如何删除“下一个”当没有更多条目可显示时分页脚本中的选项？

Question

我有这个脚本，我已经发布了很多次，但始终无法清楚地解释它是如何工作的。但现在我可以了，所以就在这里。我试图仅在有更多条目要显示时才显示查看“下一个”的选项。

  $result =   mysql_query("SELECT * FROM table WHERE field1 LIKE '%$q%' 
  OR field2 LIKE '%$q%' OR field3 LIKE '%$q%' OR field4 LIKE '%$q%'");


  $num_rows1 = mysql_num_rows($result);

//check if the starting row variable was passed in the URL or not
if (!isset($_GET['pg']) or !is_numeric($_GET['pg'])) {
  //we give the value of the starting row to 0 because nothing was found in URL
  $startrow = 0;
  //otherwise we take the value from the URL
} else {
  $startrow = (int)mysql_real_escape_string($_GET['pg']);
}

$sql = "SELECT field1, field2, field3, field4 FROM table WHERE
user_id='$id' limit $startrow, 10";
$result = mysql_query($sql);

$query = mysql_query($sql) or die ("Error: ".mysql_error());

if ($result == "") {
  echo "";
}

echo "";

$rows = mysql_num_rows($result);

if($rows == 0) {
  print("");
} elseif($rows > 0) {
  while($row = mysql_fetch_array($query)) {
    $field1 = $row['field1'];
    $field2 = $row['field2'];
    $field3 = $row['field3'];
    $field4 = $row['field4'];

    print("");
  }
}

if($num_rows1 > 10) {
  echo '<a id=pg href="'.$_SERVER['PHP_SELF'].'?pg='.($startrow+10).'&q='.($q).'"> Next</a>';
} 

$prev = $startrow - 10;

//only print a "Previous" link if a "Next" was clicked
if ($prev >= 0) {
  echo '<a id=pgnvg2 href="'.$_SERVER['PHP_SELF'].'?pg='.$prev.'&q='.($q).'">
  Previous</a>';
}

Answer 1

分页：

 Total Items
 Items per Page
 Current Page

您只需在以下情况下显示下一个导航：

 Total Items >= Items per Page * (Current Page + 1)

就这么简单。