php - mysqli 插入查询的结果不稳定

Question

我正在使用完全相同的参数运行相同的 mysqli 插入查询。大约有 1/3 的时间是成功的，我不知道哪里出了问题。

<?php
    //...
    $decrypted_session_key = 'unavailable'; // initialize
    $res = openssl_get_privatekey($priv_key, $passphrase);
    $result = openssl_private_decrypt($encrypted_session_key, $decrypted_session_key, $res, OPENSSL_PKCS1_OAEP_PADDING);

    if ($decrypted_session_key == 'unavailable') {
        mysqli_close($link);
        echo json_encode(array('result' => 'failed', 'message' => 'failed to decrypt the session key'));
        die();
    }

    if (!$decrypted_session_key) {
        echo json_encode(array('result'=>'failed', 'message'=>'decrypted session key has failed'));
        die();
    }

    $updated_at = date("Y-m-d H:i:s");

    // save this record to the database
    $result = mysqli_query($link, "INSERT INTO Session_Keys (session_id, session_key, iv, updated_at)
                                   VALUES ($session_id, '$decrypted_session_key', '$iv', '$updated_at')");

    if (!$result) { 
        $param_check = $session_id . " " . base64_encode($iv) . " " . base64_encode($decrypted_session_key) . " " . $updated_at;
        echo json_encode(array('result'=>'failed', 'message'=>$param_check));
        die();
    }

    // ...
}

每当失败时，我都会返回最后一个 echo 语句。我怀疑 php 解密例程失败了，但事实并非如此。所有的参数都是完美的，包括解密值。是我的插入语句错误的问题吗？我尝试了引用字段的各种组合，但没有任何一致的结果。

表结构是这样的:

'session_id' int(11)
'session_key' tinyblob
'iv' 小 Blob
'updated_at' 日期时间

我不明白为什么我的结果如此不一致。如果失败了，为什么不每次都失败呢？如果有效，为什么不是每次都有效？很困惑。任何帮助表示赞赏。谢谢!

Answer 1

在评论中向 OP 建议使用 or die(mysqli_error($link)) 到 mysqli_query() 让他们找到查询失败的原因.

他们的评论:

"I finally got at the error message here it is: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '<´§y?'T]î‘Á gô, EÈöJÀÊ?Xû¶T¢ÑÖ?, 2015-04-05 19:26:00)' at line 2 Must be something to do with the bytes that I am trying to store."

和

"Now I understand the problem. Your suggestion about looking at mysqli_error is what led me to the solution (after lots of difficulty on my end to gain access to it!). The real error (after I correctly quoted my fields) is Duplicate entry '2147483647' for key 'PRIMARY'. My ios app is randomly assigning an unsigned integer value (0 - 4294967294) but an int(11) field maxex out at 2147483647. So, half of my values were received as duplicates, thereby causing an error. Problem solved. Thanks for your help!"

---

来自:

https://stackoverflow.com/a/17783287/

2147483647是mysql最大的int值。只需将类型从 int 更改为 bigint。

• 解释了重复的错误。