MySQLi准备的插入语句失败 [英] MySQLi prepared insert statement fails

问题描述

我正在更新PHP以使用mysqli::而不是mysql_*，并且遇到了INSERT语句问题.我有以下说法:

I am updating my PHP to use mysqli:: instead of mysql_* and I have run into an issue with INSERT statements. I have the following statement:

$stmt = $link->prepare("INSERT INTO `table` (`a`, `b`, `c`) VALUES(?, ?, ?)");
$stmt->bind_param("sss", $a, $b, "0");
$stmt->execute();

我检查了$stmt，它是一个正确的mysqli_stmt对象.它已经准备妥当，但是由于某种原因，该语句将无法执行.我刚从服务器收到500错误.

I have checked $stmt and it is a proper mysqli_stmt object. It is prepared properly, but for some reason, the statement won't execute. I just get a 500 error from my server.

我想念什么?

修改

我确定问题出在bind_param方法上.

I've determined that the issue is coming from the bind_param method.

编辑2

好的，所以PHP给我的错误是这样的:

Okay, so the error PHP is giving me is this:

致命错误:无法在...中通过引用传递参数4.

Fatal error: Cannot pass parameter 4 by reference in...

这指向bind_param的行

This points to the line of bind_param

推荐答案

您不能将常量传递给bind_param.首先将值放在变量中:

You can't pass a constant to bind_param. Put the value in a variable first:

$status = "0";
$stmt->bind_param("sss", $oid, $cid, $status);
$stmt->execute();

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