I have a function that generates a prepared INSERT statement based on an associative array of column names and values to be inserted into that column and a table name (a simple string):

我有一个函数，它根据要插入该列的列名和值的关联数组以及一个表名（一个简单的字符串）生成一个准备好的INSERT语句：

function insert ($param, $table) {
        $sqlString = "INSERT INTO $table (".implode(', ',array_keys($param)).') VALUES ('.str_repeat('?, ', (count($param) - 1)).'?)';
        if ($statement = $this->conn->prepare($sqlString)):
            $parameters = array_merge(array($this->bindParams($param), $param));
            call_user_func_array(array($statement, 'bind_param', $parameters));
            if (!$statement->execute()):
                die('Error! '.$statement->error());
            endif;
            $statement->close();
            return true;
        else:
            die("Could Not Run Statement");
        endif;
    }

My problem is that $this->conn->prepare (it's part of a class, conn is a NEW mysqli object, which works with no issues) returns false, but does not give me a reason why!

我的问题是$ this-> conn-> prepare（它是一个类的一部分，conn是一个新的mysqli对象，它没有问题）返回false，但是没有给我一个理由！

Here is a sample $sqlString that gets built for the prepare statement:

以下是为prepare语句构建的示例$ sqlString：

INSERT INTO students (PhoneNumber, FirstName, MiddleInit, LastName, Email, Password, SignupType, Active, SignupDate) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)

Can anyone see a problem with this parameterized statement? Any reason the prepare function would return false?

任何人都可以看到这个参数化语句的问题？ prepare函数返回false的原因是什么？

1 个解决方案