donglu9445 2013-05-25 20:42
浏览 40

PHP fopen()工作不正确

I try to open a file but output of file is not correct,because output is half-part of data in a file. I don't know solutions to resolve this problem

  <?php
    $handle = fopen("data/weather.arff", "r");
            while ($buffer = fgetc($handle)) {
                $result .= $buffer;
            }
    echo $result;
  ?>
  • 写回答

2条回答 默认 最新

  • dongxinche1264 2013-05-25 21:02
    关注

    I don't know Why I couldn't comment.(maybe lack of reputation). Have you used the "b" binary mode?!
    Like this:

    <?php
    $handle = fopen("http://www.example.com/", "rb");
    $contents = '';
    while (!feof($handle)) {
      $contents .= fread($handle, 8192);
    }
    fclose($handle);
    ?>
    

    Source function.fread

    评论

报告相同问题?

悬赏问题

  • ¥15 c语言怎么用printf(“\b \b”)与getch()实现黑框里写入与删除?
  • ¥20 怎么用dlib库的算法识别小麦病虫害
  • ¥15 华为ensp模拟器中S5700交换机在配置过程中老是反复重启
  • ¥15 java写代码遇到问题,求帮助
  • ¥15 uniapp uview http 如何实现统一的请求异常信息提示?
  • ¥15 有了解d3和topogram.js库的吗?有偿请教
  • ¥100 任意维数的K均值聚类
  • ¥15 stamps做sbas-insar,时序沉降图怎么画
  • ¥15 买了个传感器,根据商家发的代码和步骤使用但是代码报错了不会改,有没有人可以看看
  • ¥15 关于#Java#的问题,如何解决?