php问答笔记2

用PHP打印出昨天的时间?

date_default_timezone_set('Asia/Shanghai');
echo date("Y-m-d H:i:s",strtotime("-1 day"));

$a="PHP";
$b=&$a;
unset($b);
$b="world";
echo $a;

结果是PHP,可以理解为a指针指向string，b变量取a的地址，所以b也指向了string，释放了b变量，不会对a造成影响，所以输出PHP

$str="cd";
$$str="php"
$$str.=" good";
echo $cd

答案：php good

实现不适用第3个变量，交换$a,$b的值，$a,$b的初始值自己定?

$a=333;
$d=444;
$d=explode("|",$a."|".$d);
$a=$d[1];
$d=$d[0];

黑客是这样写的

$a = 23;$b =25;
swap($a,$b);

function swap(&$a,&$b){
    echo $a,",",$b;
    $a = $a ^ $b;
    $b = $a ^ $b;
    $a = $a ^ $b;
    echo "
",$a,",",$b;
}

原格式:2010-11-23 转新格式: 10年11月23日 5点30分？

<?php
header("content-type:text/html;charset='utf-8'");
date_default_timezone_set('Asia/Shanghai');
echo date("y年m月d日 h点i分",strtotime("2010-11-23 17:30"));

写出两种连接mysql的模式代码？

$link = mysqli_connect("127.0.0.1",'root','password');
mysqli_select_db($link,'');

$pdo = new PDO("mysql:host=127.0.0.1;dbname=xxx",'用户名','密码');

多表查询:


左连接:select a.*,b.sex,b.age,b.state from a left join b on a.uid = b.uid;


右链接:select a.*,b.sex,b.age,b.state from a right join b on a.uid = b.uid;


内链接:select a.*,b.sex,b.age,b.state from a inner join b on a.uid = b.uid;

使用php插入一条信息，A表uid为递增，B表uid无递增需要与A表uid相等?

mysqli_query($link,"insert into a(uid,name) values('','xxx')");
$uid = mysqli_insert_id($link);
mysqli_query($link,"insert into b(uid,sex,age,state) values($uid,'xxx','xxx',xxx')");

查询state中有多少类型，并显示他们的姓名.

select a.name from a,b where a.uid = b.uid group by b.state;

select a.name from a left join b on a.uid = b.uid group by b.state;