如何对多维数组值求和，然后将日期分组为我的代码。 如果有任何PHP代码我应该尝试，请告诉我。
请参阅数组代码：
$array = array (
0 => array(
'date' => '2015-02-06 10:42:39',
'visit' => 1,
'bounce' => 0
),
1 => array(
'date' => '2015-02-06 13:23:21',
'visit' => 1,
'bounce' => 1
),
2 => array(
'date' => '2015-02-07 04:11:42',
'visit' => 1,
'bounce' => 1
),
3 => array(
'date' => '2015-02-08 11:35:28',
'visit' => 1,
'bounce' => 1
),
4 => array(
'date' => '2015-02-08 15:12:09',
'visit' => 1,
'bounce' => 1
),
5 => array(
'date' => '2015-02-09 15:12:09',
'visit' => 1,
'bounce' => 0
),
);
我期望的结果必须是我做的时候：
date visit bounce
2015-02-06 2 1
2015-02-07 1 1
2015-02-08 2 2
2015-02-09 1 0
以下是我尝试过的代码。但它只返回日期计数。
$items = array_column($array, 'date');
$preg = preg_quote('2015-02-06', '~');
$result = preg_grep('~' . $preg . '~', $items);
echo 'Date <br/>' . count($result);
请帮忙，先谢谢你。
答案 0 :(得分：0)
您可以创建一个新阵列：
$newArr = [];
foreach($array as $arr){
$d = (new DateTime($arr['date']))->format('Y-m-d');
$newArr[$d] = [
"visit" => isset($newArr[$d]['visit']) ? $newArr[$d]['visit'] += $arr['visit'] : $arr['visit'],
"bounce" => isset($newArr[$d]['bounce']) ? $newArr[$d]['bounce'] += $arr['bounce'] : $arr['bounce']
];
}
echo "<pre>";
var_dump($newArr);
echo "</pre>";
上面返回一个很好的格式化数组，您可以在发布的示例中轻松读出：
array(4) {
["2015-02-06"]=>
array(2) {
["visit"]=>
int(2)
["bounce"]=>
int(1)
}
["2015-02-07"]=>
array(2) {
["visit"]=>
int(1)
["bounce"]=>
int(1)
}
["2015-02-08"]=>
array(2) {
["visit"]=>
int(2)
["bounce"]=>
int(2)
}
["2015-02-09"]=>
array(2) {
["visit"]=>
int(1)
["bounce"]=>
int(0)
}
}
答案 1 :(得分：0)
您可以合并这样的条目：
$items = [];
foreach ($array as $value) {
$date = substr($value['date'], 0, 10);
if (!isset($items[$date]) {
$items[$date] = [
'date' => $date,
'visit' => 0,
'bounce' => 0
];
}
$items[$date]['visit'] += $value['visit'];
$items[$date]['bounce'] += $value['bounce'];
}
使用date作为$items数组中的键，我们确保为每个日期总结visit和bounce，而不是附加它们。
如果您想要删除密钥，只需使用array_values。
答案 2 :(得分：0)
如果您的数组名为$ myArray，则您创建的新数组是$ myDateArray：
$myDateArray = array();
foreach ($myArray as $value)
{
list($date_string, $other) = explode(" ", $value['date']);
if (array_key_exists($date_string, $myDateArray))
{
//adding the visits and bounces
$myDateArray[$date_string]['visit'] = $myDateArray[$date_string]['visit'] + $value['visit'];
$myDateArray[$date_string]['bounce'] = $myDateArray[$date_string]['bounce'] + $value['bounce'];
}
else
{
//putting the first values (of the key $date_string) into $myDateArray
array_push($myDateArray, array($date_string, $value['visit], $value['bounce']));
}
}
请告诉我这是否适合您！
答案 3 :(得分：0)
代码创建一个新数组，其中包含数组的总和结果。
$resultArray = [];
foreach($array as $row){
$dateObj = DateTime::createFromFormat('Y-m-d H:i:s', $row['date']);
$key = $dateObj->format('Y-m-d');
if(!array_key_exists($key, $resultArray)){
$resultArray[$key] = ['visit' => $row['visit'], 'bounce' => $row['bounce']];
}
else {
$resultArray[$key]['visit'] += $row['visit'];
$resultArray[$key]['bounce'] += $row['bounce'];
}
}
结果：
array (size=4)
'2015-02-06' =>
array (size=2)
'visit' => int 2
'bounce' => int 1
'2015-02-07' =>
array (size=2)
'visit' => int 1
'bounce' => int 1
'2015-02-08' =>
array (size=2)
'visit' => int 2
'bounce' => int 2
'2015-02-09' =>
array (size=2)
'visit' => int 1
'bounce' => int 0
答案 4 :(得分：0)
这可以解决这个问题吗？
<?php
$array = array (
0 => array(
'date' => '2015-02-06 10:42:39',
'visit' => 1,
'bounce' => 0
),
1 => array(
'date' => '2015-02-06 13:23:21',
'visit' => 1,
'bounce' => 1
),
2 => array(
'date' => '2015-02-07 04:11:42',
'visit' => 1,
'bounce' => 1
),
3 => array(
'date' => '2015-02-08 11:35:28',
'visit' => 1,
'bounce' => 1
),
4 => array(
'date' => '2015-02-08 15:12:09',
'visit' => 1,
'bounce' => 1
),
5 => array(
'date' => '2015-02-09 15:12:09',
'visit' => 1,
'bounce' => 0
),
);
$results = [];
foreach($array as $e)
{
$date = explode(' ',$e['date'])[0] ;
if(!isset($results[$date]))
$results[$date] = ['visit'=>0 , 'bounce'=>0] ;
$results[$date]['visit'] += $e['visit'];
$results[$date]['bounce'] += $e['bounce'];
}
print_r($results);
答案 5 :(得分：0)
使用array_walk，substr和array_values函数的解决方案：
$date_keys = [];
array_walk($array, function($v) use(&$date_keys){
$datePart = $v['date'] = substr($v["date"], 0, 10);
if (isset($date_keys[$datePart])) {
$date_keys[$datePart]['visit'] += $v['visit'];
$date_keys[$datePart]['bounce'] += $v['bounce'];
} else {
$date_keys[$datePart] = $v;
}
});
print_r(array_values($date_keys));
输出：
Array
(
[0] => Array
(
[date] => 2015-02-06
[visit] => 2
[bounce] => 1
)
[1] => Array
(
[date] => 2015-02-07
[visit] => 1
[bounce] => 1
)
[2] => Array
(
[date] => 2015-02-08
[visit] => 2
[bounce] => 2
)
[3] => Array
(
[date] => 2015-02-09
[visit] => 1
[bounce] => 0
)
)
答案 6 :(得分：0)
技术/机制
$result = array();
$items = array_column($array, 'date');
for($i=0; $i < count($items); $i++){
$date = date("Y-m-d", strtotime($items[$i]));
$result[$date][0] += $array[$i]['visit'];
$result[$date][1] += $array[$i]['bounce'];
}
<强>结果
请根据需要设计输出。
echo "date visit bounce<br/>";
foreach($result as $key => $value){
echo $key." ".$value[0]." ".$value[1]."<br/>";
}
<强>输出
date visit bounce
2015-02-06 2 1
2015-02-07 1 1
2015-02-08 2 2
2015-02-09 1 0