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php开发工程师程序员面试题


  1. 1.获取字母组合
    题目要求:
    做一个三位字母的组合功能,例如:”abd,ade,acc,aef”等等:
    目标
    1.字母所有的组合需要打印出来
    2.字母组合要排除,abc(三个字母连续),cba(三个字母倒叙),aaa(三个字母相同)的情况
    3.最后结果需要计算出abc情况的数量,cba情况的数量,aaa情况的数量,以及打印出来正确字母组合的数量
    ———————————————————————-
    代码:
    $b=array(a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z);
    for($i=0;$i<=25;$i++){
    for($j=0;$j<=25;$j++){
    for($k=0;$k<=25;$k++){
    if($i==$j-1 && $k==$j+1){//abc情况的数量
    $m++;
    }elseif($i==$j+1 && $k==$j-1){//cba情况的数量
    $n++;
    }elseif($i==$j && $k==$i){//aaa情况的数量
    $o++;
    }else{
    $p++;
    echo $b[$i].$b[$j].$b[$k].”
    ”;
    }
    }
    }
    }
    echo $p.”
    ”;
    echo $m.”
    ”;
    echo $n.”
    ”;
    echo $o.”
    ”;
    ?>
    代码2:
    $r=range(a,z);
    foreach ($r as $a => $x)
    {foreach ($r as $b => $y)
    {foreach ($r as $c => $z)
    {
    if($a==$b+1&&$b==$c+1)$abc++;
    if($c==$b+1&&$b==$a+1)$cba++;
    if($a==$b&&$b==$c)$aaa++;
    else $s[$a.$b.$c]=$x.$y.$z;
    }
    }
    }
    $r=$s;
    echo ‘abc:’.$abc.’
    ’;
    echo ‘cba:’.$cba.’
    ’;
    echo ‘aaa:’.$aaa.’
    ’;
    foreach ($r as $t){
    echo $t.’
    ’;
    }
    ?>


    2.通过不同的条件组装 一个sql语句
    题目要求:
    目前需要通过不同会员信息的条件,对会员进行高级查询,这个时候就需要对sql语句的组装工作。需要查询的条件有,会员编号,会员昵称,会员帐号,会员注册时间,会员性别,条件通过GET方式传递给程序,通过任意选择这些条件进行查询,得到相应的结果。
    参考:
    会员表为user_bas 及其字段结构;
    查询条件和数据库字段对应关系:
    会员编号:id ;结构:smallint(6),查询为精确查询(比如:查询编号为1 2的会员)
    会员昵称:nickname;结构:varchar(20) (比如:查询昵称里有”魅力”的昵称)
    会员帐号:uname;结构:varchar(16) 查询为模糊查询
    会员注册时间:regtime;结构:int(10) 查询为时间段查询
    会员性别:sex;结构:tingint(1) 查询为 查询为精确查询
    目标:
    写一段程序,实现这样的查询功能,”查询条件”需要的时候会增加和减少,考生需要编写一套灵活的程序
    注意:
    1。灵活性,
    用户可以选择1个条件,也可以一个都不选择,也可以选择多个。”
    2。考虑到系统的性能
    有的面试者用:
    “select * from user_bas where id=’$id’ and nickname=’$nickname’ and sex=’$sex’…”
    是不可以的,如果没有$id的条件数据库也还要去便历,在超过10万的数据库就非常慢了。
    3。不用考虑接受数据问题,和传递给数据库如何解释的问题。
    如果你怎么接受不知道如何书写,你可以更简单的思考,现在有几个变量。$id,$nickname,$uname,$sex,$regtime 可以用这些变量,但有的为空;对于数据库如何去查询,还有就是,数据库如何解释不用关心,如query相关函数语句你不用写。
    4.不允许使用”where 1″或”where 1=1″类型的语法

    代码
    _________________________________________________________________________
    if($_SERVER['REQUEST_METHOD']==’GET’){
    $id=$_GET['id'];
    $nickname=$_GET['nickname'];
    $uname=$_GET['uname'];
    $regtime=$_GET['regtime'];
    $sex=$_GET['sex'];
    }
    $ConditionsNumber=5;
    $ConditionsArray=array(“$id “,”$nickname”,”$uname”,”$regtime”,”$sex”);
    $SearchSQLArray=array(” where id =’$id’”,” where nickname = ‘$nickname’”,” where uname=’$uname’”,” where regtime=’$regtime’”,” where sex=’$sex’”);
    for($i=0;$i<$ConditionsNumber;$i++)
    {
    if($ConditionsArray[$i]==”")
    $SearchSQLArray[$i]=”";
    $haveWhere=false;
    for($j=0;$j<$i;$j++)
    {
    $wherePosition=strpos($SearchSQLArray[$j],”where”);
    if(($wherePosition==”1″)&&($haveWhere==false))
    {
    $SearchSQLArray[$i]=ereg_replace(“where”,”and”,$SearchSQLArray[$i]);
    $haveWhere=true;
    } } }}
    for($i=0;$i<$ConditionsNumber;$i++)
    $sql=$sql.$SearchSQLArray[$i];
    $query=”SELECT * FROM user_bas “.$sql.” order by id;”;
    ?>

