20

有谁知道如何计算此数组中“照片”的出现次数：

    Array ( 
    [0] => stdClass Object ( [type] => photo [id] => 1404781893036 [created_time] => 2012-03-02T07:58:23+0000 ) 
    [1] => stdClass Object ( [type] => photo [id] => 14047818930362 [created_time] => 2012-03-01T14:58:53+0000 ) 
    [2] => stdClass Object ( [type] => status [id] => 1404781893036 [created_time] => 2012-03-01T09:49:40+0000 ) 
    [3] => stdClass Object ( [type] => status [id] => 14047818930362 [created_time] => 2012-03-01T09:36:04+0000 ) 
    [4] => stdClass Object ( [type] => photo [id] => 14047818930362 [created_time] => 2012-02-28T07:03:25+0000 ) 
    [5] => stdClass Object ( [type] => photo [id] => 1404781893036 [created_time] => 2012-02-27T09:15:34+0000 ) 
    [6] => stdClass Object ( [type] => photo [id] => 14047818930362 [created_time] => 2012-02-27T07:32:13+0000 ) 
    [7] => stdClass Object ( [type] => status [id] => 1404781893036 [created_time] => 2012-02-25T09:36:57+0000 )
    [8] => stdClass Object ( [type] => photo [id] => 1404781893036 [created_time] => 2012-02-23T08:46:43+0000 ) 
    [9] => stdClass Object ( [type] => status [id] => 1404781893036 [created_time] => 2012-02-22T21:04:30+0000 ) 
    [10] => stdClass Object ( [type] => status [id] => 1404781893036 [created_time] => 2012-02-21T20:38:27+0000 )
    [11] => stdClass Object ( [type] => photo [id] => 1404781893036 [created_time] => 2012-02-21T07:22:44+0000 ) 
    [12] => stdClass Object ( [type] => status [id] => 14047818930362 [created_time] => 2012-02-20T08:32:46+0000 ) 
    [13] => stdClass Object ( [type] => status [id] => 1404781893036 [created_time] => 2012-02-17T15:00:11+0000 ) )
4

4 回答 4

57

要计算多维数组中字符串的匹配出现次数，您需要遍历每个数组元素并匹配字符串并增加计数。类似地@Dor 建议

$count = 0;
foreach ($array as $item) {
  if ($item->type === 'photo') {
    $count++;
  }
}

如果您想在一维数组中实现相同的效果，那么这非常简单。您可以使用array_count_values PHP 数组函数，如下所述。

<?php
   $array = array(1, "test", 1, "php", "test");
   print_r(array_count_values($array));
?>

上面的示例将输出：

Array
(
    [1] => 2
    [test] => 2
    [php] => 1
)
于 2012-11-20T09:55:01.343 回答
15 
$count = 0;
foreach ($array as $item) {
  if ($item->type === 'photo') {
    $count++;
  }
}
于 2012-03-07T10:08:48.110 回答
2

我想承认 Dor Shemer 的方法是（IMO）最直接、干净、可读和可靠的方法。我只想为那些喜欢使用函数式编程的人提供一些替代方案……array_reduce()对我来说紧随其后。最后，我想指出一个使用方法的小问题array_count_values()——请继续阅读……

方法组：（演示）

$photo_count=0;  // establish default value
foreach($array as $objects){
    if($objects->type==='photo') ++$photo_count;  // pre-increment
}
echo "foreach result = $photo_count";

echo "array_reduce = ",array_reduce($array,function($carry,$objects){return $carry+($objects->type==='photo'?1:0);},0);

echo "array_filter & count = ",sizeof(array_filter($array,function($objects){return $objects->type==='photo';}));

echo "array_column & array_filter & count = ",sizeof(array_filter(array_column($array,'type'),function($v){return $v==='photo';}));

echo "array_map & array_count_values & array_replace = ",array_replace(['photo'=>0],array_count_values(array_map(function($o) {return $o->type;}, $array)))['photo'];

echo "array_map & array_count_values (gives Notice) = ",array_count_values(array_map(function($o) {return $o->type;}, $array))['photo'];

使用 OP 的样本数据输入/输出（没问题）：

$array=[
    (object)['type'=>'photo','id'=>1404781893036,'created_time'=>'2012-03-02T07:58:23+0000'],
    (object)['type'=>'photo','id'=>14047818930362,'created_time'=>'2012-03-01T14:58:53+0000'],
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-03-01T09:49:40+0000'],
    (object)['type'=>'status','id'=>14047818930362,'created_time'=>'2012-03-01T09:36:04+0000'],
    (object)['type'=>'photo','id'=>14047818930362,'created_time'=>'2012-02-28T07:03:25+0000'],
    (object)['type'=>'photo','id'=>1404781893036,'created_time'=>'2012-02-27T09:15:34+0000'],
    (object)['type'=>'photo','id'=>14047818930362,'created_time'=>'2012-02-27T07:32:13+0000'],
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-02-25T09:36:57+0000'],
    (object)['type'=>'photo','id'=>1404781893036,'created_time'=>'2012-02-23T08:46:43+0000'],
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-02-22T21:04:30+0000'],
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-02-21T20:38:27+0000'],
    (object)['type'=>'photo','id'=>1404781893036,'created_time'=>'2012-02-21T07:22:44+0000'],
    (object)['type'=>'status','id'=>14047818930362,'created_time'=>'2012-02-20T08:32:46+0000'],
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-02-17T15:00:11+0000']
];

// output:
foreach result = 7

array_reduce = 7

array_filter & count = 7

array_column & array_filter & count = 7

array_map & array_count_values & array_replace = 7

array_map & array_count_values = 7

使用没有photo值的数据输入/输出（第二种array_count_values()方法有问题）：

$array=[
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-03-01T09:49:40+0000'],
    (object)['type'=>'status','id'=>14047818930362,'created_time'=>'2012-03-01T09:36:04+0000'],
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-02-25T09:36:57+0000'],
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-02-22T21:04:30+0000'],
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-02-21T20:38:27+0000'],
    (object)['type'=>'status','id'=>14047818930362,'created_time'=>'2012-02-20T08:32:46+0000'],
    (object)['type'=>'status','id'=>1404781893036,'created_time'=>'2012-02-17T15:00:11+0000']
];
// or if there are no object rows like: $array=[];
// output:
foreach result = 0

array_reduce = 0

array_filter & count = 0

array_column & array_filter & count = 0

array_map & array_count_values & array_replace = 0

array_map & array_count_values (gives Notice) = <br />
<b>Notice</b>:  Undefined index: photo in <b>[...][...]</b> on line <b>43</b><br />

array_count_values()无需费心生成具有0计数的元素。

于 2017-10-24T02:23:52.390 回答
1

尝试：

$input = array( /* your data */ );
$count = 0;
foreach ( $input as $value ) {
  if ( $value->type == 'photo' ) {
    $count++;
  }
}
于 2012-03-07T10:09:14.490 回答