如何解决 PHP 中的“无法通过引用传递参数"错误?

2021-12-25 00:00:00 php mysqli

这是我的代码:

$stmt = $conn->mysqli->prepare('INSERT INTO photos (CaseNo, ImageName, CaptureTime, UploadTime) VALUES (?,?,?,?)');
$stmt->bind_param('isss', $caseno, $index.'.'.$extension, date('Y-m-d H:i:s'), date('Y-m-d H:i:s'));

我也试过这个:

$stmt = $conn->mysqli->prepare('INSERT INTO photos (CaseNo, ImageName, CaptureTime, UploadTime) VALUES (?,?,?,?)');
$captureTime = date('Y-m-d H:i:s');
$uploadTime = date('Y-m-d H:i:s');
$stmt->bind_param('isss', $caseno, $index.'.'.$extension, $captureTime, $uploadTime);

我收到错误:

致命错误:** 无法通过 **...文件路径...行#中的引用传递参数 3

Fatal error:** Cannot pass parameter 3 by reference in **...file path...line #

请注意，CaptureTime 和 UploadeTime 具有数据类型 date.并忽略我传递的第 3 个和第 4 个参数的值相同的事实.

Please note that CaptureTime and UploadeTime have datatype date. And ignore the fact that I am passing the value of 3rd and 4th parameter same.

代码有什么问题?

推荐答案

更改 

$stmt->bind_param('isss', $caseno, $index.'.'.$extension, $captureTime, $uploadTime);

到

$isss = 'isss';
$indexExtention = $index.'.'.$extension
$stmt->bind_param($isss, $caseno, $indexExtention , $captureTime, $uploadTime);

我相信您必须传递变量而不是字符串.

I believe you have to pass variables rather than a string.

或者，如果您使用 PDO，您可以使用 bindvalue() 而不是 bindparam().

Or you could use bindvalue() instead of bindparam() if you're using PDO.

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