无法在 MySQLi 中通过引用传递参数

2021-12-25 00:00:00 binding php mysql mysqli prepared-statement

我试图将一个字符串传递到我的 MySQLi 准备好的语句中,但它给了我错误:

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

在 MySQLi 中不能通过引用传递参数

Cannot pass parameter by reference in MySQLi

相关代码如下:

$kv = json_encode(array($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.您需要使用提到的值创建一个变量并绑定它,因为 bind_param() 函数需要通过引用传递的变量.

'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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