Question

在我的模态窗口中，我有一个选项，用户可以从我的数据库中选择早餐。通过选择膳食并提交它，记录将保存在数据库中。但是，当用户进入相同的模态窗口时，我希望实现这一点，他们之前选择的任何餐点都将显示为默认选项。我创建了一个查询，检查是否已经选择了一餐（optionquery1）。任何想法实现这一目标？感谢

另一方面，我不确定我是否正在设置日期php变量。日，月和年存储在html隐藏输入中，但不确定如何提取值。

<div id="myModal" class="modal">

<!-- Modal content -->
<div class="modal-content">
<span class="close">&times;</span>
<div class="modal-body">
  <form action="my_planner.php" method="post">
  <!--<span class="test5"></span>-->
  <!--<span class="test"><input type="text" name="id_disabled" value="" disabled/>-->
  <input class="test" type="text" name="id1" value="" style="text-align:center; border:0px; font-weight:bold; font-size: 25px;"/>
  <hr class="my-4">
  <input type="hidden" name="id" value=""/>
  <input type="hidden" name="month" value=""/>
  <input type="hidden" name="year" value=""/>

<br>
<p id="breakfastTag"><br>Breakfast:</p>
<select id="breakfast" name="breakfast">
<option value="" disabled selected>Select your breakast</option>
<?
$meal='breakfast';
$date='<input type="hidden" name="id" value=""/>'.'<input type="hidden" name="month" value=""/>'.'<input type="hidden" name="year" value=""/>';
$query = $db->prepare("select * from recipe_category where categoryID=1");
$query->execute();

while ($results = $query->fetch()) {
    $recipeID=$results["recipeID"];
    $query3 = $db->prepare("select * from recipe where id=:recipeID");
    $dbParams3=array('recipeID'=>$recipeID);
    $query3->execute($dbParams3);
    $breakfast = $query3->fetchAll();

    $optionquery = $db->prepare("select * from user_planner where userID=:userID and date=:date and meal=:meal");
    $optionParams= array ('userID'=>$thisUser, 'date'=>$date,'meal'=>$meal);
    $optionquery->execute($optionParams);

    while ($results12 = $optionquery->fetch()) {
        $selectedmeal = $results12['recipeID'];
    }

    $optionquery1 = $db->prepare("select * from recipe where id=:recipeID");
    $optionParams1= array ('recipeID'=>$selectedmeal);
    $optionquery1->execute($optionParams1);

    while ($results123 = $optionquery1->fetch()) {
        $selectedrecipe = $results123['name'];
    }

    foreach($breakfast as $breakfast): ?>
        <option  value="<?= $breakfast['id']; ?>"><?= $breakfast['name']; ?></option>
    <?php endforeach;} ?>
</select>

Answer 1

当然，您可以使用要显示的selected上的<option>属性。首先从第一个selected：

中删除<option>属性

<option value="" disabled>Select your breakast</option>

然后在foreach循环中，您需要if条件：

$results123 = $optionquery1->fetch();
$selectedrecipe = $results123['id'];

foreach($breakfast as $breakfast): ?>
    <option  value="<?= $breakfast['id']; ?>" <?php if ($breakfast['id'] == $selectedrecipe) echo 'selected' ?>><?= $breakfast['name']; ?></option>
<?php endforeach;} ?>

注意我已经删除了while循环，因为我们确定Mysql只返回一条记录（WHERE ID = <ID>）。

将选项选项默认设置为先前设置的选项

1 个答案: