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

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 .= "
";

    // FOR ENABLING THE FIRST BUTTON
    if ($first_btn && $cur_page > 1) {
        $msg .= "
First
";
    } else if ($first_btn) {
        $msg .= "
First
";
    }

    // FOR ENABLING THE PREVIOUS BUTTON
    if ($previous_btn && $cur_page > 1) {
        $pre = $cur_page - 1;
        $msg .= "
Previous
";
    } else if ($previous_btn) {
        $msg .= "
Previous
";
    }
    for ($i = $start_loop; $i <= $end_loop; $i++) {

        if ($cur_page == $i)
            $msg .= "
{$i}
";
        else
            $msg .= "
{$i}
";
    }

    // TO ENABLE THE NEXT BUTTON
    if ($next_btn && $cur_page < $no_of_paginations) {
        $nex = $cur_page + 1;
        $msg .= "
Next
";
    } else if ($next_btn) {
        $msg .= "
Next
";
    }

    // TO ENABLE THE END BUTTON
    if ($last_btn && $cur_page < $no_of_paginations) {
        $msg .= "
Last
";
    } else if ($last_btn) {
        $msg .= "
Last
";
    }
    $goto = "";
    $total_string = "Page " . $cur_page . " of $no_of_paginations";
    $msg = $msg . "
" . $goto . $total_string . "";  // 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 .= "
";

    // FOR ENABLING THE FIRST BUTTON
    if ($first_btn && $cur_page > 1) {
        $msg .= "
First
";
    } else if ($first_btn) {
        $msg .= "
First
";
    }

    // FOR ENABLING THE PREVIOUS BUTTON
    if ($previous_btn && $cur_page > 1) {
        $pre = $cur_page - 1;
        $msg .= "
Previous
";
    } else if ($previous_btn) {
        $msg .= "
Previous
";
    }
    for ($i = $start_loop; $i <= $end_loop; $i++) {

        if ($cur_page == $i)
            $msg .= "
{$i}
";
        else
            $msg .= "
{$i}
";
    }

    // TO ENABLE THE NEXT BUTTON
    if ($next_btn && $cur_page < $no_of_paginations) {
        $nex = $cur_page + 1;
        $msg .= "
Next
";
    } else if ($next_btn) {
        $msg .= "
Next
";
    }

    // TO ENABLE THE END BUTTON
    if ($last_btn && $cur_page < $no_of_paginations) {
        $msg .= "
Last
";
    } else if ($last_btn) {
        $msg .= "
Last
";
    }
    $goto = "";
    $total_string = "Page " . $cur_page . " of $no_of_paginations";
    $msg = $msg . "
" . $goto . $total_string . "";  // 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