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當我嘗試在我的數據庫中插入一個值時,它返回我沒有錯誤,但它沒有插入任何東西,在sql server上它完美地工作。用sqlsrv插入查詢
<?php
$serverName = "12.15.88.81";
$connectionInfo = array("Database"=>"Dcs", "UID"=>"sa", "PWD"=>"123456");
$conn = sqlsrv_connect($serverName, $connectionInfo);
if($conn === false) {
die(print_r(sqlsrv_errors(), true));
}
if (sqlsrv_begin_transaction($conn) === false) {
die(print_r(sqlsrv_errors(), true));
}
$sql = "INSERT INTO [dbo].[Dcs_name] (ID,NAME) VALUES (?,?)";
$params = array('11','Ani');
$stmt = sqlsrv_query($conn, $sql, $params);
$stmt = sqlsrv_query($conn, $sql, $params);
if($stmt === false) {
if(($errors = sqlsrv_errors()) != null) {
foreach($errors as $error) {
echo "SQLSTATE: ".$error[ 'SQLSTATE']."<br />";
echo "code: ".$error[ 'code']."<br />";
echo "message: ".$error[ 'message']."<br />";
}
}
}
?>
我沒有使用sql-server函數,但它看起來像你正在開始一個事務,你需要提交這個更改來應用。 http://php.net/manual/en/function.sqlsrv-commit.php另外,不要執行查詢兩次,我會有'id'自動增量。 – chris85
檢查此:最佳方式找出 - https://stackoverflow.com/questions/42648/best-way-to-get-identity-of-inserted-row?rq=1 – vbRocks