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Closed 9 years ago.
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I have this template:
<ul id="warningMessages" style="text-align: left; padding-left: 4px;">
{#foreach $T as msg}
<li>{$T.msg.Value}</li>
{#/for}
</ul>
When applied look like this:
<ul>
<li>Value 1</li>
<li>Value 2</li>
</ul>
The style and id information ("id="warningMessages" style="text-align: left; padding-left: 4px;") disappears and I don't know why.
Can't reproduce your problem. Check this sample page. Works as expected
http://jsbin.com/icuce
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Closed 1 year ago.
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I'm starting a new project with Laravel 8 and I realised that there is two ways to extends a layout from a view, the first is with the "new" way, using:
New Parent Layout Design
{{-- any other *.blade.php file that wants to inherit base-view.blade.php view --}}
<x-base-view>
{{-- $slot is this everything block --}}
</x-base-view>
While the base-view is the base-view.blade.php blade file that has something like:
{{-- views/base-view.blade.php --}}
<html>
<body>
<div class="content">
{{ $slot }}
</div>
</body>
</html>
And we have this other way:
"Old" Parent Layout Design
The old way consists in using #extends(), #yield(), #section(), to extends, render a section and define a section respectively
My Question
my question is: which is the best option? I mean, what is the advantages of using the first and what is the advantages of using the second?
And more importantly, what is the pitfalls of using the first or the second way?
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I have written the following route
Route::get('index/sign', 'Auth\LoginController#showLoginForm')->name('login');
Also When I want to click the sign-in a-tag button
here a tag code.
<a class="nav-link" href="{{ route('index.sign') }}">Sign-In</a>
Then I want to open the page sign-in page according to URL http://127.0.0.1:8000/index/sign
You route name is 'login'
You may call {{ route('login') }}
From my understanding you want the A tag to redirect to your route.
You can do it like this:
<a class="nav-link" href="{{ route('login') }}">Sign-In</a>
In the A tag you want to use route('login') because that is the name you gave the route
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Closed 6 years ago.
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How to develop responsive layouts in angular 2 material using flex layouts.
In angular material we use the code below for responsive layouts on desktop & mobile view.
ex.
<div layout="row" layout-xs="column" layout-wrap="" layout-align="center center">
<div flex="33" flex-xs="100">
</div>
<div flex="33" flex-xs="100">
</div>
<div flex="33" flex-xs="100">
</div>
</div>
I am using it at this times. It is really comfortable.
To install it: Click
More info is here: Click
To using it in css with examples: Click
And an online example: Click
You just need to determine where you use flex-container and flex-item class. And you need to set fxLayout and fxLayoutAlign directives true. It handles rest of work.
I use it with #angular 2.4.3
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This is my sass structure please explain what does its mean?
actually am analysing a code structure. am unable to understad the structure.
.class-name{
some styles..
&.class-name2{
some styles..
}
}
The & tells the pre-processor to add the parent selector in-front of the nested class.
so this:
.class-name{
background: blue
&.class-name2{
background: red;
}
}
would compile too:
.class-name {background: blue;}
.class-name.class-name2 {background: red;}
The usage would be:
<div class="class-name">This background is blue</div>
and
<div class="class-name class-name2">This background is red</div>
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Need some help on getting the content (text) of the anchor that hold the email address while performing a scrape. I've tried //*[#id="panel"]/p[2]/span[2]/a but my problem is that not always the span appears as the 2nd child. I presume if I could target the element with the href mailto would work but unable to do it.
<div id="panel">
<p>Elements here</p>
<p>
<span class="label">Email:</span>
<span class="content">
<a class="link" href="mailto:someone#website.com">
someone#website.com
</a>
</span>
</p>
<p></p>
</div>
Use contains()
//div[#id="panel"]//a[contains(#href,"mailto:")]