Php MySQLiprepared statements 错误报告

mysqli的每种方法都可能失败.您应该测试每个返回值.如果一个失败了，想想继续使用一个不在你期望状态的对象是否有意义.(可能不是处于"安全"状态，但我认为这不是问题.)

因 for each 连接/语句只存储最后一个操作的错误消息，所以如果您在出错后继续操作，您可能会丢失关于导致错误的what的信息.您可能希望使用该信息让脚本决定是重试(仅是临时问题)、更改某些内容还是完全退出(并报告错误).而且它使调试变得容易得多.

$stmt = $mysqli->prepare("INSERT INTO testtable VALUES (?,?,?)");
// prepare() can fail because of syntax errors, missing privileges, ....
if ( false===$stmt ) {
  // and since all the following operations need a valid/ready statement object
  // it doesn't make sense to go on
  // you might want to use a more sophisticated mechanism than die()
  // but's it's only an example
  die('prepare() failed: ' . htmlspecialchars($mysqli->error));
}

$rc = $stmt->bind_param('iii', $x, $y, $z);
// bind_param() can fail because the number of parameter doesn't match the placeholders in the statement
// or there's a type conflict(?), or ....
if ( false===$rc ) {
  // again execute() is useless if you can't bind the parameters. Bail out somehow.
  die('bind_param() failed: ' . htmlspecialchars($stmt->error));
}

$rc = $stmt->execute();
// execute() can fail for various reasons. And may it be as stupid as someone tripping over the network cable
// 2006 "server gone away" is always an option
if ( false===$rc ) {
  die('execute() failed: ' . htmlspecialchars($stmt->error));
}

$stmt->close();

六年后的几句话...

The mysqli extension is perfectly capable of reporting operations that result in an (mysqli) error code other than 0 via exceptions, see mysqli_driver::$report_mode.
die() is really, really crude and I wouldn't use it even for examples like this one anymore.
So please, only take away the fact that each and every (mysql) operation can fail for a number of reasons; even if the exact same thing went well a thousand times before....