日期:2014-05-17 浏览次数:20690 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。