求一个判断数组值在字符串中出现的顺序

WBOY
Release: 2016-06-23 13:38:40
Original
799 people have browsed it

数组
$arr=array('A','M','N');
字符串不定,但是M和N在字符串中只能出现一次,如果出现多次则返回false
符合要求则返回数组$data['M'],$data['N']
比如:
A MAN  返回 2,4   //M为数组中第2个出现,N为数组中第4个出现 (A出现在1和3)
A BOY 返回 false //不含M或者N
MAN 返回1,3   //M为数组中第1个出现,N为数组中第3个出现 (A出现在2)
A ATM IN 返回 3,5   //M为数组中第3个出现,N为数组中第4个出现 (A出现在1和2)


回复讨论(解决方案)

本来一个简单的问题,被你已解释变得复杂了
开看这段代码的结果就知道该怎么做了

$arr = array('A','M','N');$data = array(  'A MAN',  'A BOY',  'MAN',  'A ATM IN');foreach($data as $text) {  $t = str_split($text);  $r[] = array_intersect($t, $arr);}print_r($r);
Copy after login
Copy after login
Array(    [0] => Array        (            [0] => A            [2] => M            [3] => A            [4] => N        )    [1] => Array        (            [0] => A        )    [2] => Array        (            [0] => M            [1] => A            [2] => N        )    [3] => Array        (            [0] => A            [2] => A            [4] => M            [7] => N        ))
Copy after login
Copy after login

本来一个简单的问题,被你已解释变得复杂了
开看这段代码的结果就知道该怎么做了

$arr = array('A','M','N');$data = array(  'A MAN',  'A BOY',  'MAN',  'A ATM IN');foreach($data as $text) {  $t = str_split($text);  $r[] = array_intersect($t, $arr);}print_r($r);
Copy after login
Copy after login
Array(    [0] => Array        (            [0] => A            [2] => M            [3] => A            [4] => N        )    [1] => Array        (            [0] => A        )    [2] => Array        (            [0] => M            [1] => A            [2] => N        )    [3] => Array        (            [0] => A            [2] => A            [4] => M            [7] => N        ))
Copy after login
Copy after login


多谢版主热心解答。
但 不是这个意思,你的代码显示出来的结果不是我要的结果,
数组前三个结果正确,是因为结果凑巧,第四个返回就是错误的。
可能是我表达比较混乱。

$arr = array('(title)','(url)','(img)');$str = '<li><a href="(url)" title="(title)" ><img  src=(img) / alt="求一个判断数组值在字符串中出现的顺序" ></a></li>';
Copy after login

我现在要得到的是 (title是第2个出现),url是第1个出现 img是第3个出现。
而下面的字符串
$str = '<li><a href="(url)"  title="(title)" >(title)<img  src=(img) / alt="求一个判断数组值在字符串中出现的顺序" ></a></li>';
Copy after login

我现在要得到的是 (title是第2,3个出现),url是第1个出现 img是第4个出现。

$arr = array('A','M','N');
$str = 'AMN';
$data = array();
foreach($arr as $v){
if(substr_count($str,$v) == 1){
$data[$v] = strpos($str,$v);
}
}
var_dump($data);

$arr = array('A','M','N');
$str = 'AMN';
$data = array();
foreach($arr as $v){
if(substr_count($str,$v) == 1){
$data[$v] = strpos($str,$v);
}
}
var_dump($data);


如果$str = 'AMTN'就会出错啦


$arr = array('A','M','N');
$str = 'AMN';
$data = array();
foreach($arr as $v){
if(substr_count($str,$v) == 1){
$data[$v] = strpos($str,$v);
}
}
var_dump($data);


如果$str = 'AMTN'就会出错啦



没出错啊
array(3) { ["A"]=> int(0) ["M"]=> int(1) ["N"]=> int(3) }



$arr = array('A','M','N');
$str = 'AMN';
$data = array();
foreach($arr as $v){
if(substr_count($str,$v) == 1){
$data[$v] = strpos($str,$v);
}
}
var_dump($data);


如果$str = 'AMTN'就会出错啦



没出错啊
array(3) { ["A"]=> int(0) ["M"]=> int(1) ["N"]=> int(3) }


M是第二出场,结果1,N是第3个出场,应该是2才对。

群星演唱会来了三个大歌手和一堆小歌星。
大歌星分别是:刘德华,张学友,林俊杰
举办方安排了出场顺序,林俊杰年轻点多唱了几首,其他两个没精力只唱了一首。
张学友,XXX,XXX,林俊杰,XXX,林俊杰,XXX,XXX,刘德华,XXX,林俊杰XXX
我现在需要排的是 他们三人的出场顺序
结果应该是林俊杰 2,3,5
张学友1
刘德华4
不知道这样解释清楚了没有。

$arr = array('A','M','N');$str = 'AMTN';$data = array();foreach($arr as $v){		if(substr_count($str,$v) == 1){		$data[$v] = strpos($str,$v)+1;	}}var_dump($data);
Copy after login
Copy after login


结果

array(3) { ["A"]=> int(1) ["M"]=> int(2) ["N"]=> int(4) }

$arr = array('A','M','N');$str = 'AMTN';$data = array();foreach($arr as $v){		if(substr_count($str,$v) == 1){		$data[$v] = strpos($str,$v)+1;	}}var_dump($data);
Copy after login
Copy after login


结果

array(3) { ["A"]=> int(1) ["M"]=> int(2) ["N"]=> int(4) }


我要的结果N是3
只统计 AMN的顺序 倒是你个了我思路,我把数组键和值反转一下,看能不能得到结果。

搞定了,用正则匹配出的。唉!~

$arr = array('(title)','(url)','(img)');		$str = '<li><a href="(url)" title="(title)" ><img  src=(img) / alt="求一个判断数组值在字符串中出现的顺序" >(title)</a></li>';		preg_match_all('/(\(title\)|\(url\)|\(img\))/i',$str,$data);		print_r($data);
Copy after login

source:php.cn
Statement of this Website
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
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template