小白编程求解答.
1.声明一个数组装5个会员(每个会员含有会员id,姓名,年龄,性别,注册时间信息),再将这些数据以表格形式表现出来,如果年龄在18以内显示未成年,在18~30显示青年,30~60显示中年,60以上显示老年,并要统计不同年龄段的人数
2.利用for循环做九九乘法表
3.利用while循环将1~500之间的质数找出来-->array
4.$year是闰年输出366,否则输出365
------解决思路----------------------
作业?
1.
<br /><?php<br />$arr = array(<br /> array('id'=>1,'name'=>'a','age'=>17,'addtime'=>date('Y-m-d')),<br /> array('id'=>2,'name'=>'b','age'=>19,'addtime'=>date('Y-m-d')),<br /> array('id'=>3,'name'=>'c','age'=>31,'addtime'=>date('Y-m-d')),<br /> array('id'=>4,'name'=>'d','age'=>62,'addtime'=>date('Y-m-d')),<br /> array('id'=>5,'name'=>'e','age'=>18,'addtime'=>date('Y-m-d')),<br />);<br />echo '<meta http-equiv="content-type" content="text/html; charset=utf-8">';<br />echo '<table>';<br />echo '<tr><td>id</td><td>name</td><td>age</td><td>addtime</td><td>年龄段</td></tr>';<br /><br />$agegroup = array();<br /><br />foreach($arr as $v){<br /> $agetype = checkage($v['age']);<br /> if(isset($agegroup[$agetype])){<br /> $agegroup[$agetype]++;<br /> }else{<br /> $agegroup[$agetype] = 1;<br /> }<br /> echo '<tr>';<br /> echo '<td>'.$v['id'].'</td>';<br /> echo '<td>'.$v['name'].'</td>';<br /> echo '<td>'.$v['age'].'</td>';<br /> echo '<td>'.$v['addtime'].'</td>';<br /> echo '<td>'.$agetype.'</td>';<br /> echo '</tr>';<br />}<br />echo '</table>';<br /><br />echo '统计结果<br>';<br />foreach($agegroup as $k=>$v){<br /> echo $k.':'.$v.'<br>';<br />}<br /><br /><br />function checkage($age){<br /> if($age<18){<br /> return '未成年';<br /> }elseif($age>=18 && $age<30){<br /> return '青年';<br /> }elseif($age>=30 && $age<60){<br /> return '中年';<br /> }else{<br /> return '老年';<br /> }<br />}<br /><br />?><br />
2.
<br />echo '<table>';<br />for($i=1; $i<=9; $i++){<br /> echo '<tr>';<br /> for($j=1; $j<=9; $j++){<br /> echo '<td>'.$i.'*'.$j.'='.($i*$j).'</td>';<br /> }<br /> echo '</tr>';<br />}<br />echo '</table>';<br />
3.
<br /><?php<br />$i=2;<br />$result = array();<br />while($i<=500){<br /> $j = 2;<br /> $k = sqrt($i);<br /> $flag = 1;<br /> while($j<=$k){<br /> if($i%$j==0){<br /> $flag = 0;<br /> break;<br /> }<br /> $j++;<br /> }<br /> if($flag==1){<br /> array_push($result, $i);<br /> }<br /> $i++;<br />}<br />print_r($result);<br />?><br />
4.
<br />echo checkyear('2016');<br />function checkyear($year){<br /> if($year%4==0 && $year%100!=0 <br><font color='#FF8000'>------解决思路----------------------</font><br> $year%400==0){<br /> return 366;<br /> }else{<br /> return 365;<br /> }<br />}<br />

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