我有一个表单，其中包含一个多选框，其中填充了从数据库表中提取的治疗师名称。
这是我的代码：
<div class="form-group col-sm-6 col-md-4">
<label for="therapist"><strong>Senior Practitional / SP: </strong></label>
<select name="therapist[]" id="therapist" multiple="multiple" class="form-control selectpicker" multiple data-live-search="true" data-live-search-placeholder="Search" data-actions-box="true" data-parsley-trigger="change" required-no>
<?php
require_once('include/database.php');
// read current record's data
try {
// prepare select query
$getUser = "select firstname, lastname, profession from user WHERE user_type = 'therapist'";
$stmt = $con->prepare($getUser);
// execute our query
$stmt->execute();
// store retrieved row to a variable
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
// values to fill up our form
$firstname = $row['firstname'];
$lastname = $row['lastname'];
$profession = $row['profession'];
echo '<option id="firstname_' . $row['id'] . '" value="' . $row['id'] . '">' . $row['firstname'] . ' ' . $row['lastname'] . ' - ' . $row['profession'] . '</option> ';
}
}
// show error
catch (PDOException $exception) {
die('ERROR: ' . $exception->getMessage());
}
?>
</select>
</div>
但我面临的问题是，如果我选择多位治疗师，我只会在结果表中得到1的整数。
**第二期**
我遇到的另一个问题是，我有一个time_of_visit表单字段，它是一个多选：
<?php $time_of_visit = $time_of_visit; ?>
<div class="form-group col-md-4">
<label for="tov"><strong>Time of Visit: </strong></label>
<select name="tov[]" class="form-control selectpicker" value='<?php echo $time_of_visit; ?>' multiple
data-live-search="true" data-live-search-placeholder="Search" data-actions-box="true" data-parsley-trigger="change" required-no>
<!-- <option selected>Select Visit Time...</option> -->
<option name="tov[]" <?php if (isset($time_of_visit) && $time_of_visit == "am") echo "selected"; ?> value="am" selected>Am</option>
<option name="tov[]" <?php if (isset($time_of_visit) && $time_of_visit == "lunch") echo "selected"; ?> value="lunch" selected>Lunch</option>
<option name="tov[]" <?php if (isset($time_of_visit) && $time_of_visit == "pm") echo "selected"; ?> value="pm">Pm</option>
<option name="tov[]" <?php if (isset($time_of_visit) && $time_of_visit == "am_pm") echo "selected"; ?> value="am_pm">Am or Pm</option>
<option name="tov[]" <?php if (isset($time_of_visit) && $time_of_visit == "pm_care") echo "selected"; ?> value="pm_care">Pm Care</option>
<option name="tov[]" <?php if (isset($time_of_visit) && $time_of_visit == "single") echo "selected"; ?> value="5pm" selected>>5pm</option>
</select>
</div>
提交表单后，我收到time_of_visit cannot be null。我猜我可能已经将它设置为数据库中的Yes NULL，我可以稍后更改，但无论查询失败还是成功，我似乎无法保留所选选项。
答案 0 :(得分：0)
我稍微清理了你的代码并重写了它，以便你可以测试提交表单时会发生什么。试试这个并确保表单输出您选择的正确ID。
<?php
require_once('include/database.php');
// read current record's data
try {
// prepare select query
$getUser = "SELECT id, firstname, lastname, profession FROM user WHERE user_type = 'therapist'";
$stmt = $con->prepare($getUser);
// execute our query
$stmt->execute();
// initialize options variable so that it doesn't get overwritten
$options = "";
// store retrieved row to a variable
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
// values to fill up our form
$firstname = $row['firstname'];
$lastname = $row['lastname'];
$profession = $row['profession'];
/*use ".=" so that you add to the $options, rather than overwrite the last row*/
$options .= '<option value=' . $row['id'] . '>' . $row['firstname'] . ' ' . $row['lastname'] . ' - ' . $row['profession'] . '</option>';
}
}
// show error
catch (PDOException $exception) {
die('ERROR: ' . $exception->getMessage());
}
// echo selections on form submit
if(isset($_POST['submit'])) {
foreach($_POST['therapist'] as $selection) {
echo $selection . "\n";
}
?>
<div class="form-group col-sm-6 col-md-4">
<form action="" method="POST">
<label for="therapist"><strong>Senior Practitional / SP: </strong></label>
<select name="therapist[]" multiple>
<?php echo $options; ?>
</select>
<input type="submit" name="submit" value="Submit"/>
</form>
</div>
<强>更新
您的问题的最终原因是您正在尝试将数组插入到SQL数据库中。您不能将数组直接插入到sql表中，因此必须在插入其值之前转换该数组。请参阅下面的编辑。
if(isset($_POST['submit'])) {
// Use the implode() function to convert the array into a comma-delimited string
$therapists = implode(',', $_POST['therapist']);
// echo $therapists . "<br>"; // Uncomment this line to see how the imploded array looks
// Now you can include the therapists in your INSERT statement like so:
$sql = "INSERT INTO `table` (`column_name`) VALUES ('".$therapists."')";
}
?>
<div class="form-group col-sm-6 col-md-4">
<form action="" method="POST">
<label for="therapist"><strong>Senior Practitional / SP: </strong></label>
<select name="therapist[]" multiple>
<?php echo $options; ?>
</select>
<input type="submit" name="submit" value="Submit"/>
</form>
</div>
在自己的专栏中显示每位治疗师
一旦您查询了您的表并返回了治疗师列中存储的值，您就可以使用explode()函数将字符串转换回数组，以便根据您的评论在新列中显示每位治疗师。< / p>
假设您将这些值存储在变量$therapists：
$string_to_array = explode(",", $therapists);
// print_r($string_to_array); // Uncomment this line to see how the new array looks
// Should look like this: Array ( [0] => therapist1 [1] => therapist2 [2] => therapist3 [3] => therapist4 )
// Now loop through the array to display each name
$columns = ""; // Initialize $columns to avoid being overwritten
foreach ($string_to_array as $key => $value) {
$columns .= "<td>$value</td>";
}
?>
<!--Echo $columns in table row-->
<table>
<tr><?php echo $columns; ?></tr>
</table>