小编Nit*_*ish的帖子

我有一个奇怪的问题.我的div宽度之一不起作用.我的CSS就像

.product_info
{
    margin-top:10px;
    background:#444;
    width:850px;

}
.product_image
{
    width:350px;
    text-align:center;
    float:left;
    padding:8px;
    border:1px solid #e9e9e9;
    background:#fff;
}
.product_details
{
    float:left;
    width:400px;
    margin-left:25px;
    font-size:14px;
    font-family:"Helvetica";
    background:#d71414;
}

我的HTML文件是

<div class="product_info">
                <div class="product_image">

                </div>

                <div class="product_details">
                    <div class="selection">
                    <form action="{$path_site}{$indeex_file}" method="post">
                    Size
                    {foreach name = feach item = k from = $product_size}
                        <input type="radio" name="size" value="{$k.product_size}" />{$k.product_size}
                    {/foreach}


                    </div>

                    <div class="selection">
                    No. of pieces <input type="text" name="quantity">

                    </div>
                    <div class="prc">
                        <span class="WebRupee">Rs.</span> {$product_info->product_cost}.00
                            </div>

                    <div class="buy_btn"><input type="submit" value="BUY" /></div>
                    </form>
                </div> …

我有一个类似于下面的表结构:

<tr>
    <td colspan="5">
        TEST
    </td>
</tr>
<tr id="abcd_<?php echo  $id; ?>" style="display: none">
    <td colspan="5">
        <span id="hidtb_<?php echo  $id; ?>"></span>
    </td>
</tr>

该表位于循环和$id更改的值内.第二个tr设置为display : block使用javascript.但是 <td colspan="5">并没有覆盖所有五个<td>,而只是一个..为什么我colspan不工作？

我想在我的网站的主容器中创建一个元素容器.为了使元素容器成一行,我申请float:left;了它们.但当我加入float它们时,主容器缩小了!我尝试应用于clear:both主容器,但没有任何变化.

CSS:

#main_container
{
    clear:both;
    margin-top:20px;
    padding:20px 10px 30px 15px;
    background:#ccc;
}
.element_container
{
    float:left;
    width:238px;
    height:300px;
    border:1px solid #000;
}

HTML:

<div id="main_container">

    <div class="element_container"></div>

    <div class="element_container"></div>

    <div class="element_container"></div>

</div>

我是新的,实际上这是我对android程序的第一次尝试.我正在关注android开发者培训网站.我activity_main.xml看起来如下:

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:orientation="horizontal"
android:paddingBottom="@dimen/activity_vertical_margin"
android:paddingLeft="@dimen/activity_horizontal_margin"
android:paddingRight="@dimen/activity_horizontal_margin"
android:paddingTop="@dimen/activity_vertical_margin"

tools:context=".MainActivity" >

<EditText android:id="@+id/edit_message"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content" 
    android:layout_hint="@string/edit_message"/>

现在它显示错误和警告,如:

警告 - This text field does not specify an inputType or a hint

错误 - error:Error:No resource found that matches the given name( at 'layout_hint' with the value '@string/edit_message')

我有一个名为的库文件 myMenu.php

<?php
class myMenu{
    function show_menu()
    {
        $obj = & get_instance();
        $obj->load->helper('url');

        $menu = "<ul>";

        $menu .= "<li>";
        $menu .= anchor('books/index','List of books');
        $menu .= "</li>";

        $menu .= "<li>";
        $menu .= anchor('books/input','Books entry');
        $menu .= "</li>";

        $menu .= "</ul>";

        return $menu;
    }
}

现在我在控制器中加载了这个库 books.php

function index()
    {
        $this->load->library('myMenu');
        $menu = new myMenu;
        $data['menu'] = $menu->show_menu();
        $this->load->view('main_view',$data);
    }

但页面显示错误An error occurred : Unable to load the requested class: mymenu.为什么这个错误显示类名 mymenu(全部为小写),其中我myMenu在控制器中写道

我是cakephp的新手.我需要添加一个默认值

<option value="0">--Please Select--</option>

在我的以下选择字段中:

$attributes = array("empty"=>false,"Selected" => 'Select City',"id" => "location");
echo $form->select("gal_location_id", $gal_locations,null,$attributes); 

我试着补充一下

$gal_locations[0] = "--Select City--";
$attributes = array("empty"=>false,"default" => 0,"id" => "location");

但选项位于列表的底部.添加默认选项的正确方法是什么？

我是drupal主题的新手.我创建了一个direcory mytheme并添加mytheme.info到它并从drupal的默认主题目录中复制了其他文件.编辑后page.tpl.php,drupal显示下面的错误.

