PHP preg_grep 错误？

Question

PHP preg_grep 不工作？我是PHP初学者，也可以用英语交流。该程序的执行结果由“ArrayArray”表示...

<?php
$news = fopen("news.txt", "r"); 
$keywords = fopen("keywords.txt", "r"); 

$open_news = [];
while (!feof($news)) {
    $open_news[] = fgets($news);
}

$arr_keywords = [];
while (!feof($keywords)) {
    $arr_keywords[] = fgets($keywords);
}

$count = count($arr_keywords); 


for ($i = 0 ; $i <= $count; $i++) {
    if ($x = preg_grep("/^" . $arr_keywords[$i]  . "/", $open_news)) {
        echo $x;
        }
}

fclose($news); 
fclose($keywords); 
?>

Answer 1

preg_grep returns匹配行的数组，因此您应该将代码重写为

for ($i = 0 ; $i <= $count; $i++) {
    if ($x = preg_grep("/^" . $arr_keywords[$i]  . "/", $open_news)) {
        echo implode(', ', $x), PHP_EOL;
    }
}

整个脚本可以简化:

<?php

$open_news    = file("news.txt",     FILE_IGNORE_NEW_LINES | FILE_SKIP_EMPTY_LINES);
$arr_keywords = file("keywords.txt", FILE_IGNORE_NEW_LINES | FILE_SKIP_EMPTY_LINES);

foreach ($arr_keywords as $keyword) {
    if ($x = preg_grep("/^" . preg_quote($keyword, '/') . "/", $open_news)) {
        echo implode(', ', $x) . PHP_EOL;
    }
}