    3.目前有会员登陆的log记录数组,登陆日志里面有这个会员的编号(id)、会员帐号(uname)和登陆时间(logtime)。
    通过这个数组需要得到如下统计:
    (1)得到某一天中不同会员的登陆情况,一天内会员登陆了4次,就记4次。统计结果如:
    在2006-07-20内的会员登陆情况
    会员编号 会员帐号 会员登陆次数
    1 User1 4次
    3 User3 2次
    2 User2 1次
    (2)得到某几天内不同会员的登陆情况,会员在某一天登陆了很多次只记一次(也可以说会员在几天内有几天登陆了)。统计结果如:
    在2006-07-20 到2006-07-22 的3天时间内的会员登陆情况:
    会员编号 会员帐号 会员登陆次数
    3 User3 3次
    1 User1 2次
    2 User2 2次
    附件:
    会员登陆日志数组:
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-22 19:18:02′);
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-22 18:15:03′);
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-22 13:50:12′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-22 13:12:09′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-22 11:10:08′);
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-22 10:52:54′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-22 08:16:04′);

    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 19:18:02′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 18:15:03′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 13:50:12′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-21 13:12:09′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-21 11:10:08′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 10:52:54′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 08:16:04′);

    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-20 19:18:02′);
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-20 18:15:03′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-20 13:50:12′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-20 13:12:09′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-20 11:10:08′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-20 10:52:54′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-20 08:16:04′);

    代码:
    ______________________________________________________________________________
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-22 19:18:02′);
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-22 18:15:03′);
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-22 13:50:12′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-22 13:12:09′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-22 11:10:08′);
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-22 10:52:54′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-22 08:16:04′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 19:18:02′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 18:15:03′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 13:50:12′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-21 13:12:09′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-21 11:10:08′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 10:52:54′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-21 08:16:04′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-20 19:18:02′);
    $userlog[] = array (“id” => 2, “uname” => ‘user2′, “logtime” => ’2006-07-20 18:15:03′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-20 13:50:12′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-20 13:12:09′);
    $userlog[] = array (“id” => 3, “uname” => ‘user3′, “logtime” => ’2006-07-20 11:10:08′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-20 10:52:54′);
    $userlog[] = array (“id” => 1, “uname” => ‘user1′, “logtime” => ’2006-07-20 08:16:04′);
    //第一题
    $user=”;
    $date=’2006-07-20′;
    foreach($userlog as $val)
    {
    if(!isset($user[$val['id']]))
    {
    $user[$val['id']]['id']=$val['id'];
    $user[$val['id']]['uname']=$val['uname'];
    $user[$val['id']]['logcounts']=0;
    }
    if(strstr($val['logtime'],$date))
    {
    $user[$val['id']]['logcounts']+=1;
    }
    }
    echo ‘

    ’;
    print_r($user);
    echo ‘
    ’;
    //第二题
    $users=”;
    $start=strtotime(’2006-07-20′);
    $end =strtotime(’2006-07-22′)+3600*24;
    $time=”;
    foreach($userlog as $val)
    {
    if(!isset($users[$val['id']]))
    {
    $users[$val['id']]['id']=$val['id'];
    $users[$val['id']]['uname']=$val['uname'];
    $users[$val['id']]['days']=array();
    $users[$val['id']]['logs']=0;
    }
    $time=strtotime($val['logtime']);
    if($time>=$start & $time<$end)
    {
    if(!in_array(substr($val['logtime'],0,10),$users[$val['id']]['days']))
    {
    $users[$val['id']]['days'][]=substr($val['logtime'],0,10);
    $users[$val['id']]['logs']+=1;
    }
    }
    }
    echo ‘
    ’;
    print_r($users);
    echo ‘
    ’;
    ?>

www.phpzy.comtrue/phpmst/3634.htmlTechArticlephp开发工程师程序员面试题 1.获取字母组合 题目要求: 做一个三位字母的组合功能,例如:”abd,ade,acc,aef”等等: 目标 1.字母所有的组合需要打印出来 2.字母组合要排除,abc(三个字母连续...

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