I have text string, that contains links, for example, like <a th:href="'someLink'">Download</a> .
I need to process that text and replace th:href="'someLink'" with correct links to show text with Download.
The text with links is stored in variable textThatContainsLinks.
My code to show text is <div th:utext="${textThatContainsLinks}">. I also tried to use preprocessing like <div th:utext="${__textThatContainsLinks__}">.
Currently this code shows links not as I expected, but non-preprocessed, ie, output is <a th:href="'someLink'">Download</a> now.
How to pre-process expressions in text, before showing it?
Thank you very much!
Take the context path and directly attach it to the relative path of a pure html5 attribute e.g LINK, <img src="/contextPath/relative/path/image.jpg" width="50" height="50" alt="logo"/>.
Notice how simple the accessibility to the resource is: /contextPath/relativePath, so the most important path there is the relative path. This is similar to Thymeleaf was unable to render <img> tag when sent from database table. I observed that once thymeleaf's namespace th: qualifies a href or src attribute that resides inside a text/String the absolute path is not properly resolved.
Related
I am retrieving text which contains images saved in WYSIWYG editor(Summernote). Is there a way to replace src attribute value in img tags using asset()?
Example:
<img src="images/image.jpg"/>...
To:
<img src="https://.../images.jpg"/>
I want solution which would cover all bases: spaces in image name, different extensions...
Sure, just use the curly brace syntax in your blade to render the asset()
<img src="{{ asset('whatever_you_want') }}"/>
I don't think you can do it in Blade. You could, in your model, add a function that replaces all images to full paths. This could be done through a regex pattern that looks for URLs in tags.
I would, however, make sure the full path to the image is included in the text in the database. This way, you always have access to the right path to the image, and you're not relying on a piece of code to display the right image.
I need to scrape images' source URLs from a directory's linked web pages to columns into a Google Sheet.
I think using IMPORTXML function would be the easiest solution, but I get the #N/A "Imported content is empty." error every time.
I have tried to use this extension as well to define XPath, but still the same error.
The page's source code, where image source URL is:
<div class="centerer" id="rbt-gallery-img-1">
<i class="spinner">
<span></span>
</i>
<img data-lazy="//i.example.com/01.jpg" border="0"/>
</div>
So I want to get "i.example.com/01.jpg" value to B2, followed by further images' URLs to adjacent cells.
The function I used is:
=IMPORTXML(A2,"//img[#class='centerer']/#data-lazy")
I tried using spinner instead of centerer, with the same result.
You can get the string i.example.com/01.jpg with the following XPath-1.0 expression:
substring-after(//div[#class='centerer']/img/#data-lazy,'//')
If you don't need to remove the leading //, you can only use
//div[#class='centerer']/img/#data-lazy
So, in the first case, the Google-Sheets expression could be
=IMPORTXML(A2,"substring-after(//div[#class='centerer']/img/#data-lazy,'//')")
and in the second it could be
=IMPORTXML(A2,"//div[#class='centerer']/img/#data-lazy")
Having difficulty to get the correct XPath to scrape the real URL of any image of my Scoop.it topic. Here is the code excerpt centered on one image. Other images are treated the same way.
<div class="thisistherealimage" >
<img id="Here a specific image ID" width="467" height="412"
class="postDisplayedImage lazy"
src="/resources/img/white.gif"
data-original="https://img.scoop.it/jKj7v6ojzPtACT6EaeztHTl72eJkfbmt4t8yenImKBVvK0kTmF0xjctABnaLJIm9"
alt="Here an alternative text" style="width:467; height: 412;" />
So, in this code sample, I dont want to scrape "/resources/img/white.gif" but the URL following the "data-original" attribute!
I'd like to capture the the data-original attribute, not only to capture it when it contains a URL.
As an XPath beginner, I've tried //div[contains(#class,'thisistherealimage')]/img[contains(#class,'postDisplayedImage')][contains(#class,'lazy')]!
But it's not specific to data-original attribute. Isn't it?
Any advice?
If you want the data-original, you can access like this:
//div[contains(#class,'thisistherealimage')]/img[contains(#class,'postDisplayedImage') and contains(#class,'lazy')]/#data-original
I am creating a crawler which needs to download the logo from every website it crawls.
It is quite hard to detect which image is the logo, however I don't need 100% accuracy, so I am thinking of just looking for <img> tags which fulfil any of the following conditions:
A. The name of the image in the <img> tag has the word "logo" in it, for example:
<img src="logo.gif">
<img src="site-logo.jpg">
<img src="mainlogo.png">
B. The class or id in the <img> tag has the word logo in it, for example:
<img class="logo" src="something.gif">
<img id="main-logo" src="something.gif">
<img class="background logo" src="something.gif">
I've tried following the W3C XPATH documentation, but it is not very user friendly. I've also tried using what are supposed to be wildcards (according to w3schools) but they do not appear to work as expected.
Is it possible to achieve what I want using XPATH? Could you help provide some pointers or example code?
Thank you.
You could use:
/html/body//img[contains(#src, 'logo') or contains(#id, 'logo') or contains(#class, 'logo')]
which will find all img tags that are a descendant of the body tag, where the src, id or class attribute contains the text logo.
How to set image path dynamically for following:
<img src="/Images/Model" + #item.ModelImagePath />
"/Images/Model" this path is fixed, rest of the path comes from [ModelImagePath] column of Model DBTable.
Please guide me how to set the path dynamically in View.
And is there any Html helper tag to display image available in MVC RZOR?
Note: i have similar type of problem as described in
How to use/pass hidden field value in ActionLink
<img src="#Url.Content(String.Format("~/Images/Model/{0}", item.ModelImagePath))"
If you have already saved the entire path into the table, follow it
<img src="#Url.Content(String.Format("{0}",item.ModelImagePath))"/>
or
<img src="#Url.Content(item.ModelImagePath)" style="width:100px;height:100px"/>