Notice: Undefined variable: hide_site_name in include() (line 99 of C:\wamp\www\dtest\sites\all\themes\mytheme\bartik\templates\page.tpl.php).
Notice: Undefined variable: hide_site_name in include() (line 109 of C:\wamp\www\dtest\sites\all\themes\mytheme\bartik\templates\page.tpl.php).
Notice: Undefined index: featured in include() (line 168 of C:\wamp\www\dtest\sites\all\themes\mytheme\bartik\templates\page.tpl.php).
Notice: Undefined index: highlighted in include() (line 187 of C:\wamp\www\dtest\sites\all\themes\mytheme\bartik\templates\page.tpl.php).
Notice: Undefined index: sidebar_second in include() (line 212 of C:\wamp\www\dtest\sites\all\themes\mytheme\bartik\templates\page.tpl.php).
Notice: Undefined index: triptych_first in include() (line 220 of C:\wamp\www\dtest\sites\all\themes\mytheme\bartik\templates\page.tpl.php).
Notice: Undefined index: triptych_middle in include() (line 220 of C:\wamp\www\dtest\sites\all\themes\mytheme\bartik\templates\page.tpl.php).
Notice: Undefined index: …

我试图在codeigniter中将数据插入mysql.控制器类:

class Ci_insert extends CI_Controller
    {
        function __construct()
        {
            parent::__construct();
        }

        function index()
        {
            $data = array(
                          "USN" => "TRE5rCS89G",
                          "name" => "NITISH DOLAKASHARIA",
                          "branch" => "CS"
                          );

            $this->load->model('ci_insert_model');

            $this->ci_insert_model->addToDb($data);
        }

    }

型号类:

class ci_insert_model extends CI_Model
    {
        function __construct()
        {
            parent::__construct();
        }
        function addToDb($data)
        {

            //var_dump($data);
            $this->db->insert('class_record',$data);
        }
    }

但是当我试图运行代码时,它会显示出来Fatal error: Call to a member function insert() on a non-object in C:\wamp\www\CodeIgniter\application\models\ci_insert_model.php on line 12.

上面的代码有什么问题？

我想在CodeIgniter中构建一个简单的表单.我的视图'input_view.php'是:

 <html>
    <head>
        <link rel="stylesheet" type="text/css" href="<?php echo "$base/$css"?>">
    </head>
    <body>
        <div id="header">
            <?php $this->load->view('books_header'); ?>
        </div>

        <div id="menu">
            <?php $this->load->view('books_menu'); ?>
        </div>
        <?php //echo $title; ?>
        <?php echo form_open('books/input'); ?>

            <?php echo $title; ?>
            <?php echo form_input('title'); ?>;

            <?php echo $author; ?>
            <?php echo form_input('author'); ?>

            <?php echo $publisher; ?>
            <?php echo form_input('publisher'); ?>

            <?php echo $year; ?>
            <?php echo form_dropdown('year',$years); ?>

            <?php echo $available; ?>
            <?php echo form_checkbox('available','yes',TRUE); ?>

            <?php echo $summery; ?>
            <?php echo form_textarea('summery'); ?>

            <?php …

我试图在javascript中大写字符串中的字符,我的代码是:

<button onclick="myFunction()">Try it</button>


<script>
function myFunction() {
    var str = "string";
    for(m = 0; m < str.length; m++){
    if(str[m] == "r"){
      str[m+1] = str[m+1].toUpperCase();
    }
    }
    alert(str);
}
</script>

所以我想要做的是,如果角色是r,则大写下一个角色.但是不是woking意味着它是相同的字符串警报.