我对 HTML 和 PHP 非常陌生。我正在尝试创建一个下拉菜单以将选择插入到我的数据库中。我已经尝试了几个小时才能让它发挥作用。我有一份客户需要填写的评估表格，他们只能从数据库中已有的工作中进行选择。
评估表包含 JobNo、evalDate、raterName、评分和注释(可选)。 在jobs表中找到JobNo为jobNo(jobs表中小写j，evaluation中大写)
如果您能够将数据库中的数据填充为下拉框选项，则可获得奖励积分。我也一直这样尝试。
请告诉我我遗漏了什么或做错了什么？
<!DOCTYPE html>
<html>
<body>
<body style="background-color:#FAEBD7;">
<title>CTS Employment Agency | Evaluation</title> <!-- title for tab on website -->
<center><h1 style="background-color: #FAEBD7;">CTS Employment Agency</h1></center>
<center><p>123 Anywhere St.<br>Huntsville, AL 35649<br>P: (256)555-5555<br>Fax:(256)554-5554<br>E-mail: questions@ctsemployment.com</p></center>
<center><h3>Evaluation Form</h3></center>
<center><p>Please take a moment to fill out the evaluation form concerning the worker that was provided to you.<br>
Please leave a rating from 1(unsatisfactory) to 5(excellent) for the worker along with a comment box for additional information.
<br> If you have any questions, please feel free to send us an email or call us!</p></center>
<form action="php_formStore.php" method="$Post_"> <!-- beginning of form -->
<br>
<center>
Worker First Name: <input type="text" name="wFName">
Middle: <input type="text" name="wMinit">
Last: <input type="Text" name="wLName">
SSN <input type="text" name="snn" placeholder="000-00-0000">
<br><br>
Job No:
<select name="JobNo">
<option value="46">46</option>
<option value="47">48</option>
<option value="48">48</option>
</select>
//<input type="text" name="JobNo"> //this works fine, why doesn't the select box work?
Worker's Job Title: <input type="text" name="title">
Rating: <input type="text" name="rating" placeholder="Rate from 1-5">
<br><br>
Evaluator's Name: <input type "text" name="raterName">
Evaluation Date: <input type="date" name="evalDate">
<br><br>
<!-- comment box -->
<textarea name="comment" rows="8" cols="60">
(Optional)
Leave any additional comments here.
</textarea><br>
<input type="submit" value="Submit"> <!-- The submit buttom -->
<input type="reset">
</center>
</form>
</body>
<html>
这是 PHP 文件
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db="cts employment agency"; //Database Name
$con=mysqli_connect($servername, $username, $password, $db);
//check connection
if(mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$Job_No = $_GET['JobNo'];
$raterName = $_GET['raterName'];
$evalDate = $_GET['evalDate'];
$rating = $_GET['rating'];
$comment = $_GET['comment'];
echo "<br><br><br><br>";
echo "<center><h2>Thank you for taking the time to fill out this evaluation<br>";
echo "Your feedback helps us to provide you with the highest quality employees<br>";
echo "If you have any questions, please e-mail us at: questions@ctsemployment.com or call: (256)555-5555</h2></center>";
mysqli_query($con, "INSERT INTO evaluation (JobNo, evalDate, raterName, rating, comment)
VALUES ('$Job_No', '$evalDate', '$raterName', '$rating', '$comment')");
mysqli_close($con);
?>
[1]: /image/2mJGC.png
最佳答案
在您的 php 脚本中，您尝试使用 $_GET 访问 $_POST 变量。
如果您想按照表单上的建议使用 post，请将声明更改为:
<form action="php_formStore.php" method="post"> <!-- beginning of form -->
然后在您的 php 脚本中，将 $_GET 变量更改为 $_POST，如下所示:
$Job_No = $_POST['JobNo'];
$raterName = $_POST['raterName'];
$evalDate = $_POST['evalDate'];
$rating = $_POST['rating'];
$comment = $_POST['comment'];
您的代码也面临着 SQL 注入(inject)的高风险。您应该研究使用准备好的语句来解决这个问题。下面是相关指南:
https://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php
这对于保持数据完整性和安全性、防止不必要的数据访问至关重要。
关于php - 在数据库中插入下拉菜单项，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/59148977/