从目录中读取 *.csv 文件并显示每个文件的内容失败

Question

我在读取文件夹和打开/输出 csv 数据方面需要帮助，我使用了其他人编写的示例，但 none 对我有用。

我现在有的是这个，但是没有输出文件：

$files = scandir($PathToCreate.$version."/"); //scan the folder
foreach($files as $file) { //for each file in the folder

//The following is another example I found but does not output anything I just need to open each file and be able to output / target specific data

$csv = array();
$lines = file($file, FILE_IGNORE_NEW_LINES);

foreach ($lines as $key => $value)
{
    $csv[$key] = str_getcsv($value);
} 
print_r($csv)

}

Answer 1

这应该适合你：

（这里我首先从目录中抓取扩展名为 *.csv 和 glob(). After this I loop through each file and read it with fopen() and fgetcsv() 的所有文件。）

<?php

    $files = glob("$PathToCreate$version/*.csv");

    foreach($files as $file) {

        if (($handle = fopen($file, "r")) !== FALSE) {
            echo "<b>Filename: " . basename($file) . "</b><br><br>";
            while (($data = fgetcsv($handle, 4096, ",")) !== FALSE) {
                echo implode("\t", $data);
            }
            echo "<br>";
            fclose($handle);
        } else {
            echo "Could not open file: " . $file;
        }

    }

?>

Answer 2

第一个问题可能是您必须忽略两个目录条目 . 和 ..，它们具有特殊含义并且对您没有用：

$files = scandir($PathToCreate.$version."/"); //scan the folder
foreach($files as $file) { //for each file in the folder
  if ( ! in_array($file, ['.','..'])) {
    $lines = file($file, FILE_IGNORE_NEW_LINES);
    foreach ($lines as $key => $value) {
      $csv[$key] = str_getcsv($value);
    } 
    print_r($csv)
  }
}

从目录中读取 *.csv 文件并显示每个文件的内容失败

Reading *.csv files from directory and showing the content of each file fails

php

csv

scandir

fgetcsv

opencsv