2017-06-11 40 views
-1

這是我的php代碼。你能幫我麼。我的數據庫中沒有數據。我無法將數據插入到我的數據庫中,但我沒有任何錯誤

<?php 

    require_once("dbconnect.php"); 

    session_start(); 


    for ($x = 0; x < $_POST['reimcounter']; $x++) { 

    $date = $_POST['date']; 
    $tin = $_POST['tin']; 
    $address = $_POST['address']; 
    $particulars = $_POST['particulars']; 
    $referencenumber = $_POST['refno']; 
    $total = floatval(preg_replace('/[^\d\.]/', '', $_POST['total' . $x])); 
    $nonvat = floatval(preg_replace('/[^\d\.]/', '', $_POST['nonvat' . 
    $x])); 

我的插入代碼有什麼問題嗎?

$sql = "INSERT INTO rtco_cms.dbo.Reimbursement VALUES (?, ?, ?, ?, ?, ?, 
    ?, ?, ?)"; 

    $params = array($date, $tin, $SESSION['empid'], $address, 
    $SESSION['clientid'], $particulars, $referencenumber, $nonvat, $total); 

    $stmt = sqlsrv_query($conn, $sql, $params); 
    } 

    header("location: ../reimbursement.php"); 


    ?> 
+1

你可能** **被得到一個錯誤,你只是不檢查它。如果查詢失敗,'sqlserv_query()'將返回'false'。像[文檔]的示例#1(http://php.net/sqlsrv_query)那樣檢查它們。 – rickdenhaan

+0

插入聲明是錯誤的! – CoderSam

+0

沒有錯誤顯示。問題是它沒有插入到我的數據庫中。我的代碼錯了嗎? –

回答

0

更改此

$params = array($date, $tin, $SESSION['empid'], $address, 
$SESSION['clientid'], $particulars, $referencenumber, $nonvat, $total); 

$params = array($date, $tin, $_SESSION['empid'], $address, 
$_SESSION['clientid'], $particulars, $referencenumber, $nonvat, $total); 

Session變量調用是錯誤的。希望它可以幫助

更新例如你的插入查詢看到這個查詢:

tsql= "INSERT INTO dbo.vF_events (
     username, 
     Rft, 
     Ging, 
     description, 
     date, 
     trdate) 
     VALUES 
     (?, ?, ?, ?, ?, ?)"; 
     $var = array($username, $sort, $ag,  $description, $date, $trdate); 
     if (!sqlsrv_query($conn, $tsql, $var)) 
      { 
     die('Error: ' . sqlsrv_errors()); 
      } 
     echo "record added"; 
+0

我試過了,但它仍然無效:( –

+0

$ params = array($ date,$ tin,$ _POST ['name-'。$ x],$ address,$ _POST ['client-'。$ x] ,$ particulars,$ referencenumber,$ nonvat,$ total); –

+0

我使用$ _POST編輯它,但它仍然不起作用 –

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