简单的问题 得看看register.php中是否正确接收了表单变量把register.php代码贴出来吧 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 <? $mysql_server_name = "localhost"; $mysql_username = "root"; $mysql_password = "1031hh"; $mysql_database = "lzhhdm"; //------- $sql = "INSERT INTO `user` ( `id` , `pwd` , `email` ) VALUES ('$id1', '$pwd1','$email1');"; $conn=mysql_connect( $mysql_server_name, $mysql_username, $mysql_password); mysql_select_db($mysql_database,$conn); $result = mysql_query($sql); $id = mysql_insert_id(); mysql_close($conn); ?> 没有正确接收但是数据库那段没有问题因为可以正常添加单单在register.php中echo "$id1";就没有显示谢谢 看看你的php.ini里面是不是register_globals=off, 如果是的话你应该用$_POST['id1']来获取。 <? $mysql_server_name = "localhost"; $mysql_username = "root"; $mysql_password = "1031hh"; $mysql_database = "lzhhdm"; $id = $_POST['id1']; $pwd = $_POST['pwd1']; $email = $_POST['email1']; $sql = "INSERT INTO user set id='$id',pwd='$pwd',email='$email'"; $conn=mysql_connect( $mysql_server_name, $mysql_username, $mysql_password); mysql_select_db($mysql_database,$conn); $result = mysql_query($sql); $id = mysql_insert_id(); mysql_close($conn); ?> 这样就ok了。/* 弹出变量 */if (!ini_get("register_globals")) { extract( $_POST ,EXTR_SKIP); extract( $_GET ,EXTR_SKIP);}$sql = "INSERT INTO `user` ( `id` , `pwd` , `email` ) VALUES ('$id1', '$pwd1','$email1'); 一个MYSQL 查询时间段的问题 POST提交表单插不进数据库 跪求提问php取值问题! linux下php变成应该用什么ide 在保存用户访问ip的时候 一般怎样获取,怎样保存 Mysql用什么字段好? 返回安全数据的用法 麻烦麻烦大家帮帮忙啊,关于imagecolortransparent设置透明色的问题,找了历史资料也没个结果. 谁有PHP产品发布和订购系统,相当于留言板一样的购物系统,要没有加密的,咱要.. 怎样用mysqldump导入数据库?? 用CURL去抓网页的时候有些内容抓取不到 ASP代码改写成PHP代码-挑战高手!!!!! php怎么取出access数据库中的值?
$mysql_server_name = "localhost";
$mysql_username = "root";
$mysql_password = "1031hh";
$mysql_database = "lzhhdm";
//-------
$sql = "INSERT INTO `user` ( `id` , `pwd` , `email` ) VALUES ('$id1', '$pwd1','$email1'
);
";
$conn=mysql_connect( $mysql_server_name, $mysql_username, $mysql_password);
mysql_select_db($mysql_database,$conn);
$result = mysql_query($sql);
$id = mysql_insert_id();
mysql_close($conn); ?>
因为可以正常添加单单在
register.php中
echo "$id1";
就没有显示
谢谢
如果是的话你应该用$_POST['id1']来获取。
$mysql_server_name = "localhost";
$mysql_username = "root";
$mysql_password = "1031hh";
$mysql_database = "lzhhdm";
$id = $_POST['id1'];
$pwd = $_POST['pwd1'];
$email = $_POST['email1']; $sql = "INSERT INTO user set id='$id',pwd='$pwd',email='$email'";
$conn=mysql_connect( $mysql_server_name, $mysql_username, $mysql_password);
mysql_select_db($mysql_database,$conn);
$result = mysql_query($sql);
$id = mysql_insert_id();
mysql_close($conn); ?>
/* 弹出变量 */
if (!ini_get("register_globals")) {
extract( $_POST ,EXTR_SKIP);
extract( $_GET ,EXTR_SKIP);
}$sql = "INSERT INTO `user` ( `id` , `pwd` , `email` ) VALUES ('$id1', '$pwd1','$email1'
);