Question

有人可以帮我解决返回记录导航的问题吗？我有一个mysql查询：

if(isset($_GET['record'])) {
$record_current = $_GET['record'];
}else{
$record_current = 1;
}

$useremail = $_SESSION['useremail'];

$view_sql = "SELECT * FROM missions, calendar, bookings 
WHERE missions.missions_id =  calendar.missions_id 
AND calendar.calendar_id = bookings.calendar_id 
AND bookings.booking_email ='$useremail' LIMIT $record_current,1";

$view_result = mysql_query($view_sql) or die(mysql_error());
$view_count = mysql_num_rows($view_result);
$rsView = mysql_fetch_assoc($view_result);

查询产生所需的结果。这些值用于填充动态表单。我遇到的问题是正确浏览记录。我正在使用此代码进行导航（允许用户查看他们所做的所有预订）。代码：

<span class="label"><?php if($record_current > 1) { ?><a href="view.php?record=<?php echo $record_current - 1; ?>" >Previous</a><?php } ?></span>
<span class="element"><?php if($record_current < $view_result) { ?><a href="view.php?record=<?php echo $record_current + 1; ?>" >Next</a><?php } ?></span>

问题是要显示的第一条记录实际上是查询中的第二条记录，即使显示最后一条记录，也会显示“下一条”链接。如果单击该链接，则会显示一个空白表单。我不知道我做错了什么，我已经玩了几个小时，任何帮助将不胜感激。干杯

修改

我已将代码修改为此（感谢Sean）：

$view_sql = "SELECT SQL_CALC_FOUND_ROWS * FROM missions, calendar, bookings 
WHERE missions.missions_id =  calendar.missions_id 
AND calendar.calendar_id = bookings.calendar_id 
AND bookings.booking_email ='$useremail' LIMIT $record_current,1";

$view_result = mysql_query($view_sql) or die(mysql_error());
$view_count = mysql_query("SELECT FOUND_ROWS() AS cnt");
$rsView = mysql_fetch_assoc($view_result);
$adj_count = (mysql_result($view_count, 0, "cnt") - 1);

显然，“SELECT FOUND_ROWS”返回从零开始的集合。这现在正在运作。感谢所有贡献者。干杯，Spud

Answer 1

注意：在提醒LIMIT $record_current,1计算机逻辑时每个计数从零开始，所以如果你想要获得第一个数据，你必须将您的初始值$record_current设置为0

Answer 2

if()行中的NEXT是错误的，因为您正在针对查询$record_current检查$view_result而不是行数 - $view_count 。尝试更改为 -

if($record_current < $view_count)

修改
在http://dev.mysql.com/doc/refman/5.0/en/information-functions.html#function_found-rows处查看SQL_CALC_FOUND_ROWS / FOUND_ROWS()。

$view_sql = "SELECT SQL_CALC_FOUND_ROWS * FROM missions, calendar, bookings WHERE missions.missions_id = calendar.missions_id AND calendar.calendar_id = bookings.calendar_id AND bookings.booking_email ='$useremail' LIMIT $record_current,1"; $view_result = mysql_query($view_sql) or die(mysql_error()); $view_count = mysql_query("SELECT FOUND_ROWS()"); $rsView = mysql_fetch_assoc($view_result);

Answer 3

if(isset($_GET['record'])) {
$record_current = $_GET['record'];
}else{
$record_current = 1;
}

$view_sql = "SELECT * FROM missions, calendar, bookings 
WHERE missions.missions_id =  calendar.missions_id 
AND calendar.calendar_id = bookings.calendar_id 
AND bookings.booking_email ='$useremail' LIMIT $record_current,1";

问题是要显示的第一条记录实际上是来自查询的第二条记录

请注意，当您使用LIMIT时，第一行的偏移量将始终为0而不是1 ...

无法浏览mysql查询结果

3 个答案: