下面是视频教程“how to create a database website with php and mysql 07-add-in input form”的最后一部分：
https://www.youtube.com/watch?v=mgig00d1xzc&list=plhpyefl5u-i0zeadf0iplyvm8zoknz70r&index=7
最后是我的代码，用于在数据库中插入新的\u jokes.php脚本的部分（到目前为止，本系列的所有内容都可以很好地工作）
基本上，我得到的似乎是经典的“插入到”不工作，虽然我所有的语法看起来正确。我是不是漏掉了什么明显的东西？我没有得到错误，只是行没有添加。

<?php

include "db_connect.php";

$new_joke_question = $_GET["newjoke"];
$new_joke_answer = $_GET["newanswer"];

// Search the database for the word chicken
echo "<h2>Trying to add a new joke and answer: $new_joke_question  
$new_joke_answer </h2>";

$sql = "INSERT INTO Jokes_table (JokeID, Joke_question, Joke_answer) VALUES 
(NULL, '$new_joke_question', '$new_joke_answer' )";
$result = $mysqli->query($sql);

include "search_all_jokes.php";
?>
<a href="index.php">Return to the main page</a>

以下是所需的db\u connect.php代码：

<?php

// four variables to connect the database
$host = "localhost";
$username = "root";
$user_pass = "usbw";
$database = "test";

// create a database connection instance
$mysqli = new mysqli($host, $username, $user_pass, $database);

?>

这里是search\u all\u jokes.php（它有小错误检查）：

// if there are any values in the table, select them one at a time 
if ($mysqli->connect_errno) {
    echo "Connection to MySQL failed: (" . $mysqli->connect_errno . ") " . 
$mysqli->connect_error;
} 
echo $mysqli->host_info . "<br>";
$sql = "SELECT JokeID, Joke_question, Joke_answer FROM Jokes_table";
$result = $mysqli->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "JokeID: " . $row["JokeID"]. " - Joke_question: " . 
$row["Joke_question"]. " " . $row["Joke_answer"]. "<br>";
    }
} else {
    echo "0 results";
}

?>

下面是在myphpadmin中查看的表结构屏幕截图：

我在new_jokes.php中添加了错误捕获，灵感来自于这个堆栈溢出帖子：insert into syntax error
并得到以下错误：错误：您的sql语法有错误；检查与您的mysql服务器版本对应的手册，以通过tcp/ip在第1localhost行的“t jump.”）附近使用正确的语法