我可以将多个链接传递给fopen，fread和feof吗？

I have the following function :

    function doParse($parser_object) {
    $links=file("./fullsoccer.TXT", "r");
    $i=0;
    while(!empty($links[$i]))
    {
        set_time_limit(0);
        if (!($fp = fopen($links[$i], "r")));
        {
            //loop through data
            while ($data = fread($fp, 4096)) {
                //parse the fragment
                xml_parse($parser_object, $data, feof($fp));
            }
        }
        $i++;
    }
}

This code save a list of 507 links of XML data from fullsocce.txt into $links then I read the content of each file (link: online links) using fread and pass the $data to my main function which is xml_parse to parse and save the data using SAX parser . my problem is: just that last file of the array $links is passed to the function and parse the data , I want your help to know why it is working just with one file? please It is an emergency case

!($fp = fopen($links[$i], "r")

fopen returns FALSE on error so this check is only true at eof or an invild line. Maybe changing it to

 if (($fp = fopen($links[$i], "r") != FALSE)

I also noticed an extra ; and ) in if (!($fp = fopen($links[$i], "r"))); Maybe I'm just not matching properly.

you are missing fclose, but other than that I think you need to do some debugging, try comment out the "xml_parse" function and add echo $links[$i]."<br /> ";

function doParse($parser_object) {
  $links=file("./fullsoccer.TXT", "r");
  $i=0;
  foreach($links as $link)
  {
      set_time_limit(0);
      echo 'reading '.$link."
";
      $fp = fopen($link, "r");
      if ($fp!==false)
      {
          //loop through data
          while ($data = fread($fp, 4096)) {
              //parse the fragment
              //xml_parse($parser_object, $data, feof($fp));
              echo $data;
          }
          echo "
";
          fclose($fp);
      } else {
          echo 'Cannot Open Link '.$link."
";
      }
      $i++;
  }
}