使用fgetcsv时PHP返回错误

Question

可能重复:
mysql_fetch_array()期望参数1是资源,在select中给出boolean

嗨,我正在尝试使用fgetcsv处理csv,但我所有我要去的是一个无限循环的错误

警告:第5行的testcsv.php中的fopen("Tile","User"...)

警告:fgetcsv()期望参数1是资源,在第6行的/home/ratena/public_html/proc_files/testcsv.php中给出布尔值

<?php
$uploadcsv = "/home/ratena/public_html/temp/files/BatchLoadPM15.csv";
$filecsv = file_get_contents($uploadcsv);
         //$batchid = $_POST['batchid'];
         $handle = fopen("$filecsv", "r");
         while (($data = fgetcsv($handle, 100000, ",")) !== FALSE) {
           print_r($data);   
        }
?>

Answer 1

这部分有问题

/* this is un-necessary */
$filecsv = file_get_contents($uploadcsv);
/* this expecting a file in */
$handle = fopen("$filecsv", "r");

尝试用这个替换3行

$handle = fopen($uploadcsv, 'r');

跳过第一行

$column_headers = array();
$row_count = 0;
while (($data = fgetcsv($handle, 100000, ",")) !== FALSE) 
{
  if ($row_count==0)
  {
    $column_headers = $data;
  }
  else
  {
    print_r($data);
  }
  ++$row_count;
}