Mysqli准备声明 - 返回错误,但为什么?

[英]Mysqli Prepare Statement - Returning False, but Why?


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

#1


44  

I'm copying the solution into this answer so this can be given an upvote, otherwise the question will appear in the "unanswered questions" forever. I'm marking this answer CW so I won't get any points.

我正在将解决方案复制到这个答案中,因此可以给予一个upvote,否则问题将永远出现在“未答复的问题”中。我将这个答案标记为CW,所以我不会得到任何积分。

@Andrew E. says:

@Andrew E.说:

I just turned on mysqli_report(MYSQLI_REPORT_ALL) to get a better understanding of what was going on - turns out that one of my field names was incorrect - you'd think that prepare() would throw an exception, but it fails silently.

我只是打开mysqli_report(MYSQLI_REPORT_ALL)来更好地理解发生了什么 - 事实证明我的一个字段名称是不正确的 - 你认为prepare()会抛出异常,但它会无声地失败。


注意!

本站翻译的文章,版权归属于本站,未经许可禁止转摘,转摘请注明本文地址:https://www.itdaan.com/blog/2009/08/02/c0b854d542ccd4bc6eb38324d096f4a9.html



 
  © 2014-2022 ITdaan.com 联系我们: