prepared语句:列不能为null

2w3rbyxf  于 2021-06-21  发布在  Mysql
关注(0)|答案(1)|浏览(318)

我一直在想为什么我在准备好的陈述中会出现这个错误,但我无法解决这个问题。我得到一个错误:

Notice: Undefined index: tipimage in C:\xampp\htdocs\Matt\addtip.php on line 68
ERROR: Could not execute query: INSERT INTO tips (tiptitle, tiptext, tiplink, tipimage) VALUES (?, ?, ?, ?). Column 'tipimage' cannot be null

我的数据库中的所有值都设置为文本类型,id除外。

....
else {
        /* Attempt MySQL server connection. Assuming you are running MySQL
        server with default setting (user 'root' with no password) */
        $mysqli = new mysqli("localhost", "paul", "pass", "yourcomp");

        // Check connection
        if($mysqli === false){
            die("ERROR: Could not connect. " . $mysqli->connect_error);
        }

        // Prepare an insert statement
        $sql = "INSERT INTO tips (tiptitle, tiptext, tiplink, tipimage) VALUES (?, ?, ?, ?)";

        if($stmt = $mysqli->prepare($sql)){
            // Bind variables to the prepared statement as parameters
            $stmt->bind_param("ssss", $tiptitle, $tiptext, $tiplink, $tipimage);

            // Set parameters
            $tiptitle = $_REQUEST['tiptitle'];
            $tiptext = $_REQUEST['tiptext'];
            $tiplink = $_REQUEST['tiplink'];
            $tipimage = $_REQUEST['tipimage'];

            // Attempt to execute the prepared statement
            if($stmt->execute()){
                $successmsg = "<div class='alert alert-success'>Tip Added Successfully!</div>";
            } else{
                echo "ERROR: Could not execute query: $sql. " . $mysqli->error;
            }
        } else{
            echo "ERROR: Could not prepare query: $sql. " . $mysqli->error;
        }
acruukt9

acruukt91#

您正在查找未随请求一起发送的post变量。这就是未定义索引通知试图告诉您的内容。在将它传递到数据库之前,不检查它是否存在,因此它最终被传递为null。只要给一个空字符串的默认值(这里使用null coalesce操作符),它就可以正常工作了。

$tipimage = $_REQUEST['tipimage'] ?? "";

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