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PHP printf 的有关问题

热度:717   发布时间:2012-02-08 19:52:21.0
PHP printf 的问题
想把number   of   record   的内容print出来,可是在页面上无法显示,无错误提示信息。
      $link   =   new   mysqli( 'localhost ', 'root ', ' ', 'book ');
     
      if(mysqli_connect_errno())
      {
          echo   '   Connection   failed '.mysqli_connect_error();
      }
     
      $query   =   "insert   into   detail   values        
( ' ".$title. " ', ' ".$author. " ', ' ".$isbn. " ', ' ".$price. " ') ";
      $result   =   $link-> query($query);
      $num   =   $link-> affected_rows;
     
      if   ($result)
      {
        //   echo   $num.   'record   inserted   into   database ';
        //就是这一行想显示出来,可是无法在页面上显示//
        printf( "%c   record   stored   in   the   database ",$num);
      }
     
      $link-> close();
}
请教到底怎么回事?谢谢

------解决方案--------------------
c - the argument is treated as an integer, and presented as the character with that ASCII value.
d - the argument is treated as an integer, and presented as a (signed) decimal number.

printf( "%d record stored in the database ",$num);
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