无法在MySQLi中通过引用传递参数 [英] Cannot pass parameter by reference in MySQLi

问题描述

我正在尝试将字符串传递到MySQLi准备好的语句中，但这给了我错误:

I am trying to pass a string into my MySQLi prepared statement but it gives me the error:

在MySQLi中无法通过引用传递参数

Cannot pass parameter by reference in MySQLi

以下是相关代码:

$kv = json_encode(array($key => $value));
$stmt->prepare("insert into rules (application_id, ruletype, rule_name, rule_info) values (?, ?, ?, ?);");
$stmt->bind_param('iiss', $application_id, 1, $config_name, $kv);

推荐答案

'1'不能通过引用传递，因为它不是变量，而是文字.您需要创建一个具有上述值的变量并将其绑定，因为 bind_param()函数希望变量通过引用传递.

'1' cannot be passed by reference because it's not a variable but a literal. You need to create a variable with mentioned value and bind it instead because bind_param() function expects variables passed by reference.

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