Question

我在使用Web表单将数据输入到SQL中的表时遇到问题。现在，下拉菜单从SQL表中提取数据，允许用户选择一个选项，然后提交表单以将信息添加到“passedexams”中。表。到目前为止，我得到了一个未知的索引＆＃34;第14行和第15行的错误（标有＆＃34; *＆＃34;）。 html网站和php脚本之间的变量都没有完成。

这是HTML：

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Add your passed Exam!</title>
</head>
<body>
<form action="cert.php" method="post">

    <p>
        <label for="name">Name:</label>
<?php

$conn = new mysqli('localhost', '*****', '******', '*******') 
or die ('Cannot connect to db');

    $result = $conn->query("SELECT UID, UName from users");
    echo 
    echo "<select name='Name'>";

    while ($row = $result->fetch_assoc()) {

                  unset($name);
                  $name = $row['UName']; 
                  $UID = $row['UID'];
                  echo "<option value=\"$UID\">$name</option>\n";

}

    echo "</select>";

    echo "<br>";
    echo "<br>\n\n";
    echo "<label for='vendor'>Certification:</label>\n\n";



    $result = $conn->query("SELECT CertName, CertID from certs");


    echo "<select id='CertName' name='CertName'>";

    while ($row = $result->fetch_assoc()) {

                  unset($name);
                  $certname = $row['CertName']; 
                  $certid = $row['CertID'];
                  echo "<option value=\"$certid\">$certname</option>\n";

}

    echo "</select>";
?>
  <br>
  <br>
    <input type="submit" value="Submit">
</form>
</body>
</html>

这是cert.php：

<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "********", "********", "********");

// Check connection
if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

// Escape user inputs for security
//$Uname = mysqli_real_escape_string($link, $_POST['Name']);
//$certname = mysqli_real_escape_string($link, $_POST['CertName']);
"*"$UID = $_POST['UID'];
"*"$certid = $_POST['certid'];

echo $certname;
// attempt insert query execution
$sql = "INSERT INTO passedexams (CertID, UserID) 
VALUES ('$certid', '$UID')";
if(mysqli_query($link, $sql)){
    echo "Records added successfully.";
    // Redirect to home table
    //header( "refresh:5; url=wherever.php" ); 
} else{
    echo "ERROR: Unable to execute $sql. " . mysqli_error($link);
}

// close connection
mysqli_close($link);

?>

Answer 1

您必须使用选择框的名称属性来获取帖子值。更改以下行

$UID = $_POST['UID'];
$certid = $_POST['certid'];

要

$UID = $_POST['Name'];
$certid = $_POST['CertName'];

将Web表单输入的PHP菜单下拉到SQL

1 个答案: