日期:2014-05-17 浏览次数:20824 次
<?php
#adddate.php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
// some code
mysql_select_db("survey", $con);
#处理add1
if(isset($_GET['date']) && $_GET['date']==1){
$sql="insert INTO information (t1) VALUES('".trim($_POST['t1'])."')";
if (!mysql_query($sql,$con)){
die('Error: ' . mysql_error());
}
echo "<script>alert('ok! 1 record added') </script>";
mysql_close($con);
}
#处理add2
if(isset($_GET['date']) && $_GET['date']==2){……}
#处理add3
if(isset($_GET['date']) && $_GET['date']==3){……}
?>
#add.php
<form action="adddate.php?date=1" method="POST" id="form1" name="form1" onSubmit="return check()">
<input name="t1" type="text" size="50" maxlength="100" id="tel" />
</form>
#add2.php
<form action="adddate.php?date=2" method="POST" id="form2" name="form1" onSubmit="return check()">
<input name="t2" type="text" size="50" maxlength="100" id="tel" />
</form>
#add3.php
<form action="adddate.php?date=3" method="POST" id="form3" name="form1" onSubmit="return check()">
<input name="t3" type="text" size="50" maxlength="100" id="tel" />
</form>
------解决方案--------------------
if(isset($_POST[t1]))
$sql="insert INTO information (t1) VALUES('{$_POST[t1]}')";
if(isset($_POST[t1]))
$sql="insert INTO information (t2) VALUES('{$_POST[t2]}')";
if(isset($_POST[t1]))
$sql="insert INTO information (t3) VALUES('{$_POST[t3]}')";
不需要什么函数。