小编car*_*ium的帖子

不确定我的问题的标题措词是否正确，如果错误，请更正我。

我正在使用TypeIt来显示对话框。立即使用其上下文示例：

new TypeIt('#myElement', {
  // options
})
.type("This is a sentence.")
.pause(500)
.type(" A second sentence after waiting 500 milliseconds")
.go();

Codepen上的示例预览：https ://codepen.io/carpenumidium/pen/yLBawrZ

我试图显示一个数组中的随机对话，但是我的方法导致从第一个对话开始同时从列表中一次显示每个对话。

我的密码（单击蓝色栏上的对话框出现）

Javascript：

var dialogue = new TypeIt('#dialogues', {
    speed: 50,
    waitUntilVisible: true,
    cursor: false
});

var dialogueText = [
    dialogue.type("First line of a dialogue. ").pause(500).break().type("Second line of dialogue."),
    dialogue.type("First line of second dialogue. ").pause(500).break().type("Second line of dialogue."),
    dialogue.type("First line of third dialogue. ").pause(500).break().type("Second line of dialogue."),
    dialogue.type("First line of fourth dialogue. …

我正在尝试创建一个用户可以更新其个人资料详细信息的表单,但它似乎无法正常工作.

我是服务器端编程的初学者,所以我正在拼凑不同教程的代码.来自http://www.codingcage.com/2015/04/php-login-and-registration-script-with.html

该class.user.php文件,最初只有进行登录,并注册代码.我复制了注册函数并更改了一些内容而不是更新:

public function update($id,$uname,$umob,$uaddr,$uacc,$upass) {
    try {
        $upass = password_hash($upass, PASSWORD_DEFAULT);

        $stmt = $this->conn->prepare(
            "UPDATE users
                SET
                    id       = :id, 
                    name     = :uname, 
                    mobile   = :umob,
                    address  = :uaddr,
                    accNo    = :uacc, 
                    password = :upass
              WHERE id = :id"
        );

        $stmt->bindParam(":id", $id);
        $stmt->bindParam(":upass", $upass);
        $stmt->bindParam(":uacc", $uacc);                                         
        $stmt->bindParam(":uname", $uname);
        $stmt->bindParam(":uaddr", $uaddr); 
        $stmt->bindParam(":umob", $umob);

        $stmt->execute();   

        return $stmt;   
    }
    catch(PDOException $e) {
        echo $e->getMessage();
    }               
}

并在view_account.php中 :(编辑3,整个文件包括@e_i_pi的代码更正):

<?php
ini_set("error_log", "/path/to/error.log");
    require_once("session.php");

    require_once("class.user.php");

    $auth_user = new USER(); …

我有十个兄弟元素,有类.post,标记是这样的:

<div class='book'>
    <div class='post text'>one</div>
    <div class='post photo'>two</div>
    <div class='post quote'>three</div>
    <div class='post video'>four</div>
    <div class='post audio'>five</div>
    <div class='post link'>six</div>
    <div class='post text'>seven</div>
    <div class='post video'>eight</div>
    <div class='post quote'>nine</div>
    <div class='post link'>ten</div>
</div>

我想.post div一次封装两个div(比如类.page),所以标记看起来像:

<div class='book'>
    <div class='page'>
        <div class='post text'>one</div>
        <div class='post photo'>two</div>
    </div>
    <div class='page'>
        <div class='post quote'>three</div>
        <div class='post video'>four</div>  
    </div>
    <div class='page'>  
        <div class='post audio'>five</div>
        <div class='post link'>six</div>
    </div>
    <div class='page'>    
        <div class='post text'>seven</div>
        <div class='post …