Question

我有4个php文件，它们几乎都是相同的，只是表名不同，数组中的值也不同，我想将所有这些值放在另一个php文件中的一个数组中。

此文件为Germany.php

foreach($array as $key => $val) {
    $Home=0;
    $Draw=0;
    $Away=0;
    $sql = "SELECT * FROM Germany where B = '{$val['B']}' AND E = '{$val['E']}' AND F = '{$val['F']}' AND O ='{$val['O']}' AND A = '*' ";
    $result = $conn->query($sql);
    if ($result->num_rows > 0) {
    $rowcount=mysqli_num_rows($result);
        // output data of each row 
        while($row = $result->fetch_assoc()) {
            if($row['R'] == 1){
                $Home++;
            }elseif($row['R'] == 0){
                $Draw++;
            }else{
                $Away++;
            }
        }

        //We use an array rather than overriding everytime
        $AwayPrediction[$key] = round(($Away/$rowcount )*100);
    } else {
        echo "0 results";
    }
}
$conn->close();

我认为我必须在index.php中做类似的事情

<table class="table">   
<thead> 
      <tr>
        <th>Home</th>
         <th>Away</th>
         <th>Result</th>
      </tr>
</thead>

    <?php






$array_of_teams=array();
        include 'germany.php';
        foreach($array as $key => $data) {
    $array_of_teams+=merge($AwayPrediction[$key]);
        echo "<tr>";
        echo "<td>".$data['H']."</td>";
        echo "<td>".$data['M']."</td>";

    echo "<td>".$AwayPrediction[$key]."</td>";
    }

include 'france.php';
foreach($array as $key => $data) {

    $array_of_teams+=merge($AwayPrediction[$key]);

echo "<tr>";
echo "<td>".$data['H']."</td>";
echo "<td>".$data['M']."</td>";

echo "<td>".$AwayPrediction[$key]."</td>";
}

如何合并$ AwayPrediction [$ key]或将其添加到$ array_of_teams = array（）;对于每个文件？

Answer 1

您的PHP“显示”部分

foreach($array as $key => $data) {
    echo "<tr>";
    echo "<td>".$data['H']."</td>";
    echo "<td>".$data['M']."</td>";

echo "<td>".$AwayPrediction[$key]."</td>";
}

仅显示include 'england.php';中的最后一个数组因此，在每个数组之后放置一个显示部分，例如

include 'germany.php';
foreach($array as $key => $data) {
        echo "<tr>";
        echo "<td>".$data['H']."</td>";
        echo "<td>".$data['M']."</td>";

    echo "<td>".$AwayPrediction[$key]."</td>";
    }

include 'france.php';
foreach($array as $key => $data) {
echo "<tr>";
echo "<td>".$data['H']."</td>";
echo "<td>".$data['M']."</td>";

echo "<td>".$AwayPrediction[$key]."</td>";
}

Answer 2

首先，您应该遵循DRY principle并尝试最小化代码库中的代码重复。就您而言，您可以为重复的功能创建一个功能：

function predictionForCountry($countryName, $array, $conn)
{
    foreach ($array as $key => $val) {
        $Home = 0;
        $Draw = 0;
        $Away = 0;
        $sql = "SELECT * FROM {$countryName} where B = '{$val['B']}' AND E = '{$val['E']}' AND F = '{$val['F']}' AND O ='{$val['O']}' AND A = '*' ";
        $result = $conn->query($sql);
        if ($result->num_rows > 0) {
            $rowcount = mysqli_num_rows($result);
            // output data of each row 
            while ($row = $result->fetch_assoc()) {
                if ($row['R'] == 1) {
                    $Home++;
                } elseif ($row['R'] == 0) {
                    $Draw++;
                } else {
                    $Away++;
                }
            }

            //We use an array rather than overriding everytime
            $AwayPrediction[$key] = round(($Away / $rowcount) * 100);
        } else {
            echo "0 results";
        }
    }
    return $array;
}

另一种可能的解决方案，如果表的列数相同，则可以使用UNION ALL运算符在SQL中合并所有结果，并根据需要调整WHERE条件（如果每个国家/地区的条件不同）

$sql = "SELECT * FROM Germany where B = '{$val['B']}' AND E = '{$val['E']}' AND F = '{$val['F']}' AND O ='{$val['O']}' AND A = '*' 
        UNION ALL
        SELECT * FROM France where B = '{$val['B']}' AND E = '{$val['E']}' AND F = '{$val['F']}' AND O ='{$val['O']}' AND A = '*' 
        UNION ALL
        SELECT * FROM Holland where B = '{$val['B']}' AND E = '{$val['E']}' AND F = '{$val['F']}' AND O ='{$val['O']}' AND A = '*' 
        UNION ALL
        SELECT * FROM England where B = '{$val['B']}' AND E = '{$val['E']}' AND F = '{$val['F']}' AND O ='{$val['O']}' AND A = '*' ";
$result = $conn->query($sql);

Answer 3

$array并没有在那里更改，并且对于所有国家来说似乎都是相同的，$key和$val或$data也是如此。每个收录都将覆盖之前的$AwayPrediction[$key]，并且您不会在国家/地区之间看到任何差异。

如果要获得全局结果，可以将每个$AwayPrediction[$key]与其前一个值相加，与行数相同，然后在显示前计算百分比。

在国家/地区文件中：

$rowcount += mysqli_num_rows($result);
$AwayPrediction[$key] += $Away;

在index.php中

echo "<td>".round(($AwayPrediction[$key]/$rowcount )*100)."</td>";

如果要获得每个国家/地区的结果，则可以向$AwayPrediction[$key]['Germany']添加1个维度

然后您可以执行以下操作：

foreach ($array as $key => $data) {
  foreach ($AwayPrediction[$key] as $country => $prediction) {
    echo "<tr>";
    echo "<td>".$data['H']."</td>";
    echo "<td>".$data['M']."</td>";
    echo "<td>".$country."</td><td>".$prediction."<td>\n";
  }
}

如何合并来自不同文件的一个数组中的两个或多个数组值

3 个答案: