关于php中利用sql语句创建表的问题
<!-- Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ --><?php $con=mysql_connect("localhost","root",""); if($con) { mysql_select_db("text",$con); mysql_query("set name gtf8"); $sql="create table user_date ( id int(5) not null auto_increment primary key, name char(10) not null default '', password char(12) not null default '', age int(3) not null default 0, sex char(10) not null default 'man', mail char(50) not null default '', qq char(10) not null default '', gedree char(10) not null default '', fav char(50) not null default '' )"; $do=mysql_query($sql,$con); if($do) { echo "create user data successful!!"; } else echo "create error!!!!!"; } else { echo "connect error!!!!"; } ?>
<!-- Code highlighting produced by Actipro CodeHighlighter (freeware) http://www.CodeHighlighter.com/ --> <?php echo "<style type=\"text/css\"> <!-- body {color:#000099;font-size:10pt; text-align:center} --> "; if($_POST) { $user=$_POST["user"]; $pass=$_POST["pass"]; $age=$_POST["age"]; $sex=$_POST["sex"]; $mail=$_POST["mail"]; $qq=$_POST["qq"]; $degree=$_POST["degree"]; $fav=$_POST["fav"]; $len=count($fav); $fav_z=""; for($i=0;$i0) { echo "have the same name!try another one."; } else { $sql="insert into user2(name,password,age,sex,mail,qq,degree,fav)value('$user','$pass','$age','$sex','$mail',$qq','$degree','$fav_z')"; $re=mysql_query($sql); if($re) echo "insert successful!"; else echo "insert error~!"; echo "<p>"; } } else { echo "nothing upload!<br>"; } echo "<br>click<a href="reg.html">there</a> return"; ?> </p>