I'm tasked with making some of the images on a website appear in Google Sheets via the -IMPORTXML function. I have a basic knowledge of Xpath, but here I am just trying to show the current image (because it changes often) instead of pulling in the image URL.
On this link: https://www.kissusa.com/nails/best-sellers
I'm using this, returns "N/A": =importxml(A2,"//*[#id='layer-product-list']/div[2]/ol/li1","src")
To get this image in the first row of products: enter image description here
Any suggestions on how this can happen are greatly appreciated.
Related
Good morning,
a customer of ours asked us if it was possible to change the image that Google shows next to his site in Google search results.
After several searches, we tried using different techniques all followed by re-indexing the page in order to instantly see the results.
We tried using structured data (both with ld+json and using microdata) and also of the attributes "og:image" and "og:title" in the "meta" tags, but none of these tests changed the image displayed on the right side next to the site in Google results.
We expected that with one of these methods would have changed the image, but nothing happened
Therefore, we wondered whether it was possible to change that image or whether Google chose the best image based on its search parameters.
Thank you for your valuable help,
Best regards
I'm making an image sitemap for my site I'm trying to extract the image link from the post and I'm using the TinyMCE text editor Need Help pls.
I use this code but the problem is I do this in the controller and I can't do it in the view This code being in the Controller leads to two problems, first, that I will not be able to all the images for one post, but I have to fetch the images of all posts together at once. The second thing or the second problem is that this code extracts the images twice: the link I want and the link inside So I want a solution so that I can extract images for one post
Below is a picture of the code in the controller
I am trying to reproduce a RSS reader like Feedly with Google Sheet and displaying with Glide as an app on my mobile phone.
Everything's fine with IMPORTFEED() function with titles, description, URL.
But it seems this function doesn't allow pictures to be displayed even if they are in the feed (which is not all the time).
So I am looking for a way to extract the main image from a blog post... the one displayed when you hover on a link in a Google Sheet cell.
I would like to get the link of that image displayed in the link preview and put that link in another cell.
Here is an example:
I tried IMAGE()
and also IMPORTXML when there is an image in the RSS feeds (but not all of them do... so I stopped)
Is it possible in Google Sheet to get the main image from the one displayed in the link preview ?
For instance, one of the blog I want to extract the main picture of a blog article would be Creajv (URL : https://creajv.com/ ; Feed : https://creajv.com/feed/)
So the IMPORTFEED() function I did in Google Sheets was :
=IMPORTFEED("https://creajv.com/feed/";"items";FALSE;3)
Which stands for :
=IMPORFEED(...) the function to import feeds from an URL
"items" the way to pull every data there is in the feed (you can use other parameters and can see all the possibilities on the GoogleFormulas documentation)
FALSE because I don't want the headers to be included
and the number 3 because I want only the last 3 results displayed.
And it displays perfectly : author, description, URL, date
But I did a little digging in Google and found that basically IMPORTFEED() cannot get images from feeds, even if it is added by the author of the blog (he has to add a feature to do it).
So I am now trying to see if there is another way which is not IMPORTFEED() to get every time the main image of a blog post.
And I saw Google Sheet is able to pull instantly it when I copy paste the URL of a blog article within a cell for instance for Creajv :
Print screen of the image I get when I click in the cell which contains the post URL
So my thoughts would be that I can pull the author, date, description etc. with IMPORTFEED (which works perfectly every time) and use a formula on the cell with the URL to get in another cell the URL of the picture pulled from the one in the link preview.
Two other possibilities might also be with Google App Script :
creating with the App Script a custom function
or creating a script pulling the image in a cell every time a new row is added via the IMPORTFEED() function.
Functions only, as Apps Script doesn't run on mobile Apps
How about this solution? I checked the website and inspected the image from the thumbnail.
Luckily, the structure is simple:
<div class="article-image">
<img src="https://creajv.com/wp-content/uploads/2020/11/HighresScreenshot00000.png" alt="Concours de Level Design avec Unreal Engine, du 11/11 au 05/12/2020">
</div>
You can get the url with IMPORTXML, and apply IMAGE to it:
=IMAGE(ImportXml("https://creajv.com/2020/11/08/concours-de-level-design-avec-unreal-engine/", "//div[#class='article-image']//img/#src"))
Since you are already retrieving the post url with your previous formula, change the source url by the correspondent cell:
=IMAGE(ImportXml(C1, "//div[#class='article-image']//img/#src"))
For example:
There is a google sheet containing a list of MPN's (manufacturer part numbers). Trying to scrape a site called wikiarms for the UPC Codes when I have the MPN for an item.
I have the correct formula for doing this on another site.
=IMPORTXML("http://gun.deals/search/apachesolr_search/"&B1,"//dd/a[../../dt[contains(text(),'UPC')]]|//dd/span[../../dt[contains(text(),'UPC')]]")
Trying to figure out what the correct xpath to complete this formula. Some videos I have watch said to open the page in Chrome and use inspector to select and copy the xpath to complete the importxml function. I tried this with no luck.
Sample
Visit https://www.wikiarms.com/guns?q=20071
In the table there is a button "available in 6 stores" click that to reveal the list. The UPC should be listed after the MPN.
If I copy the xpath in Chrome this is the result
/html/body/div[1]/div/div/div[2]/div/div/div[2]/div[2]/table/tbody/tr[2]/td[5]
=IMPORTXML("https://www.wikiarms.com/guns?q="&B2,"xpath here")
What do I have to add at the end of this formula to pull in the UPC code? I will be using this formula to pull in UPC code for about 1000 items.
Thank you for your help.
Using your sample link, try
=IMPORTXML("https://www.wikiarms.com/guns?q=20071","//td[#class='upc']/a/#title")
and see if it works for you.
I have a website made to provide free web-based tools for making indie games. Currently, it only supports artists contributing to games. The features for helping artists consist of a set of artist community tools that allow artists to upload images based on a description, then we post that image in a gallery page. Other artists can upload their images and each image can have several revisions.
The way I chose to implement the image upload and display feature is by serializing uploaded images to a byte array and storing it in the database. When I need to display the image in the UI I just call a controller action I named "GetScaledGalleryImage" and pass in the image ID. That controller action takes the binary from the database and converts it back into an image, returning the requested image back.
This works very well functionally, but the problem I realized later is that the google crawler thinks all of my images are named "GetScaledGalleryImage" so if someone searches for "sylph" on google images, nothing comes up from my site, but if someone searches for site:watermintstudios.com getscaledgalleryimage, all of my images come up.
Here is an example of the URL that is being output in my HTML http://watermintstudios.com/EarnAMint/GetScaledMedia/68?scale=128
In the past, pre-MVC I would handle 404 errors and return content based on what was requested even if the page didn't actually exist. This would of course allow me to have the images pulled back by the image name (or description).
Is that the best way to do this? Or is there a better option? Something simpler would be better like if I could just do http://watermintstudios.com/EarnAMint/GetScaledMedia/Iris%20Doll?id=68&scale=128, but based on how google indexes images, would that give me what I need? Or do I need to provide image file extensions for maximum indexability?
Thanks all
It is important when doing Search Engine Optimization to always use alt="this is a crazy robot" for your images. This will help the crawler identify them. Note: always use alt, don't always name your images this is a crazy robot.