这里是我的代码:如何在PHP中解决'无法通过引用传递参数'错误?
$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'));
我已经试过这也:
$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:
Fatal error:** Cannot pass parameter 3 by reference in **...file path...line #
请注意,CaptureTime和UploadeTime有数据类型日期。而忽略了我传递第三和第四参数的值相同的事实。
代码有什么问题?
[PHP mysqli包装:通过引用传递__call()和call_user_func_array()]的可能重复(http://stackoverflow.com/questions/2566289/php-mysqli-wrapper-passing-by-reference-with- call-and-call-user-func-array) – ajreal
参考为:此问题有类似问题 http://stackoverflow.com/questions/13105373/php-error-cannot-pass-parameter-2-通过引用 – Yoshi
[PHP错误:“无法通过引用传递参数2”](https://stackoverflow.com/questions/13105373/php-error-cannot-pass-parameter-2-by-reference) –