我想遍历目录text files并回显word count每个文件。例如，如果目录包含两个具有以下内容的文本文件：
file1.txt -> this is file1.
file2.txt -> this is another file called file2.
然后输出应为：
wordcount: 3
wordcount: 6
我有以下代码：
$directory = "C:\\dir";
$files = scandir($directory);
foreach($files as $file) {
$fh = fopen($file, "r");
$contents = fread($fh, filesize($file));
fclose($fh);
echo "wordcount: "; //this should be modified to display the wordcount for each file..
}
回声应修改为回声每个文件的字数。
这应该为您工作：
$directory = "C:\\xampp\\htdocs\\Sandbox";
foreach (glob("$directory\\*.txt") as $file) {
$fh = fopen($file, "r");
if(filesize($file) > 0) {
$contents = fread($fh, filesize($file));
$count = str_word_count($contents, 0);
} else {
$count = 0;
}
fclose($fh);
echo "File: " . basename($file) . " Wordcount: $count<br />";
}
输出可能如下所示：
File: test - Kopie (2).txt Wordcount: 3
File: test - Kopie.txt Wordcount: 2
File: test.txt Wordcount: 7
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我来说两句