Php 如何组合3个阵列

你好社区

我只是想问一下我的代码。我只想合并我的3个变量

$result = mysqli_query($con, "SELECT disease,age,SUM(CASE WHEN gender = 'm' THEN 1 ELSE 0 END) AS `totalM`, SUM(CASE WHEN gender = 'f' THEN 1 ELSE 0 END) AS `totalF` FROM mdr where disease = '$diseaseselection' GROUP BY disease , age");
$chart_data = '';
while($row = mysqli_fetch_array($result))
{
    $tabx[]=$row['age'];
    $taby[]=$row['totalM'];
    $tabz[]=$row['totalF'];

}
$tableau=array_combine($tabx,$taby,$tabz);

foreach($tableau as $key=>$value){

    $string[]=array('age'=>$key,'totalM'=>$value,'totalF'=>$value);

}

echo json_encode($string);

它可以很好地使用此代码。具有2个变量。我希望它是由树变量完成的

$result = mysqli_query($con, "SELECT disease,age,SUM(CASE WHEN gender = 'm' THEN 1 ELSE 0 END) AS `totalM`, SUM(CASE WHEN gender = 'f' THEN 1 ELSE 0 END) AS `totalF` FROM mdr where disease = '$diseaseselection' GROUP BY disease , age");
$chart_data = '';
while($row = mysqli_fetch_array($result))
{
    $tabx[]=$row['age'];
    $taby[]=$row['totalM'];

}
$tableau=array_combine($tabx,$taby);

foreach($tableau as $key=>$value){

    $string[]=array('age'=>$key,'totalM'=>$value);

}

echo json_encode($string);

这是我的预期输出

{ age:'0-1', totalM:2, totalF:1},

{ age:'1-4', totalM:1, totalF:0},

{ age:'10-14', totalM:0, totalF:1},

{ age:'15-19', totalM:0, totalF:1},

{ age:'5-9', totalM:0, totalF:3},

{ age:'55-59', totalM:6, totalF:0}

---

不需要使用多个数组，使用多维数组可以获得所需的结果

$key = 0;
$output = [];
while($row = mysqli_fetch_array($result)){
 $output[$key]['age'] = $row['age'];
 $output[$key]['totalM'] = $row['totalM'];
 $output[$key]['totalF'] = $row['totalF'];
 $key++;
}
echo json_encode($output);

---

现在还不清楚你到底想要实现什么。请发布输入和预期输出以及错误/问题Array combine正在合并2个数组以形成一个键值对。在这种情况下，您真的不需要使用。你想创建一个JSON字符串3个或更多的键吗？我已经在上面添加了我的预期输出。我只是想组合我的3个数组。上面的代码是一个例子，你必须用这种方法来解决。是的，我同意我在数组变量中使用了适当的名称，我应该更改that@B001ᛦ 好吧，我猜提问者会明白的，我应该给出完整的例子：）。

$result = mysqli_query($con, "SELECT disease,age,SUM(CASE WHEN gender = 'm' THEN 1 ELSE 0 END) AS `totalM`, SUM(CASE WHEN gender = 'f' THEN 1 ELSE 0 END) AS `totalF` FROM mdr where disease = '$diseaseselection' GROUP BY disease , age");

$chart_data = '';

$data = [];

while($row = mysqli_fetch_array($result)) {
    $data[] = [
        'age' => $row['age'],
        'totalM' => $row['totalM'], 
        'totalF' => $row['totalF']
    ];
}

echo json_encode($data);