我有一张下表
- ITEM_NUMBER QTY_FROM QTY_TO
- ABC 10
- ABC 20
- ABC 30
- DEF 100
- DEF 250
- DEF 400
我需要编写一个plsql程序，该程序将基于QTY_to下一行的qty_from列填充相同的项目编号。
例如，如果项目编号为ABC且qty_from在第一行为10，在下一行为20 .... id期望该过程查看第二行...并填充qty_from在第一个为19（next_row_qty_from - 1）。
答案 0 :(得分：1)
select ITEM_NUMBER, QTY_FROM,
lead(QTY_FROM, 1) over (partition by ITEM_NUMBER, order by QTY_FROM) - 1 as QTY_TO
from table;
作为lead功能的第三个参数，您可以添加默认值。当给定的ITEM_NUMBER不存在下一行时将使用此选项，因此如果您- ABC 30行lead(QTY_FROM, 1, QTY_FROM) 29 QTY_TO，则会<?php
require 'vendor/autoload.php';
use net\authorize\api\contract\v1 as AnetAPI;
use net\authorize\api\controller as AnetController;
define("AUTHORIZENET_LOG_FILE", "phplog");
function authorizeCreditCard($amount){
// Common setup for API credentials
$merchantAuthentication = new AnetAPI\MerchantAuthenticationType();
$merchantAuthentication->setName(\SampleCode\Constants::MERCHANT_LOGIN_ID);
$merchantAuthentication->setTransactionKey(\SampleCode\Constants::MERCHANT_TRANSACTION_KEY);
$refId = 'ref' . time();
// Create the payment data for a credit card
$creditCard = new AnetAPI\CreditCardType();
$creditCard->setCardNumber("4111111111111111");
$creditCard->setExpirationDate("1226");
$creditCard->setCardCode("123");
$paymentOne = new AnetAPI\PaymentType();
$paymentOne->setCreditCard($creditCard);
//create a transaction
$transactionRequestType = new AnetAPI\TransactionRequestType();
$transactionRequestType->setTransactionType( "authOnlyTransaction");
$transactionRequestType->setAmount($amount);
$transactionRequestType->setPayment($paymentOne);
$request = new AnetAPI\CreateTransactionRequest();
$request->setMerchantAuthentication($merchantAuthentication);
$request->setRefId( $refId);
$request->setTransactionRequest( $transactionRequestType);
$controller = new AnetController\CreateTransactionController($request);
$response = $controller->executeWithApiResponse( \net\authorize\api\constants\ANetEnvironment::SANDBOX);
if ($response != null)
{
if($response->getMessages()->getResultCode() == \SampleCode\Constants::RESPONSE_OK)
{
$tresponse = $response->getTransactionResponse();
if ($tresponse != null && $tresponse->getMessages() != null)
{
echo " Transaction Response code : " . $tresponse->getResponseCode() . "\n";
echo " Successfully created a transaction with Auth code : " . $tresponse->getAuthCode() . "\n";
echo " TRANS ID : " . $tresponse->getTransId() . "\n";
echo " Code : " . $tresponse->getMessages()[0]->getCode() . "\n";
echo " Description : " . $tresponse->getMessages()[0]->getDescription() . "\n";
}
else
{
echo "Transaction Failed \n";
if($tresponse->getErrors() != null)
{
echo " Error code : " . $tresponse->getErrors()[0]->getErrorCode() . "\n";
echo " Error message : " . $tresponse->getErrors()[0]->getErrorText() . "\n";
}
}
}
else
{
echo "Transaction Failed \n";
$tresponse = $response->getTransactionResponse();
if($tresponse != null && $tresponse->getErrors() != null)
{
echo " Error code : " . $tresponse->getErrors()[0]->getErrorCode() . "\n";
echo " Error message : " . $tresponse->getErrors()[0]->getErrorText() . "\n";
}
else
{
echo " Error code : " . $response->getMessages()->getMessage()[0]->getCode() . "\n";
echo " Error message : " . $response->getMessages()->getMessage()[0]->getText() . "\n";
}
}
}
else
{
echo "No response returned \n";
}
return $response;
}
if(!defined('DONT_RUN_SAMPLES'))
authorizeCreditCard( 23.32);
?>
在你的问题中定义边缘情况。
答案 1 :(得分：0)
无需手续，只需简单更新：
update t
set qty_to =
(select min(qty_from)-1 from t t2
where item_number = t.item_number and qty_from > t.qty_from)
结果：
SQL> select * from t order by item_number, qty_from;
ITEM_NUMBER QTY_FROM QTY_TO
----------- -------- -------
ABC 10 19
ABC 20 29
ABC 30
DEF 100 249
DEF 250 399
DEF 400
6 rows selected