I have the following array from a count query:
我有一个计数查询中的以下数组:
array = (
0=>array(
'date'=>2017-09-01,
'total'=>4
),
1=>array(
'date'=>2017-09-03,
'total'=>6
),
# and so on..
);
I want to fill the date even if my query has no records on that date! How can I add the missing date index to the array. The dates are to be sequential. so I want the following output:
即使我的查询在该日期没有记录,我想填写日期!如何将缺少的日期索引添加到数组中。日期是顺序的。所以我想要以下输出:
array = (
0=>array(
'date'=>2017-09-01,
'total'=>4
),
1=>array(
'date'=>2017-09-02,
'total'=>0
)
2=>array(
'date'=>2017-09-03,
'total'=>6
),
3 个解决方案
#1
2
Extended solution with DateTime
object, array_map
and range
functions:
使用DateTime对象,array_map和范围函数的扩展解决方案:
$arr = [
['date' => '2017-09-01', 'total' => 4],
['date' => '2017-09-07', 'total' => 6],
['date' => '2017-09-09', 'total' => 7]
];
$result = [];
foreach ($arr as $k => $item) {
$d = new DateTime($item['date']);
$result[] = $item;
if (isset($arr[$k+1])) {
$diff = (new DateTime($arr[$k+1]['date']))->diff($d)->days;
if ($diff > 1) {
$result = array_merge($result , array_map(function($v) use($d){
$d_copy = clone $d;
return [
'date' => $d_copy->add(new DateInterval('P' . $v. 'D'))->format('Y-m-d'),
'total' => 0
];
}, range(1, $diff-1)));
}
}
}
print_r($result);
The output:
Array
(
[0] => Array
(
[date] => 2017-09-01
[total] => 4
)
[1] => Array
(
[date] => 2017-09-02
[total] => 0
)
[2] => Array
(
[date] => 2017-09-03
[total] => 0
)
[3] => Array
(
[date] => 2017-09-04
[total] => 0
)
[4] => Array
(
[date] => 2017-09-05
[total] => 0
)
[5] => Array
(
[date] => 2017-09-06
[total] => 0
)
[6] => Array
(
[date] => 2017-09-07
[total] => 6
)
[7] => Array
(
[date] => 2017-09-08
[total] => 0
)
[8] => Array
(
[date] => 2017-09-09
[total] => 7
)
)
#2
1
Here I create new arrays as reference.
One for all dates from first item in your array to last (range).
And one array with only the dates from your array so that I can search it.
This should handle more than one missing date and multiple missing dates.
在这里,我创建新的数组作为参考。一个适用于所有日期,从数组中的第一个项目到最后一个(范围)。一个数组只包含数组中的日期,以便我可以搜索它。这应该处理多个丢失日期和多个丢失日期。
$arr = array (
0=>array(
'date'=>'2017-09-01',
'total'=>4
),
1=>array(
'date'=>'2017-09-07',
'total'=>6
)
);
// array with only dates to search in
$dates = array_column($arr, "date");
// Create an array with all dates from first item to last item
$start = new DateTime($arr[0]["date"]);
$end = new DateTime(end($arr)["date"]);
$range = new DatePeriod($start, new DateInterval('P1D'), $end);
// $range is now all dates from start to end minus last one.
// Loop through the range
foreach($range as $date){
//See if the current date exist is first array
$find = array_search($date->format("Y-m-d"), $dates);
If($find !== false){
$result[] = $arr[$find]; // if it does copy it to result array
}Else{
// If not add it and create a total = 0
$result[] = array('date' => $date->format("Y-m-d"), 'total' => 0);
}
}
// Since the loop does not loop all dates we need to add the last item to result.
$result[] = end($arr);
Var_dump($result);
https://3v4l.org/Hf73Z
Edit; forgot date ->format()
https://3v4l.org/Hf73Z编辑;忘了日期 - >格式()
#3
0
Using an if statement try this :
使用if语句试试这个:
foreach($array as $ar){
if(!$ar["date"]){
$ar["date"] = date("j- n- Y"); ;
}
}
#1
2
Extended solution with DateTime
object, array_map
and range
functions:
使用DateTime对象,array_map和范围函数的扩展解决方案:
$arr = [
['date' => '2017-09-01', 'total' => 4],
['date' => '2017-09-07', 'total' => 6],
['date' => '2017-09-09', 'total' => 7]
];
$result = [];
foreach ($arr as $k => $item) {
$d = new DateTime($item['date']);
$result[] = $item;
if (isset($arr[$k+1])) {
$diff = (new DateTime($arr[$k+1]['date']))->diff($d)->days;
if ($diff > 1) {
$result = array_merge($result , array_map(function($v) use($d){
$d_copy = clone $d;
return [
'date' => $d_copy->add(new DateInterval('P' . $v. 'D'))->format('Y-m-d'),
'total' => 0
];
}, range(1, $diff-1)));
}
}
}
print_r($result);
The output:
Array
(
[0] => Array
(
[date] => 2017-09-01
[total] => 4
)
[1] => Array
(
[date] => 2017-09-02
[total] => 0
)
[2] => Array
(
[date] => 2017-09-03
[total] => 0
)
[3] => Array
(
[date] => 2017-09-04
[total] => 0
)
[4] => Array
(
[date] => 2017-09-05
[total] => 0
)
[5] => Array
(
[date] => 2017-09-06
[total] => 0
)
[6] => Array
(
[date] => 2017-09-07
[total] => 6
)
[7] => Array
(
[date] => 2017-09-08
[total] => 0
)
[8] => Array
(
[date] => 2017-09-09
[total] => 7
)
)
#2
1
Here I create new arrays as reference.
One for all dates from first item in your array to last (range).
And one array with only the dates from your array so that I can search it.
This should handle more than one missing date and multiple missing dates.
在这里,我创建新的数组作为参考。一个适用于所有日期,从数组中的第一个项目到最后一个(范围)。一个数组只包含数组中的日期,以便我可以搜索它。这应该处理多个丢失日期和多个丢失日期。
$arr = array (
0=>array(
'date'=>'2017-09-01',
'total'=>4
),
1=>array(
'date'=>'2017-09-07',
'total'=>6
)
);
// array with only dates to search in
$dates = array_column($arr, "date");
// Create an array with all dates from first item to last item
$start = new DateTime($arr[0]["date"]);
$end = new DateTime(end($arr)["date"]);
$range = new DatePeriod($start, new DateInterval('P1D'), $end);
// $range is now all dates from start to end minus last one.
// Loop through the range
foreach($range as $date){
//See if the current date exist is first array
$find = array_search($date->format("Y-m-d"), $dates);
If($find !== false){
$result[] = $arr[$find]; // if it does copy it to result array
}Else{
// If not add it and create a total = 0
$result[] = array('date' => $date->format("Y-m-d"), 'total' => 0);
}
}
// Since the loop does not loop all dates we need to add the last item to result.
$result[] = end($arr);
Var_dump($result);
https://3v4l.org/Hf73Z
Edit; forgot date ->format()
https://3v4l.org/Hf73Z编辑;忘了日期 - >格式()
#3
0
Using an if statement try this :
使用if语句试试这个:
foreach($array as $ar){
if(!$ar["date"]){
$ar["date"] = date("j- n- Y"); ;
}
}