小编use*_*405的帖子

将多个类组合成一个css规则

我试图在这里做一些桶,一般来说我会一次做一个类元素.这似乎很愚蠢,因为类可以共享属性.

HTML

<div id = "outerBuckets">
        <div class = "bucket1">
            <div class ="bucketIcon">
                <p></p>
            </div>
        </div>
        <div class = "bucket2">
            <div class ="bucketIcon">
                <p></p>
            </div>
        </div>
        <div class = "bucket3">
            <div class ="bucketIcon">
                <p></p>
            </div>
        </div>
        <div class = "bucket4">
            <div class ="bucketIcon">
                <p></p>
            </div>
        </div>
        <div class = "bucket5">
            <div class ="bucketIcon">
                <p></p>
            </div>
        </div>
        <div class = "bucket6">
            <div class ="bucketIcon">
                <p></p>
            </div>
        </div> 
    </div>
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所以我想像这样做我的css规则:

.bucket 1 .bucket 2 . bucket 3 {
}

.bucket 4 .bucket …
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css rules elements

5
推荐指数
1
解决办法
3万
查看次数

找不到对象!在此服务器上找不到请求的URL。本地主机

大家好。我有一个典型的设置A,M,P,我只是在本地服务器上进行一些测试以设置网页。我对php和动态网站有些陌生,所以我有点困惑。所以我在这里陷入僵局。每当我尝试在浏览器中查看页面并单击链接时,都会出现此错误:

找不到对象!在此服务器上找不到请求的URL。推荐页面上的链接似乎有误或已过时。请将该错误告知该页面的作者。如果您认为这是服务器错误,请与网站管理员联系。错误404本地主机Apache / 2.4.3(Win32)OpenSSL / 1.0.1c PHP / 5.4.7

my url looks like this: http://localhost/test/content/home
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这是我的index.php页面和nav.php页面的代码

第一个index.php

<?php
include('Config/setup.php');
?>
<?php
if(isset($_GET['page']) && $_GET['page'] == !'') {
    $pg = $_GET['page'];
} else {
    $pg = 'home';
}
#var_dump($pg);
#exit;
?>

<!DOCTYPE HTML>
<html>
<head>
    <meta content="text/html; charset=utf-8" http-equiv="Content-Type">

    <title>FTS</title>
    <link href="css/Styles.css" rel="stylesheet" type="text/css">
</head>

<body>
<div class="header temp_Block">
    <?php include('templates/header.php');?>
</div>

<div class="main_nav temp_Block">
    <?php include('templates/main_nav.php');?>
</div>

<div class="main_Content temp_Block">
    <?php include('Content/'.$pg.'.php');?>
</div>

<div class="footer temp_Block">
    <?php include('templates/footer.php');?>
</div>
</body> …
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php localhost http-status-code-404

0
推荐指数
1
解决办法
8万
查看次数

Python:当我运行代码时,为什么我的字符串不会返回或打印

def the_flying_circus(Question = raw_input("Do you like the Flying Circus?")):
    if Question == 'yes':
        print "That's great!" 
    elif Question == 'no':
        print "That's too bad!"
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我试图获取if表达式来运行代码并返回基于原始输入的字符串.每次我运行它时,问题会提示但是当我尝试输入"是或否"时它会给我这个错误:

    Traceback (most recent call last):
  File "C:\Users\ftidocreview\Desktop\ex.py", line 1, in <module>
    def the_flying_circus(Question = input("Do you like the Flying Circus?")):
  File "<string>", line 1, in <module>
NameError: name 'yes' is not defined
>>> 
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python input

0
推荐指数
1
解决办法
203
查看次数

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css ×1

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input ×1

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