Question

我正在尝试在MySQL中使用AES_DECRYPT来解密成功加密的SSN。在输出中，我得到单词“Array”而不是该字段的实际数据。我的PHP和MySQL知识有点生疏，所以我确定这是我忽略的傻事。任何帮助将不胜感激。

输出：
verify_name other_names ssn dob
：test：test：Array：test

CODE：
$key="88b871WZ3SntWK67rN3l2J1SvMqsOjyk"; $SQLstring = "SELECT * FROM applications"; $QueryResult = @mysql_query($SQLstring, $conn) or die("Query Problem - " . mysql_error($conn) . " - Error Number - " . mysql_errno($conn)); echo "verify_name other_names ssn dob"; $num_result = mysql_num_rows($QueryResult); for ($i = 0; $i < $num_result; $i++) { $row = mysql_fetch_array($QueryResult); $SQLstring2 = "SELECT AES_DECRYPT(ssn,'$key') FROM applications WHERE name='" . $row["name"] . "'"; $QueryResult2 = @mysql_query($SQLstring2, $conn) or die("Query Problem - " . mysql_error($conn) . " - Error Number - " . mysql_errno($conn)); $num_result2 = mysql_num_rows($QueryResult2); for ($j = 0; $j < $num_result; $j++){ $ssndecrypt = mysql_fetch_array($QueryResult2); echo $ssndecrypt[0]; } echo $row["verify_name"]; echo $row["other_names"]; echo $ssndecrypt; echo $row["dob"];

Answer 1

这是因为您将结果作为数组提取。

$ssndecrypt = mysql_fetch_array($QueryResult2);
...
echo $ssndecrypt;

它回应Array，因为您永远不会重新分配$ssndecrypt变量。

然而，问题的核心似乎是您不必要地使查询复杂化。你可以做的就是没有理由两次查询表格：

SELECT verify_name,
       other_names,
       dob,
       AES_DECRYPT(ssn,'$key') AS ssn
  FROM applications

这简化了代码：

$stmt = "SELECT verify_name, other_names, dob, AES_DECRYPT(ssn,'$key') AS ssn FROM applications";
$result = @mysql_query($stmt, $conn) or die("Query Problem - " . mysql_error($conn) . " - Error Number - " . mysql_errno($conn));
echo "verify_name other_names ssn dob";
while ($row = mysql_fetch_assoc($result))
{
    echo $row["verify_name"];
    echo $row["other_names"];
    echo $row["ssn"];
    echo $row["dob"];
}

但理想情况下，您应该使用PDO而不是mysql_*函数：

try {
    $dbh = new PDO('mysql:host=localhost;dbname=mydb', $user, $pass);
    foreach($dbh->query("SELECT verify_name, other_names, dob, AES_DECRYPT(ssn,'$key') AS ssn FROM applications") as $row) {
        echo $row["verify_name"];
        echo $row["other_names"];
        echo $row["ssn"];
        echo $row["dob"];
    }
    $dbh = null;
} catch (PDOException $e) { die("ERROR: " . $e->getMessage()); }

AES_DECRYPT返回“Array”而不是解密数据

1 个答案: