[英]PHP to report status of MSSQL query table select into
我有一個按鈕:
<a href="#" onclick="loadPage('admin_manualsend.php?refresh=true'); return false;">
<i class="fa fa-refresh"></i> 1. Refresh Now
</a>
並在單擊時加載同一頁面,它是帶有refresh = true的頁面,因此現在在頁面加載時將觸發以下內容:
if(isset($_GET["refresh"]) && $_GET["refresh"]=="true"){
$res = db_query("begin try drop table [TmpTable] end try begin catch end catch");
$res = db_query("SELECT * INTO [TmpTable] FROM [dbo].[ViewLive]");
}
需要一些時間,我希望它報告表刪除是否成功,並且還報告它現在正在傳輸的記錄的計數,然后報告完成的時間,如下所示:
if(isset($_GET["refresh"]) && $_GET["refresh"]=="true"){
$res = db_query("begin try drop table [TmpTable] end try begin catch end catch");
Show modal, echo 'Drop done'; (or drop failed)
WHILE ( $res = db_query("SELECT * INTO [TmpTable] FROM [dbo].[ViewLive]");
echo 'Count of current record'
count of record +1
)
echo 'Update done';
(user can then close modal)
}
我該怎么做? 我使用:PHP,MSSQL,腳本src =“ // ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js”>
謝謝
更新:
function db_query($query) {
global $conn;
$res = sqlsrv_query($conn, $query);
return $res;
}
function db_fetch_array($res) {
$result = array();
while( $array = sqlsrv_fetch_array($res) ) {
$result[count($result)] = $array;
}
sqlsrv_free_stmt($res);
return $result;
}
function db_fetch_row($res) {
$array = sqlsrv_fetch_array($res);
sqlsrv_free_stmt($res);
return $array;
}
function db_fetch_one($res) {
$array = sqlsrv_fetch_array($res);
sqlsrv_free_stmt($res);
return $array[0];
我不確定您要做什么,但是如果您使用ajax,則可以在不離開頁面的情況下執行sql並獲得成功或錯誤的回調!
// put this in your document ready function for index.html $("#submit").on("click", function() { $.ajax({ type: "post", url: "process.php", data: "action=SomeActionHere", success: function(data) { alert("SUCCESS"); }, error: function(e) { alert("ERROR"); } }); };
<!--index.html --> <button id="submit">GOGO</button> <!--process.php--> <?php header("Content-type: application/json; charset=utf-8"); //connect to the server $action = $_POST['action']; if($action == "SomeActionHere"){ // do you sql stuff here // when done sql echo json_encode("SUCCESS"); // whatever data you want to send back } ?>
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.