我正在尝试将具有外键的 id 更新为另一个名称表。我有一个下拉菜单，在下拉菜单中，我有表 NAME_TEST 中的名字。我需要选择名称，但我想要的插入是：
INSERT INTO (test) values (the value that i need is the ID of selected name)
法典：
<html> <h1>Update form</h1></html>
<?php
if (isset($_POST['submit'])) {
$connect = mysqli_connect('localhost', 'root', '', 'test');
$query = "UPDATE test SET location_name='".$_POST['new_location']."' WHERE id='".$_POST['location']."' LIMIT 1";
$res = mysqli_query($connect, $query);
if (!$res) {
die("Something went wrong");
}
}
// This is the code to load the select options
$connect = mysqli_connect('localhost', 'root', '', 'test');
$query = 'SELECT * FROM name_test';
$res = mysqli_query($connect, $query);
echo "Choose setup";
$options = array();
while($row = mysqli_fetch_assoc($res)) {
$options[] = $row;
}
?>
<form method="POST" action="<?= $_SERVER['PHP_SELF'] ?>"><BR><BR>
<select name="location">
<option value="0">--- Select an option ---</option>
<?php foreach ($options as $option): ?>
<option value="<?= $option['id'] ?>"><?= $option['name'] ?></option>
<?php endforeach; ?>
</select><br /><BR><BR>
<B> New name:</B> <BR> <input type="text" name="new_location"><br /><BR><BR>
<input type="submit" name="submit" value="Update" />
</form>
您可以使用join。 这个想法是这样的：
INSERT INTO t(nameid)
select nameid
from names n
where n.name = ?;
t是要插入到的表。 names是具有名称 ID 和名称的表。