Question

我遇到了创建一些csv文件的PHP脚本问题。 PHP脚本如下：

<?php

    $inputFile = "/var/www/vhosts/pecso.it/httpdocs/test/export30gg.txt";

    $csvData = file_get_contents($inputFile);

    $rows = explode(PHP_EOL, $csvData);

    $rowsArray = array();
    foreach ($rows as $row) {
        $rowsArray[] = str_getcsv($row);
    }

    $csvFileName = "/var/www/vhosts/pecso.it/httpdocs/graphs/export30gg.csv";

    if (file_exists($csvFileName)){
        unlink($csvFileName);
    }

    $csvFile = fopen($csvFileName, "w");
    $csvFileForGraph = fopen($csvFileNameForGraph, "w");    

    for ($i = 0; $i < count($rowsArray); $i++) {
        $dateTime = DateTime::createFromFormat('d/m/Y', $rowsArray[$i][0]);
        $d = $dateTime->format('Y-m-d');
        $rowsArray[$i][0] = $d;
        $rowForGraph = $rowsArray[$i];
        unset($rowForGraph[1]);
        $row = implode(',',$rowsArray[$i]);
        $rowForGraph = implode(',',$rowForGraph);
        file_put_contents($csvFileName, $row.PHP_EOL , FILE_APPEND);
    }

    fclose($csvFileName);
?>

此脚本正常运行并且正确创建了csv文件export30gg.csv但是，每次运行此脚本时，我都会出现以下错误：

fclose() expects parameter 1 to be resource

请帮帮我吗？

Answer 1

应该是

fclose($csvFile);

因为您在$ csvFile中存储了追索链接，而不是在$ csvFileName

中

Answer 2

flose()接受参数作为文件指针资源，在代码执行时会找到它，例如在你的情况下是$csvFile = fopen($csvFileName, "w"); 所以，它应该是

fclose($csvFile);

而不是fclose（$ csvFileName）;

fclose（）期望参数1是资源

2 个答案: