现在我一直试图找到一种方法,使用PHP将用户在Web界面上的下拉菜单选择插入到MySQL数据库table_2中。问题是下拉列表项是从另一个table_2从MySQL数据库中检索的。有人可以帮帮我吗?先感谢您!下面显示了我正在使用的代码。
<?php
$con = mysqli_connect("localhost","root","");
$myDB = mysqli_select_db($con, "database");
$sqlSELECT = mysqli_query($con, 'SELECT disastergroup FROM disastergroups');
if (isset($_POST['group']))
{
$group = $_POST['group'];
$test = "SELECT disastergroupid FROM disastergroups WHERE disastergroup = '$group'";
mysqli_query($con, $test);
$test_store = "INSERT INTO events (groupid_FK) VALUES ($test);"
mysqli_query($con,$test_store);
}
else
{
echo "An option must be selected!";
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Title of Page</title>
<meta http-equiv="content-type" content="text/html; charset=utf-8"/>
</head>
<body>
<form action = "detailslog.php" method = "POST">
<label for="groups">Disasters:</label>
<select name = "groups">
<option value = "">Select...</option>
<?php while($row = mysqli_fetch_assoc($sqlSELECT)):;?>
<option><?php $row1['disastergroup'];?></option>
<?php endwhile;?>
</select>
<input type="submit" value="Submit Data">
</form>
</body>
</html>
所以我想做的就是从选项&#34;群组&#34;中选择用户。并使用该值从表灾难组中获取该值的ID,然后将该ID存储到表中&#34; events&#34;作为外键。这让我很难理解。任何帮助将不胜感激!谢谢!
答案 0 :(得分:0)
$sql_quer = mysqli_query($con, $test);
$FK_id = mysqli_fetch_assoc($sql_quer);
$test_store = "INSERT INTO events (groupid_FK) VALUES ($FK_id[disastergroupid]);"
mysqli_query($con,$test_store);
这解决了查看此帖子的任何人的问题!