如果模式不匹配，如何使preg_match_all返回一个空数组值？

问题描述：

Let's say we have the following data, with the code to match the pattern we'd like, in this case we're catching all numbers and unicode fractions.

$array = array('1 ½ cups','¼ cup','2 tablespoons', '½ cup', '1/3 cup',  '2 large', '1 ½ teaspoons', '2 tablespoons', 'Large egg', '1 teaspoon', '¼ teaspoon');

foreach($array as $arr){
    preg_match_all("/^(?:[\p{Pd}.\/\s-]*[\d↉½⅓⅔¼¾⅕⅖⅗⅘⅙⅚⅐⅛⅜⅝⅞⅑⅒⅟])+/um", $arr, $output);

    foreach($output[0] as $data){
        $try[] = $data;
    }
}

If we print_r($try) we get:

Array
(
    [0] => 1 ½
    [1] => ¼
    [2] => 2
    [3] => ½
    [4] => 1/3
    [5] => 2
    [6] => 1 ½
    [7] => 2
    [8] => 1
    [9] => ¼
)

There are 11 items in the array, one of them being complete text, Large egg in this example.

What I'm trying to do is make preg_match_all return an empty value for that iteration, so we'd get this instead:

Array
(
    [0] => 1 ½
    [1] => ¼
    [2] => 2
    [3] => ½
    [4] => 1/3
    [5] => 2
    [6] => 1 ½
    [7] => 2
    [8] =>
    [9] => 1
    [10] => ¼
)

What I've tried?

I looked over the preg_match_all manual but I wasn't able to find anything that could lead me to my answer, at this point I'm thinking it might have to be done over at the regex pattern, but I'm not sure at all at this point.

https://eval.in/885289

假设我们有以下数据，代码与我们想要的模式相匹配，在这种情况下我们 '捕获所有数字和unicode分数。 p>

  $ array = array（'1½cups'，'¼can'，'2汤匙'，'½杯'，'  1/3杯'，'2大'，'1½茶匙'，'2汤匙'，'大蛋'，'1茶匙'，'¼茶匙'）; 
 
foreach（$ array as $ arr）{  
 preg_match_all（“/ ^（？：[\ p {Pd}。\ / \ s  - ] * [\d↉½⅓⅔¼¾⅕⅖⅗⅘⅙⅚⅐⅛⅜⅝⅞⅑⅒⅟]）+ / um”，$ arr，$ output）; 
 
 foreach（$ output [  0] as $ data）{
 $ try [] = $ data; 
} 
} 
  code>  pre> 
 
 如果我们 print_r（$ try）   code>我们得到： p> 
 
 
  Array 
（
 [0] =＆gt;1½
 [1] =＆gt;¼
 [2] =  ＆gt; 2 
 [3] =＆gt;½
 [4] =＆gt; 1/3 
 [5] =＆gt; 2 
 [6] =＆gt;1½
 [7] =＆gt;  2 
 [8] =＆gt; 1 
 [9] =＆gt;¼
）
  code>  pre> 
 
 数组中有11个项目，其中一个 在这个例子中是完整的文本，大蛋 em>。   p> 
 
 
我要做的是让 preg_match_all  code>为该迭代返回一个空值，所以我们得到这个： p> 
  
 
  Array 
（
 [0] =＆gt;  1½
 [1] =＆gt;  ¼
 [2] =＆gt;  2 
 [3] =＆gt;  ½
 [4] =＆gt;  1/3 
 [5] =＆gt;  2 
 [6] =＆gt;  1½
 [7] =＆gt;  2 
 [8] =＆gt; 
 [9] =＆gt;  1 
 [10] =＆gt;  ¼
）
  code>  pre> 
 
  我尝试了什么？ strong>  p> 
 
 
我查看了preg_match_all  手册，但我找不到任何可能的内容 引导我回答，此时我认为可能必须在正则表达式模式下完成，但我现在还不确定。 p> 
 
 

  https://eval.in/885289   p> 
  div>

答

It looks like each iteration can only return a maximum of one match, so preg_match_all with the inner foreach should not be needed. You can use preg_match and append a default blank value if there isn't a match.

foreach($array as $arr){
    preg_match("/^(?:[\p{Pd}.\/\s-]*[\d↉½⅓⅔¼¾⅕⅖⅗⅘⅙⅚⅐⅛⅜⅝⅞⅑⅒⅟])+/um", $arr, $output);

    $try[] = $output[0] ?? '';

}

答

Modify the pattern to this:

/^(?:[\p{Pd}.\/\s-]*[\d↉½⅓⅔¼¾⅕⅖⅗⅘⅙⅚⅐⅛⅜⅝⅞⅑⅒⅟]|)+/um

Notice the "OR nothing" at the end.

<?php
$array = array('1 ½ cups','¼ cup','2 tablespoons', '½ cup', '1/3 cup',  '2 large', '1 ½ teaspoons', '2 tablespoons', 'Large egg', '1 teaspoon', '¼ teaspoon');

foreach($array as $arr){
    preg_match_all("/^(?:[\p{Pd}.\/\s-]*[\d↉½⅓⅔¼¾⅕⅖⅗⅘⅙⅚⅐⅛⅜⅝⅞⅑⅒⅟]|)+/um", $arr, $output);
    foreach($output[0] as $data){
        $try[] = $data;
    }
}
var_dump($try);

Demo

Result:

array(11) {
  [0]=>
  string(4) "1 ½"
  [1]=>
  string(2) "¼"
  [2]=>
  string(1) "2"
  [3]=>
  string(2) "½"
  [4]=>
  string(3) "1/3"
  [5]=>
  string(1) "2"
  [6]=>
  string(4) "1 ½"
  [7]=>
  string(1) "2"
  [8]=>
  string(0) ""
  [9]=>
  string(1) "1"
  [10]=>
  string(2) "¼"
}

Disclaimer: this looks like a dirty hack that is probably brittle, but it can get you in the right direction

如果模式不匹配，如何使preg_match_all返回一个空数组值？

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