看看陣列並從重複的值計算真或假

Question

嗨，有人可以給我一個想法如何編碼。

感謝

Answer 1

有使用循環的簡單（不推薦）溶液：

$resultq = mysql_query('select value, user_id from answers'); 
$answers_per_user = array(); // positive answers per user 
$totals_per_user = array(); // total answers per user 
while($result = mysql_fetch_assoc($resultq)){ 
if($result['answer']) 
$answers_per_user[$result['user_id']] += $result['answer']; // increment positive answer  counter for user 
$totals_per_user[$result['user_id']]++; 
} 

你將有一個數組每用戶，然後可以用它來計算負保持每個用戶和總的答案肯定回答答案

推薦的解決方案是使用一組由sql語句給出的所有計算信息。

$result = mysql_query('select sum(value) as positivecount, count(*) as total, user_id from answers group by user_id'); 
while($data = mysql_fetch_assoc($result)){ 
// $data will hold positivecount,total, and user_id giving you all the data you need for calculating negative answer values. 
} 
// alternatively, use a query like this for counting the answers that were 'beans': 
// select sum(if(value = "beans", 1, 0)) as answered_beans, count(*) as total, user_id from answers group by user_id 

見：http://dev.mysql.com/tech-resources/articles/wizard/page3.html

Answer 2

這個問題最優雅的解決方案實際上有兩個SQL表;每個用戶有一行（用戶ID，用戶名等）和每個投票一個，每個用戶可以有多個。

以下示例將回顯有關數據的一些信息。

<?php 
$sqlusers = mysql_query("SELECT userid FROM user_table")//This line grabs all users from the database. 
$users = mysql_fetch_array($sqlusers);//This line creates an array containing all users. 
foreach($users as $key=>$currentuser){ 
    $sqlvotes = mysql_query("SELECT userid, vote FROM vote_table WHERE userid = $currentuser[userid]"); 
    $votes = mysql_fetch_array($sqlvotes);//obtain an array of votes the current user has submitted 
    $votefrequency = array_count_values($votes)//counts the amount of trues and falses in the $votes array, and returns an array with the [true] and [false] indexes containing their respective frequency. 
    echo "user ".$userid." has voted ".$votefrequency[true]." times true and ".$votefrequency[false]." times false/n"; 
    echo "average vote:". (($votefrequency[true] - $votefrequency[false] > 0) ? "true" : "false"); 
}