我有一个WebForm,从中我想记录应当提交到下相同的数据库名称的两个表。插入从单一形式代码2个表是给出问题----请大家帮忙
我的代码是
<?php
$con = mysql_connect("localhost","************","***********");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
$db=mysql_select_db("qserves1_uksurvey", $con);
$sql="INSERT INTO forms (date, Receivingsky, Title, Firstname, Lastname, House, Street, Town, County, Postcode, Number, WarrantyCoverForSky, Tvmake, Warrantycover, Payingmonthly, Agentnotes, Agentname)
VALUES
(NOW(),'$_POST[Receivingsky]','$_POST[Title]','$_POST[Firstname]','$_POST[Lastname]','$_POST[House]','$_POST[Street]','$_POST[Town]','$_POST[County]','$_POST[Postcode]','$_POST[Number]','$_POST[WarrantyCoverForSky]','$_POST[Tvmake]','$_POST[Warrantycover]','$_POST[Payingmonthly]','$_POST[Agentnotes]','$_POST[Agentname]')";
$sql_result = mysql_query($sql, $con) or die (mysql_error());
$con2 = mysql_connect("localhost","*******8","*********8");
if (!$con2)
{
die('Could not connect: ' . mysql_error());
}
$db2=mysql_select_db("qserves1_uksurvey", $con2);
$sql2="INSERT INTO dupphones (date, Number)
Values
(NOW(),'$_POST[Number]')";
$sql_result = mysql_query($sql2, $con2) or die (mysql_error());
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo '<html>
<head>
<title>Lead Submitted successfully!!!</title>
</head>
<body>
<center>
<strong><a href="http://q2serves.com/uksurvey.html">Lead Submitted ---- Click Here To Enter New Lead</a></strong>
</center>
</body>
</html>!';
mysql_close($con)
?>
这被提交3根引线1是在表dupphones和2根引线在表形式。
我想这只有每个表提交1分领先。
请帮
感谢
为什么要创建到同一数据库的两个连接?您还可以执行mysql_query多次,有时使用$ sql2,然后再使用$ sql。 – halfdan 2011-04-09 10:01:48
只是几句建议。在提交之前清理您的$ _POST数据。并考虑使用主键。另外,为什么你建立一个新的连接来运行第二个查询? – JohnP 2011-04-09 10:04:50