如何使用explode函数来找出这样的结果？并且请解释一下在分解函数中使用的每个$delimiter的含义

Question

我的输入是$text = '( ( LED AND DIODE ) OR ( "LEE power" and system ) ) '

我正在对此输入应用分解函数

function multiexplode ($delimiters,$string) 
{
    $ready = str_replace($delimiters, $delimiters[0], $string);
    $launch = explode($delimiters[0], $ready);
    return  $launch;
}               
$exploded = multiexplode(array(' ',":" ), $text);
for($i=0;$i<(count($exploded));$i++)
    echo "<br> $exploded[$i]";

我的输出如下所示

0 - (   
1 - (  
2 - LED  
3 - AND  
4 - DIODE  
5 - )  
6 - OR  
7 - (  
8 - "LEE  
9 - power"  
10 - and  
11 - system  
12 - )  
13 - )  

但我希望输出如下所示

0 - (  
1 - (  
2 - LED  
3 - AND  
4 - DIODE  
5 - )  
6 - OR  
7 - (  
8 - LEE power  
9 - and  
10 - system  
11 - )  
12 - )  

我希望将“”中的所有单词存储在数组的一个元素中。

Answer 1

检查以下代码：

$text = '( ( LED AND DIODE ) OR ( "LEE power" and system ) ) ';

preg_match_all('/"(?:\.|[^"])*"|\S+/', $text, $exploded);
for($i=0;$i<(count($exploded[0]));$i++)
    echo "<br>". $exploded[0][$i];

基于comment的更新

对于$text = '((LED AND DIODE) OR ("LEE power" and system)) ';的输入

$text = '((LED AND DIODE) OR ("LEE power" and system)) ';
$re = '/"(?:\.|[^"])*"|\(?|\)|(\'(.*?)\'|(\w+))/';
preg_match_all($re, $text, $exploded);

for($i=0;$i<(count($exploded[0]));$i++)
    if($exploded[0][$i]!='')
        echo "<br>". $exploded[0][$i];

Answer 2

虽然这不是一个经过全面测试的答案，但它回答了您所描述的当前问题。祝你好运！

function concatenateQuotedElements($input_array)
{
  //As per the question, we'll be exploding our array by spaces " ".
  $pieces = explode(" ", $input_array);

  //This array will be returned as our final array of concatenated elements.
  $output_array = array();


  /*When we happen upon an element containing a 
  parenthesis ('"') as the first character, we'll 
  temporarily store so that we may concatenate 
  future elements with it.*/
  $stored_element = null;
  $stored_position = null; 

  /*Parse through our array and look for any character 
  that contains a " and see whether it's at the beginning 
  or the end of the string element.*/
  for ($i = 0; $i < count($pieces); $i++)
  {   

    //If we detect a parenthesis in our element...  
    if (strpos($pieces[$i], '"') !== false)
    { 

      //See if it's at the beginning, and if it is, store it or future use.      
      if (substr($pieces[$i], 0, 1) === '"')
      {
        $stored_element = $pieces[$i];
        $stored_position = $i;
      }

      /*Or, see if it's at the end of the string. If it is, we need to see 
       if there's been a previous element with an opening parenthesis that 
       we've stored and concatenate this element onto our stored element.*/
      if (
          substr($pieces[$i], -1) === '"' && 
          $stored_element !== null && 
          $stored_position !== null
         )
      {
        $output_array[$stored_position] = $stored_element . " " . $pieces[$i];

        /*Reset our temporary storage so that it hold 
        any forthcoming elements with parenthesis.*/  
        $stored_element = null;
        $stored_position = null;

        /*Finally, [Continue] to the next element without 
        pushing this as a non-concatenated, standalone element.*/
        continue;
      }
    }

    array_push($output_array, $pieces[$i]);
  }

  return $output_array;
}

对于与给定数组相关的特定输出，您可以使用：

//Our example array given in the question:
$array = '( ( LED AND DIODE ) OR ( "LEE power" and system ) )';

//Our output:
$concatenate_quotes_array = concatenateQuotedElements($array);

//"Pretty print" the array result of the function
echo "<pre>";
print_r($concatenate_quotes_array);
echo "</pre>";

输出：

Array
(
    [0] => (
    [1] => (
    [2] => LED
    [3] => AND
    [4] => DIODE
    [5] => )
    [6] => OR
    [7] => (
    [8] => "LEE power"
    [9] => and
    [10] => system
    [11] => )
    [12] => )
)

Answer 3

这里

<?php
    $text = str_replace('LEE power','LEE-power',"( ( LED AND DIODE ) OR ( 'LEE power' and system ) ) ");
    $newArray = explode(' ',$text);
    foreach($newArray as $row) $data[] = str_replace('-',' ',$row);
    echo "<pre>";print_r($data);die;
?>