Question

准备好的陈述给我带来了麻烦。我可以从数据库中读取数据，但无法写入。这是我遇到麻烦的准备好的陈述。没有错误，但代码执行时没有任何反应。感谢。

 /*Prepared statement option 1*/

    $stmt = mysqli_prepare($con, "INSERT USERS (userEmail, userPassword, userFname, userLname) VALUES (?, ?, ?, ?)");

/*prepared statement option 2*/

    $query = "INSERT INTO users (userEmail, userPassword, userFname, userLname) VALUES (?, ?, ?, ?)";
    $stmt = mysqli_prepare($con, $query);

/ 绑定声明 /

    mysqli_stmt_bind_param($stmt, 'ssss', $userEmail, $userPassword1, $euserFname, $userLname); 

/* execute prepared statement */
    mysqli_stmt_execute($stmt); 

/* close statement and connection */
    mysqli_stmt_close($stmt);

Answer 1

始终在连接到mysqli之前使用此行：

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

mysqli准备的声明没有做任何事情

1 个答案: