Question

我可以读取本地主机的csv文件，一旦我转移到在线显示错误警告：fopen（boo1.csv）：无法打开流：第28行/home/ashishco/public_html/group/fetchCsv.php中没有此类文件或目录，

我不知道如何给出路径。

<?php
include("includes/connection.php");
include("includes/mail.php");
if(isset($_POST["Import"]))

        {

    echo $filename=$_FILES["file"]["name"]; 
    //echo $ext=substr($filename,strrpos($filename,"."),(strlen($filename)-strrpos($filename,".")));

    mysql_query("CREATE TABLE IF NOT EXISTS mytable (
      `MId` bigint(20) NOT NULL AUTO_INCREMENT,
      `Batch` varchar(50) NOT NULL,
      `Department` varchar(50) NOT NULL,
      `RegisterNo` varchar(50) NOT NULL,
      `Name` varchar(200) NOT NULL,
      `EmailId` varchar(250) NOT NULL,
      `Password` varchar(50) NOT NULL,
      `MobileNo` int(50) NOT NULL,
      `Status` tinyint(4) NOT NULL DEFAULT '0',
      PRIMARY KEY (`MId`)
    )");
     if($_FILES["file"]["size"] > 0)
     {

    $file = fopen($filename, "r");
             $count=1;
             while (($emapData = fgetcsv($file, 10000, ",")) !== FALSE)
             {
                //print_r($emapData);
                if($count>1)
                {
                    $sql = "INSERT into myTable(Batch,Department,RegisterNo,Name,EmailId,Password,MobileNo) values('$emapData[0]','$emapData[1]','$emapData[2]','$emapData[3]','$emapData[4]','$emapData[5]','$emapData[6]')";
                    //echo $sql; 
                    mysql_query($sql);
                }
                $count++;
             }

         //die();
         fclose($file);
         echo "CSV File has been successfully Imported";

             fnSendMail();

}

    else
    {
    echo "Invalid File:Please Upload CSV File";
    }

}
?>

Answer 1

你应该像这样给出你的道路

$file = fopen("/ your path/".$filename, "r");

假设您的csv文件的路径为/home/group/file.csv，而您的php文件的路径为/home/yourphp.php，那么您将调用

$file = fopen("/group/".$filename, "r");

在你的情况下只是用作

$file = fopen($filename, "r");

上传csv文件时出错

1 个答案: