mysql_query不返回任何東西

-2

我使用下面的代碼，它不顯示任何東西，只是一個新行。什麼可能是錯的？謝謝。mysql_query不返回任何東西

$result = mysql_query($sql); 
echo($result."<br>");

整個代碼，

<?php 
include 'dbc.php'; 
$LeadFirstName = $_POST['LeadFirstName']; 
$LeadFirstName = $_POST['LeadFirstName']; 
$LeadEmail =$_POST['LeadEmail']; 
$LeadEmail2 =$_POST['LeadEmail2']; 
$LeadPhone =$_POST['LeadPhone']; 
$LeadCity =$_POST['LeadCity']; 
$LeadAddress =$_POST['LeadAddress']; 
$LeadPostcode =$_POST['LeadPostcode']; 
$LeadUserId =$_POST['LeadUserId']; 
$LeadLeadStatusId =$_POST['LeadLeadStatusId']; 
$LeadMonth =$_POST['LeadMonth']; 
$LeadAreas =$_POST['LeadAreas']; 
$LeadMinPrice =$_POST['LeadMinPrice']; 
$LeadMaxPrice =$_POST['LeadMaxPrice']; 
$LeadMinBedrooms =$_POST['LeadMinBedrooms']; 
$LeadMaxBedrooms =$_POST['LeadMaxBedrooms']; 
$LeadMinBathrooms =$_POST['LeadMinBathrooms']; 
$LeadMaxBathrooms =$_POST['LeadMaxBathrooms']; 
$LeadMinYear =$_POST['LeadMinYear']; 
$LeadNextFollowup_mm =$_POST['LeadNextFollowup_mm']; 
$LeadNextFollowup_dd =$_POST['LeadNextFollowup_dd']; 
$LeadNextFollowup =$_POST['LeadNextFollowup']; 
$sql="INSERT INTO 'realtorl_leads'.`data (`LeadFirstName`, `LeadLastName`, `LeadEmail`, `LeadEmail2`, `LeadPhone`, `LeadCity`, `LeadAddress`, `LeadPostcode`, `LeadUserId`, `LeadLeadStatusId`, `LeadMonth`, `LeadAreas`, `LeadMinPrice`, `LeadMaxPrice`, `LeadMinBedrooms`, `LeadMaxBedrooms`, `LeadMinBathrooms`, `LeadMaxBathrooms`, `LeadMinYear`, `LeadNextFollowup_mm`, `LeadNextFollowup_dd`, `LeadNextFollowup`) VALUES ($LeadFirstName, $LeadLastName, $LeadEmail, $LeadEmail2, $LeadPhone, $LeadCity, $LeadAddress, $LeadPostcode, $LeadUserId, $LeadLeadStatusId, $LeadMonth, $LeadAreas, $LeadMinPrice, $LeadMaxPrice, $LeadMinBedrooms, $LeadMaxBedrooms, $LeadNextFollowup_mm, $LeadNextFollowup_dd, $LeadNextFollowup)"; 
$result = mysql_query($sql); 
echo($result."<br>"); 
if (mysql_affected_rows($result)){ 
    echo("worked"); 
}else { 
    echo("does not work"); 
} 
?>

來源

2013-09-01 Fahad Uddin

嘗試用'mysql_affected_rows（）'而不是'mysql_affected_rows（$ result）'。 – sarwar026

@ sarwar026：試過了。不起作用。 –

請在下面看到我的回答 – sarwar026

你已經錯過了在查詢中`跡象。

$sql="INSERT INTO 'realtorl_leads'.`data (`LeadFirstName`, 
             ^

正確的查詢將

$sql="INSERT INTO 'realtorl_leads'.`data` (`LeadFirstName`,

現在應該工作。

來源

2013-09-01 09:15:18 sarwar026

如果'realtorl_leads'是一個數據庫名稱，它也應該包含反引號，而不是撇號。 –

mysql_query不返回任何東西

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