';
        }
    }

?>

如何插入从数据库中选择的选项的记录？

0 个答案: