PHP 中 DATETIME 字段 MS-SQL 的问题

Question

我在 MSSQL 中有以下查询：

SELECT DataFeed.AccountTranID, DataFeed.Datetime as MyDT
FROM admin_all.DataFeed
where cast(DataFeed.Datetime as Date) = cast(getdate() as Date)

这给出了以下输出：

AccountTranID   MyDT
124552  2019-07-31 00:00:04.660
124553  2019-07-31 00:00:07.933
124554  2019-07-31 00:00:25.623
124555  2019-07-31 00:00:29.013
124556  2019-07-31 00:00:29.206
124557  2019-07-31 00:00:44.893
124558  2019-07-31 00:00:56.796
124559  2019-07-31 00:01:11.353
124560  2019-07-31 00:01:12.260
124561  2019-07-31 00:01:19.413
124562  2019-07-31 00:01:19.510
124563  2019-07-31 00:01:28.596
124564  2019-07-31 00:01:30.710
124565  2019-07-31 00:01:46.976
124566  2019-07-31 00:01:49.823
124567  2019-07-31 00:01:57.340

尝试使用此 PHP 代码获取 MyDT 字段时，我无法获得输出（空白）。使用 AccountTranID，没问题。

这是 PHP 代码：

$tsql = "SELECT DataFeed.AccountTranID, DataFeed.Datetime as MyDT 
FROM admin_all.DataFeed where cast(DataFeed.Datetime as Date) = cast(getdate() as Date) ";
// Executes the query
$stmt = sqlsrv_query($conn, $tsql);

// Error handling
if ($stmt === false) {
    die(formatErrors(sqlsrv_errors()));
}
?>

<h1> Results : </h1>

<?php

while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {


    echo  $row['AccountTranID'].'-'. $row['MyDT'].'<br>';

}

sqlsrv_free_stmt($stmt);
sqlsrv_close($conn);

Answer 1

正如我们发现的，查询的第二个选定元素是 DateTime 类型，因此您必须将其解析为字符串。

您可以使用以下代码：

while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {
    echo  $row['AccountTranID'].'-'. $row['MyDT']->format('Y-m-d H:i:s').'<br>';
}