PHP query voting exercise (0508)

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Release: 2016-07-29 08:50:41
Original
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Practice questions:

php 之 查询 投票练习(0508)php 之 查询 投票练习(0508)php 之 查询 投票练习(0508)

Solution:

Method 1:

1. Main voting page:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" c />
<title>投票</title>
<style>
#list
{
	width:400px;
	height:300px;}
#jieguo
{
	width:400px;
	height:300px;}
.x
{
	
	float:left;}

</style>
</head>

<body>

<form action="TouChuLi.php" method="post">
<?php
include ("DBDA.class.php");
$db=new DBDA();
//从调研题目表中找出题目代号和名称
$sql="select * from diaoyantimu limit 0,1";
$arr=$db->query($sql);
$tmmc=$arr[0][1];
$tmdh=$arr[0][0];
echo "<div><h2>{$tmmc}:</h2></div>";
//从调研选项表中输出选项内容:
$sqlxx="select * from diaoyanxuanxiang where timudaihao='{$tmdh}'";
$arrxx=$db->query($sqlxx);
echo "<div>";
foreach ($arrxx as $v)
{
	echo "<div><input type='checkbox' value='{$v[0]}' name='xx[]'>{$v[1]}</div><br />";
}

?>
<input type="submit" value="提交">
<input type="button" value="查看结果" >
<?php
//计算总人数:
$sqlzs="select sum(numbers) from diaoyanxuanxiang where timudaihao='{$tmdh}'";
$zrs=$db->query($sqlzs);

foreach ($arrxx as $v)
{
	
	$name=$v[1];
	$number=$v[2];
	if($zrs[0][0]==0)
     {
         $bfb = 0;
     }
     else
     {
          $bfb = ($number/$zrs[0][0])*100;
     }
	$bfb=round($bfb,2);	
	echo "<div>
		<span>{$name} </span>
		<div >
		<div> </div>
		</div>
		<span>{$number} </span>
		<span>{$bfb}% </span>
		</div>
		<br />
		";	
}
?>
<input type="button" value="返回" ><div class="code" style="position:relative; padding:0px; margin:0px;"><pre class="brush:php;toolbar:false"><?php
$arr=$_POST["xx"];
include ("../DBDA.class.php");
$db=new DBDA();


foreach($arr as $v)
{
	$sql = "update diaoyanxuanxiang set numbers = numbers+1 where ids = '{$v}'";
	$db->Query($sql,1);//1代表$sql的类型
}
header ("location:TouPiao.php");
?>
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 3. Establish access database Class, encapsulation for reference:

<?php
//执行一个sql语句,返回相应的结果
class DBDA
{
	public $host="localhost";//数据库服务器地址
	public $uid="root";//数据库用户名
	public $password="";//数据库密码
	//执行SQL语句的方法
	//参数里面:$sql代表要执行的sql语句;$type是sql语句的类型,0代表查询,1代表其他(增删改);$db代表要操作的数据库
	function Query($sql,$type=0,$db="mydb")
	{
		//造连接对象
		$dbc MySQLi($this->host,$this->uid,$this->password,$db);
		//判断连接是否出错
		!mysqli_connect_error() or die("连接失败!");
		//执行sql语句
		$result=$dbconnect->query($sql);
		//判断SQL语句类型
		if($type==0)
		{
			 //如果是查询语句返回结果集的二维数组
			return $result->fetch_all();
			}
		else
		{
			//如果是其他语句,返回true或false
			return $result;
			}		
		}
		
	}
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Method 2:

1. Voting main page:

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" c />
<title>投票</title>
<style>
.x
{
	
	float:left;}

</style>
</head>

<body>

<form action="TouChuLi.php" method="post">
<?php
include ("../DBDA.class.php");
$db=new DBDA();
//从调研题目表中找出题目代号和名称
$sql="select * from diaoyantimu limit 0,1";
$arr=$db->query($sql);
$tmmc=$arr[0][1];
$tmdh=$arr[0][0];
echo "<div><h2>{$tmmc}:</h2></div>";
//从调研选项表中输出选项内容:
$sqlxx="select * from diaoyanxuanxiang where timudaihao='{$tmdh}'";
$arrxx=$db->query($sqlxx);
echo "<div>";
foreach ($arrxx as $v)
{
	echo "<div><input type='checkbox' value='{$v[0]}' name='xx[]'>{$v[1]}</div><br />";
}

?>
<input type="submit" value="提交">
<a href="ChaKan.php"><input type="button" value="查看结果" ></a>
</form>
</body>
</html>
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2. Processing voting page:

<?php
$arr=$_POST["xx"];
include ("../DBDA.class.php");
$db=new DBDA();


foreach($arr as $v)
{
	$sql = "update diaoyanxuanxiang set numbers = numbers+1 where ids = '{$v}'";
	$db->Query($sql,1);//1代表$sql的类型
}
header ("location:TouPiao.php");
?>
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3. View voting results page:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" c />
<title>查看结果</title>
<style>
.x
{
	float:left;}
</style>
</head>
<body>

<?php
include ("../DBDA.class.php");
$db=new DBDA();
//从调研题目表中找出题目代号和名称
$sql="select * from diaoyantimu limit 0,1";
$arr=$db->query($sql);
$tmmc=$arr[0][1];
echo "<div><h2>{$tmmc}:</h2></div>";
//从调研选项表中输出选项内容:
$sqlxx="select * from diaoyanxuanxiang where timudaihao='{$arr[0][0]}'";
$arrxx=$db->query($sqlxx);

//计算总人数:
$sqlzs="select sum(numbers) from diaoyanxuanxiang where timudaihao='{$arr[0][0]}'";
$zrs=$db->query($sqlzs);

foreach ($arrxx as $v)
{
	 $name=$v[1];//调研项目名称
	 $number=$v[2];//选择该项的人数
	 //判断总人数是否为0
	 if($zrs[0][0]==0)
     {
         $bfb = 0;
     }
     else
     {
          $bfb = ($number/$zrs[0][0])*100;//求百分比
     }
	
	$bfb=round($bfb,2);	//取小数点后两位
	echo "<div>
	<span>{$name} </span>
	<div >
	<div> </div>
	</div>
	<span>{$number}  </span>
	<span>{$bfb}%</span><br />
	</div><br />";	
}
?>
<br />
<a href="TouPiao.php"><input type="button" value="返回"></a>
</body>
</html>
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Web page display results:

php 之 查询 投票练习(0508)php 之 查询 投票练习(0508)

The above introduces the PHP query voting exercise (0508), including PHP and exercise content. I hope it will be helpful to friends who are interested in PHP tutorials.

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