Php fclose()要求参数1是资源
我对创建csv文件的PHP脚本有问题。 PHP脚本如下所示:Php fclose()要求参数1是资源,php,fopen,fclose,Php,Fopen,Fclose,我对创建csv文件的PHP脚本有问题。 PHP脚本如下所示: <?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 a
<?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);
?>
你能帮帮我吗?应该是这样的
fclose($csvFile);
因为您已将追索权链接存储在$csvFile中,而不是$csvFileName中
fclose($csvFile);
因为您在$csvFile中存储了追索权链接,而不是在$csvFileName中。接受参数作为文件指针资源,该资源将在代码执行时找到,例如在您的情况下,它是$csvFile=fopen($csvFileName,“w”)代码>
所以,应该是这样
fclose($csvFile);
而不是fclose($csvFileName) 接受参数作为文件指针资源,该资源将在代码执行时找到,例如,在您的示例中,它是$csvFile=fopen($csvFileName,“w”)代码>
所以,应该是这样
fclose($csvFile);
而不是fclose($csvFileName) 资源是$csvFile
$csvFileName
是一个字符串资源,$csvFile
$csvFileName
是一个字符串