我正在尝试构建一个简单的下拉表单,并将提交的内容放在我的数据库中,这样以后我就可以检查结果了。
每当我提交带有submit按钮的表单时,什么都不会发生。这页保持不变..。
Chrome控制台返回一个错误500问题。我做错了什么,因为我现在正在失去它。;-)
index.php
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>V1</title>
</head>
<body>
<form method="post" action="">
<label >I'm feeling...</label>
<select name="subject_names">
<option value="Angry">Angry</option>
<option value="Anxious">Anxious</option>
<option value="Bored">Bored</option>
<option value="Cranky">Cranky</option>
<option value="Crappy">Crappy</option>
<option value="Depressed">Depressed</option>
<option value="Energetic">Energetic</option>
<option value="Exhausted">Exhausted</option>
<option value="Frustrated">Frustrated</option>
<option value="Happy">Happy</option>
<option value="Motivated">Motivated</option>
<option value="Okay">Okay</option>
<option value="Relaxed">Relaxed</option>
<option value="Sad">Stressed</option>
<option value="Lazy">Lazy</option>
</select>
<button type="submit" name="submit" >Submit</button>
</form>
<?php
if(isset($_POST["submit"]))
{
//Including dbconfig file.
include 'db-connect.php';
$subjectName=$_POST["feeling"];
mysql_query("INSERT INTO thefeels VALUES ('$feeling')");
echo " Added Successfully ";
}
?>
</body>
</html>db-connect.php:
$hostname = "localhost";
$username = "USERNAME";
$password = "PASSWORD";
$dbname = "DBNAME";
$conn = mysql_connect($hostname, $username, $password);
if (!$conn)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($dbname, $conn);
?>发布于 2018-05-28 14:12:06
您正在访问未设置的POST值:
$subjectName=$_POST["feeling"];你可能想:
$subjectName=$_POST["subject_names"];不过,您应该停止使用mysql_query。它被废弃,并在PHP7中被删除。使用mysqli或PDO代替。
编辑:错误500可能是由错误的文件权限引起的,例如。你查过许可了吗?
https://stackoverflow.com/questions/50568212
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