Question

这是我桌子的内容：

如何计算不同表格单元格中的所有名称以获得这样的结果？

鲍勃-7

亚历-3

伊凡-5

尼娜-5-
  .........

使用此代码：

<?php
$q= mysqli_query($db, 'SELECT * FROM names');
    while ($all = mysqli_fetch_array($q)) {
        $k1= $all['Name1'];
        $k2= $all['Name2'];
        $k3= $all['Name3'];
        $k4= $all['Name4'];
        $k5= $all['Name5'];
        $k6= $all['Name6'];
        $arr[]= $k1;
        $total_values = array_count_values($arr);
    }
    foreach ($total_values as $key => $value) {
        echo $key .'-'. $value .'<br>';
    }

我的输出结果是：

鲍勃-3

亚历-1

伊凡-1

尼娜-2

当我将代码更改为：

时

<?php
$q= mysqli_query($db, 'SELECT * FROM names');
    while ($all = mysqli_fetch_array($q)) {
        $k1= $all['Name1'];
        $k2= $all['Name2'];
        $k3= $all['Name3'];
        $k4= $all['Name4'];
        $k5= $all['Name5'];
        $k6= $all['Name6'];
        $arr=array($k1, $k2);
        $total_values = array_count_values($arr);
        foreach ($total_values as $key => $value) {
            echo $key .'-'. $value .'<br>';
        }
    }

我的输出结果是：

鲍勃-1

亚历-1

伊凡-1

尼娜-1

尼娜-1

鲍勃-1

鲍勃-1

伊凡-1

尼娜-1

尼娜-1
  .......

在数组$ arr中添加$ k2，$ k3 .... $ k6有什么问题？我需要做什么？

Answer 1

这将计算单个名称的出现次数：

$q= mysqli_query($db, 'SELECT * FROM names');
$arr = array();

// 1) loop through rows
while ($all = mysqli_fetch_array($q, MYSQLI_NUM)) {
 // 2) loop through cells in a row
 foreach($all as $val) {
  // 3) 'roll out' the values into a one-dimensional array
  if(empty($val)) { continue; } // if you don't want to count empty cells
  $arr[] = $val;
 }    
}
// 4) count the number of occurences
$names_qty = array_count_values($arr);

// optional loop that shows the results
foreach($names_qty as $name=>$qty) {
 echo $name.'-'.$qty.'<br />';
}

此解决方案的好处是您只能调用一次计数函数而不是每次迭代。

Answer 2

您必须在循环中累积数据并在处理完所有数据后将其打印出来。

$total = array();
$result = mysqli_query($db, 'SELECT * FROM names', MYSQLI_USE_RESULT);
while (mysqli_fetch_assoc($result) as $row) {
    foreach ($row as $col => $value)
        total[$col]++;
    }
}

print_r($total);

Answer 3

此代码已经过测试，应该可以解决问题。

<?php
$arr = array();
$q = mysqli_query($db, 'SELECT * FROM names');
while ($all = mysqli_fetch_assoc($q)) {
    foreach($all as $val) {
        if(empty($val)) {
            continue;
        }
        // saves all names with the name as key
        $arr[$val][] = $val;
    } 
}

// Output
foreach($arr as $k => $v) {
    // count how many times the name was pushed into the array
    echo $k.' - '. count($v);
}

PHP计算在许多列中包含相同值的MYSQL行

3 个答案: