javascript - PHP和Ajax做的分页跳转,可是出不来。望大神能看看

时间:2016-12-05 13:44:11 来源:互联网 作者: 神秘的大神 字体:

想做一个ajax局部刷新分页跳转,可是不知道为什么用ajax接受的数据在页面
上显示出来总是这么一句话MySql ErrorNo database selected 然后我做了几个测试
发现echo"111111";echo"22222";echo"3333"echo"4444";只有4444输出!PHP我不是很懂
所以想麻烦大神们看下是怎么回事!小弟在此谢过了!



  

  " . $row['id'] . " " . $htmlmsg . "";
    }
    $msg = "
  
    " . $msg . "
"; // Content for Data **echo "3333333333333333333333";** /* --------------------------------------------- */ $query_pag_num = "SELECT COUNT(*) AS count FROM content"; $result_pag_num = mysql_query($query_pag_num); $row = mysql_fetch_array($result_pag_num); $count = $row['count']; $no_of_paginations = ceil($count / $per_page); **echo "2222222222";** /* ---------------Calculating the starting and endign values for the loop----------------------------------- */ if ($cur_page >= 7) { $start_loop = $cur_page - 3; if ($no_of_paginations > $cur_page + 3) $end_loop = $cur_page + 3; else if ($cur_page <= $no_of_paginations && $cur_page > $no_of_paginations - 6) { $start_loop = $no_of_paginations - 6; $end_loop = $no_of_paginations; } else { $end_loop = $no_of_paginations; } } else { $start_loop = 1; if ($no_of_paginations > 7) $end_loop = 7; else $end_loop = $no_of_paginations; } /* ----------------------------------------------------------------------------------------------------------- */ $msg .= " "; // Content for pagination **echo $msg;** **echo "1111111111111";** } ?>

以上是代码

回复内容:

想做一个ajax局部刷新分页跳转,可是不知道为什么用ajax接受的数据在页面
上显示出来总是这么一句话MySql ErrorNo database selected 然后我做了几个测试
发现echo"111111";echo"22222";echo"3333"echo"4444";只有4444输出!PHP我不是很懂
所以想麻烦大神们看下是怎么回事!小弟在此谢过了!



  

  " . $row['id'] . " " . $htmlmsg . "";
    }
    $msg = "
  
    " . $msg . "
"; // Content for Data **echo "3333333333333333333333";** /* --------------------------------------------- */ $query_pag_num = "SELECT COUNT(*) AS count FROM content"; $result_pag_num = mysql_query($query_pag_num); $row = mysql_fetch_array($result_pag_num); $count = $row['count']; $no_of_paginations = ceil($count / $per_page); **echo "2222222222";** /* ---------------Calculating the starting and endign values for the loop----------------------------------- */ if ($cur_page >= 7) { $start_loop = $cur_page - 3; if ($no_of_paginations > $cur_page + 3) $end_loop = $cur_page + 3; else if ($cur_page <= $no_of_paginations && $cur_page > $no_of_paginations - 6) { $start_loop = $no_of_paginations - 6; $end_loop = $no_of_paginations; } else { $end_loop = $no_of_paginations; } } else { $start_loop = 1; if ($no_of_paginations > 7) $end_loop = 7; else $end_loop = $no_of_paginations; } /* ----------------------------------------------------------------------------------------------------------- */ $msg .= " "; // Content for pagination **echo $msg;** **echo "1111111111111";** } ?>

以上是代码

看一下文档啊 少年 点击链接
既然你是用的mysqli,三楼说的没错,选择一下数据库,mysqli是mysqli_select_db()这个,链接里有,你看下。

mysql 和mysqli混用!
上面用的$conn=mysqli_connect("localhost","root","","info");
下面用mysql_query($query_pag_data)

要不改成

$conn=mysql_connect("localhost","root","","info");
mysql_query($conn,"SET NAMES 'utf8'");

这个报错是因为没有选择数据库呀,你在连接之后执行一下mysqli_query('use database')database是你的数据库,另外像楼上说的,别mysqli和mysql混杂用,php官方已经放弃mysql扩展了,用pdo或者mysqli