mysql_query不返回任何东西

-2

我使用下面的代码，它不显示任何东西，只是一个新行。什么可能是错的？谢谢。mysql_query不返回任何东西

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

整个代码，

<?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"); 
} 
?>

来源

2013-09-01 Fahad Uddin

尝试用'mysql_affected_rows（）'而不是'mysql_affected_rows（$ result）'。 – sarwar026

@ sarwar026：试过了。不起作用。 –

请在下面看到我的回答 – sarwar026

你已经错过了在查询中`迹象。

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

正确的查询将

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

现在应该工作。

来源

2013-09-01 09:15:18 sarwar026

如果'realtorl_leads'是一个数据库名称，它也应该包含反引号，而不是撇号。 –

mysql_query不返回任何东西

回答

相关问题