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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

php

0

Muratcan · 技术社区 · 6 年前

我有4个php文件,它们几乎都是一样的,只是表名不同,数组中的值也不同,我想把所有这些值放在一个数组中的另一个php文件中。

这个文件是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();

我想在index.php中必须这样做

<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>";
}

如何合并或添加$AWAYPROCED[$KE]到$ARYAYOFOFSARS=ARARYY();对于每个文件?

0 回复 | 直到 6 年前

1

uxmal 6 年前

你的php“display”部分

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

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

仅显示上一个数组 include 'england.php'; 所以在每个数组后面放置一个显示区域,比如

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>";
}

2

0

Andrii Filenko 6 年前

首先,你应该跟着 DRY principle 尽量减少代码库中的代码重复。在您的情况下,可以为重复功能创建一个函数:

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;
}

另一个可能的解决方案,如果您的表具有相同数量的列,则可以使用 UNION ALL 运算符将所有结果合并到SQL中,根据需要调整where条件(如果每个国家的情况不同):

$sql = "SELECT * FROM Germany where B = '{$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);

3

0

Tuckbros 6 年前

$array 没有改变,似乎所有国家都一样 $key 和 $val 或 $data . 每个include都将覆盖上一个 $AwayPrediction[$key] 你不会发现国家之间有什么不同。

如果你想要一个全局结果,你可以每个 $awaypredition[$key] 对于它的前一个值,与rowcount相同,在显示它之前,您要计算百分比。

在国家档案中:

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

在ph.php中

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

如果您想要每个国家的结果,可以将1维添加到 $AwayPrediction[$key]['Germany']

然后你可以做这样的事情:

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";
  }
}