Rendering HTML from database in vuejs component in laravel - laravel

I'm trying to fetch data from my database and display it as HTML content but it only turns out as a string, I'm using vuejs component in laravel. This is the code
<div id="main-window" v-for="post in posts">
<div class="post">
<div class="content">
{{post.content}}
</div>
</div>
How can I use the post.content as HTML?

You need to use v-html binding. Docs
<div id="main-window" v-for="post in posts">
<div class="post">
<div class="content" v-html="post.content">
</div>
</div>

Related

How can I display vue.js data in blade

I am doing a small project with vue.js and laravel. I have a list of products, when I click on a specific product on "Quick view" button I want to show all data about that product in a modal but the data is not showing in the modal.
Can you help me please. Here is my code:
#extends('app')
#section('content')
<div id ="crud" class="row">
<div class="col-xs-12">
<h1 class="page-header">Lista de Articulo</h1>
</div>
<div class="wrap-icon right-section col-md-12" data-toggle="modal" data-target="#list" >
<div class="wrap-icon-section wishlist">
<a href="#" class="link-direction">
<i class="fa fa-heart" aria-hidden="true"></i>
<div class="left-info">
<span class="index">0 item</span>
<br>
<span class="title">Wishlist</span>
</div>
</a>
</div>
</div>
<div class="row">
<div v-for="keep in keeps" class="col-md-4" style="width:25%;" #mouseover="showByIndex = keep" #mouseout="showByIndex = null">
<div class="container">
<img :src="keep.foto" style="width:100%; max-width:150px;">
<button class="btn" v-show="showByIndex === keep" v-on:click.prevent="showKeep(keep)">Quick view</button>
</div>
<h3>
<b> #{{keep.nombre}}</b>
</h3>
<p> #{{keep.descripcion}}</p>
I fixed it. The error was because I called the modal file outside the div. Thanks to everybody.
you can't send data directly from vue in blade structure, you need to write a component for that.
https://v3.vuejs.org/examples/modal.html#modal-component

Can't seem to get Col-4 Working on Laravel

So I am trying to create a car sales website using Laravel. The code presented shows 20 Cars from the database being shown but all one after the other. I want it to appear in columns of 3 or 4 if possible. I've been trying to use col-4 but it doesn't make any difference. Any help appreciated. The only answer I could find online was that the container was within the row but I don't have this issue so unsure.
<div class="container text-center">
<h2>Cars</h2>
<div class="row">
#foreach ($allcars as $car)
<div class="col-4">
<div class="card">
<img class="card-img-top" src="{{asset('qashqai.jpg')}}" alt="Card
image cap">
<div class="card-body">
<h4 class="card-title">{{$car->name}}</h4>
<p class="card-text">{{$car->description}}</p>
</div>
<div class="card-body">
Card link
Another link
</div>
</div>
</div>
#endforeach
</div>
</div>

How to get last child while parsing HTML with Goutte in Laravel

<div class="table">
<div class="table-head">
<div class="table-head-title">Ranking Equipos</div>
</div>
<div class="table-body">
<div class="table-body-row active">
<div class="col-key">Mark</div>
<div class="col-value">9233</div>
</div>
<div class="table-body-row">
<div class="col-key">Amanda</div>
<div class="col-value">7216</div>
</div>
<div class="table-body-row">
<div class="col-key">Mark</div>
<div class="col-value">6825</div>
</div>
<div class="table-body-row">
<div class="col-key">Paul</div>
<div class="col-value">6184</div>
</div>
<div class="table-body-row">
<div class="col-key">Amanda</div>
<div class="col-value">5866</div>
</div>
</div>
</div>
This is my HTML and I want to get last child of .table-body.
I tried to use JavaScript like logic and used indexing like this
$lastChild = $node->filter('.table-body .table-body-row')[4]; but it shows error. Cannot use object of type "Symfony\Component\DomCrawler\Crawler" as array
I was stuck in similar situation recently and I resolve this by using last() method. Syntax is here: $node->filter('.table-body .table-body-row')->last();

Laravel 5.0 paginate jump to page section

I'm using Laravel 5.0 and paginate along with the render() function in my view. Rather than go back to the top of the screen after I select the next page I want it to jump to #msgs.
Here is some of my code:
<h1 class="centered" id="msgs">Announcements</h1>
#foreach($messages->getCollection()->all() as $message)
<div class="row">
<div class="row">
<div class="col-sm-9">
<h3>{{$message->title}}</h3>
</div>
<div class="col-sm-3">
<h4>From: {{$message->author_name}}</h4>
</div>
</div>
<div class="row">
<p>{{$message->text}}</p>
</div>
</div>
<hr class="featurette-divider">
#endforeach
<div class="centered">
{!! $messages->render() !!}
</div>
I'm still getting a grasp on the flow of things in laravel so I'm not really sure what to do. I tried to use append->('#msgs') but realized that its appending an array in the parameters, not appending a path to the url. I there a simple way to achieve this without overriding anything?
You might want to use fragment() method:
{!! $messages->fragment('msgs')->render() !!}
Here is the link to the documentation: Pagination - Laravel

How to add images in codeigniter using Bootstrap

I have started designing a website using codeigniter tutorial from tuts plus and encountered problem on how to add carousel images using Bootstrap in codeigniter. My images are found at /images/...
How can I solve this problem?
<div class="intro-block">
<div class="container">
<div class="row">
<div class="span3">
<div class=" t">
</div>
</div>
<div class="span9">
<div id="artcoursel" class="carousel slide">
<div class="carousel-inner">
<img src= "images/test-1.png" alt="">
</div>
</div>
</div>
</div>
</div>
</div>
Try this.
<img src=<?php echo base_url('images/test-1.png')?>>

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