Home >Backend Development >PHP Tutorial >请问一下该如何解析出这个数组呢?谢谢了。

请问一下该如何解析出这个数组呢?谢谢了。

WBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWB
WBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOriginal
2016-06-23 14:23:13881browse

Array(    [doctorDate] => Array        (            [0] => Array                (                    [doctorOfDate] => 2013-10-02                    [orderingCount] => 12                )             [1] => Array                (                    [doctorOfDate] => 2013-10-03                    [orderingCount] => 8                )         ) )


回复讨论(解决方案)

什么意思?你想做什么

假设数组名为:test
test['doctorDate'][0]['doctorOfDate']=2013-10-02
test['doctorDate'][0]['orderingCount']=12
test['doctorDate'][1]['doctorOfDate']=2013-10-03
test['doctorDate'][1]['orderingCount']=8
这个数组的定义是:

    test=array('doctorDate'=>                    array(0=>array('doctorDate'=>'2013-10-02',                                'orderingCount'=>12),                          1=>array('doctorDate'=>'2013-10-03',                                'orderingCount'=>8)                           )               )

我就想把这个数组的数据显示成:
2013-10-02  12
2013-10-02   8

我就想把这个数组的数据显示成:
2013-10-02  12
2013-10-02   8

Array(    [doctorDate] => Array        (            [0] => Array                (                    [doctorOfDate] => 2013-10-02                    [orderingCount] => 12                )             [1] => Array                (                    [doctorOfDate] => 2013-10-03                    [orderingCount] => 8                )         ) )

<?php$test=array('doctorDate'=>                    array(0=>array('doctorDate'=>'2013-10-02',                                'orderingCount'=>12),                          1=>array('doctorDate'=>'2013-10-03',                                'orderingCount'=>8)                           )               );foreach ($test as $arr){foreach ($arr as $value){echo $value['doctorDate'].'   '.$value['orderingCount'].'<br>';}}?>

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn