Question

我有一个问题。我想制作一个历史时间表，在这个时间轴上用户可以添加＆＃39;元素＆＃39; （例如每年的元素。）。

在表单1中，用户创建时间轴并说明时间线上的元素数量，这些元素将被发送到数据库（这是有效的）。在表单2中，用户可以填充元素。（这不起作用）。

在表单1（如果有人想要5个元素）发布后，数据库会创建5条记录（每个元素一条）。

现在我想要5种不同的形式来填充这5个元素（标题，描述等），但是有1个提交按钮。

现在，如果我尝试这个（见下面的代码），只有第5个表格被发布... 我如何让它工作？

提前致谢，

<?php

error_reporting(E_ALL);
ini_set('display_errors', 1);

require_once('classes/database.class.php');

$DB = Database::getInstance();

$titel = '';
$beschrijving = '';
$afbeeldingURL = '';
$jaar = '';
$element_id = '';

$velden = '';

$sql3 = "SELECT `id` , `aantal_elementen` FROM `tijdlijn` ORDER BY id DESC LIMIT 0, 1";
$last_id = $DB->_query($sql3);

if ($last_id->num_rows > 0) {
    while($row = $last_id->fetch_assoc()) {
        $last_id2 = $row["id"]; 
        $aantal_elementen3 = $row["aantal_elementen"];
    }
}

$sql4 = "SELECT * FROM `elementen` WHERE `tijdlijn_id` = ".$last_id2."";

$sql5 = "SELECT `id` FROM `elementen` WHERE `tijdlijn_id` = ".$last_id2." ORDER BY id ASC LIMIT 0,1";

$sql6 = "SELECT `id` FROM `elementen` WHERE `tijdlijn_id` = ".$last_id2." ORDER BY id DESC LIMIT 0,1";

$sql7 = "SELECT `id` FROM `elementen` WHERE `tijdlijn_id` = ".$last_id2." ORDER BY id ASC";

$last_id3 = $DB->_query($sql4);

$min_id = $DB->_query($sql5);
$min_id = mysqli_fetch_assoc($min_id);
$min_id = $min_id['id'];

$max_id = $DB->_query($sql6);
$max_id = mysqli_fetch_assoc($max_id);
$max_id = $max_id['id'];

$dingid[] = $DB->_query($sql7);

echo $min_id."</br>";
echo $max_id."</br>";
$current_id = $min_id;

if (isset($_POST) && !empty($_POST)) {

    $boolError = false;

    foreach ($_POST as $k => $v) {
        $_POST[$k] = trim($v);
    }

    if (!isset($_POST['titel']) || trim($_POST['titel']) == '') {
        $titel    = 'error';
       $boolError = true;
    }

    if (!isset($_POST['beschrijving']) || trim($_POST['beschrijving']) == '') {
        $beschrijving = 'error';
        $boolError = true;
    } 

    if (!isset($_POST['afbeeldingURL']) || trim($_POST['afbeeldingURL']) == '') {
    } 

    if (!isset($_POST['jaar']) || trim($_POST['jaar']) == '') {
        $jaar_start  = 'error';
        $boolError = true;
    } 

    if (
        !isset($_POST['element_id'])
        || trim($_POST['element_id']) == ''
        || !preg_match('/^\d+$/', $_POST['element_id'])
    ) {
        $element_id  = 'error';
        $boolError = true;
    } 


    if ($boolError === false) {
        $sql = "UPDATE `elementen`
                SET
                `titel` = '" . $_POST['titel'] . "',
                `beschrijving` = '" . $_POST['beschrijving'] . "',
                `afbeelding_url` = '" . $_POST['afbeeldingURL'] . "',
                `jaar` = '" . $_POST['jaar'] . "'
                WHERE `id` = ".$_POST['element_id'].";
        ";


        if ($DB->_query($sql)) {
         //$current_id++;
         //exit;


        } else {
            header('Location: ?oops2');
           exit;

        }

    } else {
        $velden = "Ff alles invullen he";
    }
?> <p>Vul het formulier hier onder in voor de elementen en maak een tijdlijn!</p><?php
} 

     if ($last_id3->num_rows > 0) {
        while($row2 = $last_id3->fetch_assoc()) {
            $last_id4 = $row2["id"]; 
?>


        <form class="form-nieuws" method="post">

            </br></br>
            <?php echo $last_id2; ?>
            <?php echo $current_id; ?>
            <input id="titel2" class="<?= $titel ?> form-control" type="text" placeholder="Titel tijdlijn" name="titel" value="<?= isset($_POST['titel']) ? $_POST['titel'] : '' ?>">
            <br>
            <textarea id="beschrijving2" class="<?= $beschrijving ?> form-control" rows="3" placeholder="Beschrijving over tijdlijn" name="beschrijving" value="<?= isset($_POST['beschrijving']) ? $_POST['beschrijving'] : '' ?>"></textarea>
            <br>
            <input id="afbeeldingURL2" class="<?= $afbeeldingURL ?> form-control" type="text" placeholder="URL voor afbeelding" name="afbeeldingURL" value="<?= isset($_POST['afbeeldingURL']) ? $_POST['afbeeldingURL'] : '' ?>">
            <br>            
            <input id="jaar" class="<?= $jaar ?> form-control" type="text" placeholder="Start jaar tijdlijn" name="jaar" value="<?= isset($_POST['jaar']) ? $_POST['jaar'] : '' ?>">
             <input id="element_id" class="<?= $element_id ?> form-control" type="text" placeholder="<?php echo $last_id4 ?>" name="element_id" value="<?= isset($_POST['element_id']) ? $_POST['element_id'] : '' ?>">
            <div class="foutlabel">
                <em class="<?= ($velden) ? 'error' : '' ?>"><?= ($velden) ? $velden : '' ?>
                </em>
            </div>
            </br>    


                     <?php $current_id++;} 

                     } ?> <div class="gabutton">
                <input type="submit" value="Creeer tijdlijn">
            </div>


        </form>

Answer 1

最好的方法是只为每个元素提供一个输入表单。您可以根据元素生成每个表单输入。

通过这种方式，您只有一个按钮和一个调用来将数据发送到您的服务器。

在服务器端，您只需要解析和检查数据，并注意您发送的元素数量。

在PHP中循环表单，使用1按钮提交所有表单

1 个答案: