是這樣的，我設(shè)計(jì)的數(shù)據(jù)庫userID是自增的，所以不用用戶填寫，我在一個頁面中插入數(shù)據(jù)，然后判斷，如果插入成功，那么我用sql 語句再查詢出我插進(jìn)去的數(shù)據(jù)的，自增的拿個屬性userID，可是查不出來啊。我試著吧插入完成后把數(shù)據(jù)庫重新斷開重連，還是不行。求解。代碼是這樣的。<?php$name=$_POST[name];$password=$_POST[password];$gender=$_POST[gender];$nickname=$_POST[nickname];$age=$_POST[age];//緊接著 開始鏈接數(shù)據(jù)庫$conn=mysql_connect("localhost","root","123456") or die("數(shù)據(jù)庫服務(wù)器鏈接失敗".mysql_error());mysql_select_db("travelspace",$conn) or die("數(shù)據(jù)庫訪問錯誤".mysql_error());mysql_query("set names gb2312");$sql=mysql_query("insert into usertable(username,password,gender,nickname,age) values('$name','$password','$gender','$nickname','$age')");if($sql){//插入成功就輸出信息mysql_close($conn);//暫時(shí)關(guān)閉數(shù)據(jù)庫，重新打開。$conn=mysql_connect("localhost","root","123456") or die("數(shù)據(jù)庫服務(wù)器鏈接失敗".mysql_error());mysql_select_db("travelspace",$conn) or die("數(shù)據(jù)庫訪問錯誤".mysql_error());mysql_query("set names gb2312");$sql2=mysql_query("select * from usertable where username =='$name'");if(!$sql2){echo "沒有查詢到數(shù)據(jù)！";}@ $row =mysql_fetch_object($sql2);echo "恭喜".$name."注冊成功請記住登陸賬號";echo $row->userID;echo "和密碼！";echo "<br>";echo "<a href='login.php'>點(diǎn)擊登陸</a>";}else{echo "對不起，該賬號已經(jīng)注冊！請重新注冊";echo "<a href ='register.php'>點(diǎn)擊再次注冊</a>";}mysql_close($conn);?>