PHP stmt准备失败但没有错误

2022-12-04 16:44:09

我正在尝试准备一个mysqli查询,但它无声地失败而没有给出任何错误.

 $db_hostname  = "test.com";
 $db_database   = "dbname";
 $db_username  = "db_user";
 $db_password   = "password";
 $db = new mysqli($db_hostname,$db_username,$db_password,$db_database);

 $q = "INSERT INTO Members (`wp_users_ID`,`MemberID`,`Status`,`MiddleName`,`Nickname`,`Prefix`,`Suffix`,`HomeAddress`,`City`,`State`,`Zip`,`ExtendedZip`,`BadAddress`,`SpouseFirstName`,`SpouseMiddleName`,`HomePhone`,`CellPhone`,`WorkPhone`,`WorkPhoneExt`,`OfficePhone`,`OfficePhoneExt`,`Pager`,`Fax`,`Company`,`CompanyType`,`OfficeAddress`,`OfficeAddress2`,`OfficeCity`,`OfficeState`,`OfficeZip`,`OTYPECO`,`OSTAG`,`UPCODE`,`Region`,`Department`,`Classification`,`Retired`,`Industry`,`Comments`,`Officer`,`OfficerType`,`OfficerTitle`,`OUNIT`,`ReceiveEMagazine`,`CD`,`SD`,`AD`,`isOrganization`,`DEL`,`Dues`,`DataSource`) VALUES ((?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?),(?));";
 $stmt = $db->prepare($q);
 if ( false === $stmt ) {
      echo "<pre>";
      print_r( $db );
      echo "</pre>";
      mysqli_report(MYSQLI_REPORT_ALL);
      echo mysqli_error();
      }

实际显示任何内容的唯一部分是print_r($db)：

 mysqli Object
 (
      [affected_rows] => -1
      [client_info] => 5.1.73
      [client_version] => 50173
      [connect_errno] => 0
      [connect_error] => 
      [errno] => 0
      [error] => 
      [error_list] => Array
      (
      )
      [field_count] => 1
      [host_info] => dbhost.com via TCP/IP
      [info] => 
      [insert_id] => 919910
      [server_info] => 5.1.73-log
      [server_version] => 50173
      [stat] => Uptime: 1924325  Threads: 8  Questions: 642600129  Slow queries: 28158  Opens: 24168750  Flush tables: 1  Open tables: 403  Queries per second avg: 333.935
      [sqlstate] => 00000
      [protocol_version] => 10
      [thread_id] => 9939810
      [warning_count] => 0
 )

有没有人看到任何会导致这种情况的东西没有任何错误,很难看出有什么问题…我尝试将生成的查询直接复制并粘贴到phpmyadmin中并且运行得很好(手动用测试值替换问号后).

谢谢！

UPDATE

看来自添加mysqli_report(MYSQLI_REPORT_ALL);在页面顶部,插入查询的查询现在失败,但仍然没有给出错误.这个执行失败：

 echo "1";
 $idDataSources = "";
 echo "2";
 $q = "SELECT idDataSources FROM DataSources WHERE `description`=(?);";
 echo "3";
 $stmt = $db->prepare($q);
 echo "4";
 $stmt->bind_param('s',$description);
 echo "5";
 $description = "File - 01/10/2015";
 echo "6";
 $stmt->execute() or die( mysqli_stmt_error( $stmt ) );
 echo "7";
 $stmt->bind_result($idDataSources);
 echo "8";
 $stmt->fetch();
 echo "9";
 unset($params);

OUTPUT：

它到达$stmt-> execute()并失败.我再次尝试输出错误,但没有任何显示.这真是令人费解.我想知道我是否应该恢复到旧的mysql(非面向对象)方法……它是不安全的,但至少它工作一致并在出现问题时显示错误.

更新2

好吧,我只是使用mysql(非面向对象)而不是mysqli重写了整个脚本…就像一个梦想.我希望我可以切换到更新的标准,但是随机故障和这样的错误报告很差,这肯定很难.我将保留“更好”的版本,直到我能弄清楚它失败的原因.

更新3

我注意到mysqli有一个有趣的行为.在同一代码的其他地方,我有两个查询依次运行STMT.这偶尔会失败.失败并不一致,因为我可以提交相同的数据50次,其中,它可能会失败20次……相同的数据,相同的功能.

为了确定脚本错误输出的确切位置,我在两个查询中的每个语句之间放入了echo命令,只是吐出一个数字以查看计数停止的位置 – 结果是使用不相关的命令,它减慢了STMT的速度足以让它始终如一地运作.这让我想知道STMT连接是否未正确关闭.

$q = "";
$stmt = $this->db->prepare( "SELECT ID FROM Members WHERE MemberID='5' LIMIT 1;" );
$stmt->execute();
$stmt->store_result();
if ( $stmt->num_rows > 0 ) {
    $q = "UPDATE Members SET Name='Test' WHERE MemberID=(?) LIMIT 1;";
    }
$stmt->close();

// here if we continue, it has a chance of erroring out. However, 
// if we run just the following command instead, everything works perfect.
//  
// mysql_query( "UPDATE Members SET Name='Test' WHERE MemberID='5' LIMIT 1;" );

if ( $q != "" ) {
    $stmt = $this->db->prepare($q);
    $stmt->bind_param('i',$params['ID']);
    $params['ID'] = 5;
    $stmt->execute();
    $stmt->close();
    unset($params);
    }

任何人都可以解释这种行为吗？它似乎不应该是冲突的,因为我在开始一个新查询之前使用close()命令,并且它在某些时候工作……似乎很奇怪.

解决方法:

这是来自php.net的一个稍微改编的示例脚本,具有错误处理：

<?php
$mysqli = new mysqli("example.com", "user", "password", "database");
if ($mysqli->connect_errno) {
   echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}

/* Prepared statement, stage 1: prepare */
if (!($stmt = $mysqli->prepare("SELECT idDataSources FROM DataSources WHERE `description`=(?)"))) {
     echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
}

/* Prepared statement, stage 2: bind and execute */
$description = "File - 01/10/2015";
if (!$stmt->bind_param('s', $description)) {
    echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
}

if (!$stmt->execute()) {
    echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}

/* explicit close recommended */
$stmt->close();
?>

请注意,$mysqli或$stmt可以保存错误描述.

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