Firstly, I'm a long time user of stackoverflow for existing answers, first time asker, so go easy if this is too basic or has been answered elsewhere.
I'm trying to use Drupal 7's RESTful web services module to serve content to an Ionic app. I have the content flowing through fine (e.g. title and body text) but am struggling to access the images. The relevant part of the JSON response is:
"field_cover_image": {
"file": {
"uri": "http://example.com/drupal/file/9",
"id": "9",
"resource": "file"
}
},
How can I translate that information for use inside an image tag, from what I can see I am only being given the ID of the image at this point where what I need is the url/filename.
Thanks
You can create path to original image file (the one user uploaded) from uri with function file_create_url() :
https://api.drupal.org/api/drupal/includes%21file.inc/function/file_create_url/7.x
And you can create path to image file in specific image style with function image_style_url(). Second parameter of this function is actually uri, not path.
https://api.drupal.org/api/drupal/modules%21image%21image.module/function/image_style_url/7.x
I didn't use Restful Web Service, so I can't tell you how to do that exactly, but you must somehow pass of this uri to one of those 2 functions to get url to image file.
Related
I am currently working on a 1 page HTML app that uses URL parameters to do an API call. the URL parameters are set and used in QR codes so its necessary that they are able to change dynamically. A example URL would be something like app.com/index.html/?environment=demo&location=Kiosk
I currently have this app deployed in AWS Amplify, but I cant get other keywords to chain together. I have the following redirects in place:
These redirects make sure that every URL parameter I pass in the link works EXCEPT some keywords like the "location" keyword, next to some others. Using this keyword as a URL param gives a 502 server error, or if the redirects are not used an access denied error.
does anyone know how to get the location keyword to work? Thanks in advance!
You should be able to use a single rule that will forward everything to index.html EXCEPT urls with a "file extension" from the list below. That lets all your links work, but assets like images, fonts, code will pass through.
Doc for: Using Redirects - Single Page Apps
</^[^.]+$|\.(?!(css|gif|ico|jpg|js|png|txt|svg|woff|woff2|ttf|map|json)$)([^.]+$)/>
Problem:
I need to be able to remove all link decoration from the download URL that is generated for images in Firebase Storage.
However, when all link decoration is stripped away, the resulting link currently would return a JSON document of the image's metadata.
Context:
The flow goes as follows:
An image is uploaded to Firebase from an iOS app. Once that is done the download URL is then sent in a POST request to an external server.
The server that the URL is being sent to doesn't accept link decoration when submitting image URLs.
Goal:
Alter the Firebase Storage download URL such as it is stripped of all link decoration like so:
https://firebasestorage.googleapis.com/v0/b/example.appspot.com/o/[FOLDER_NAME]%[IMAGE_NAME].jpg
Notes:
The problem is twofold really, first the link needs to be manipulated to remove all the link decoration. Then the behavior of the link needs to changed, since in order to return an image, you need ?alt=media following the file extension, in this case .jpg. Currently, without link decoration, using the link with my desired structure would return a JSON document of the metadata.
The current link structure is as follows:
https://firebasestorage.googleapis.com/v0/b/example.appspot.com/o/[FOLDER_NAME]%[IMAGE_NAME].jpg?alt=media&token=[TOKEN]
Desired link structure:
https://firebasestorage.googleapis.com/v0/b/example.appspot.com/o/[FOLDER_NAME]%[IMAGE_NAME].jpg
The token is necessary for accessing the image depending security rules in place, but can be ignored with the proper read permissions. I can adjust the rules as needed, but I still need to be able to remove the ?alt=media and still return an image.
Building up on Frank's answer, if you access to your associated Google Cloud Platform project, find the bucket in the Storage tab and make this bucket public, you will be able to get the image from here with the format you wish. That is, you will not be accessing through Firebase
https://firebasestorage.googleapis.com/v0/b/example.appspot.com/o/[FOLDER_NAME]%[IMAGE_NAME].jpg
but through Google Cloud Storage, with a link like
https://storage.googleapis.com/[bucket_name]/[path_to_image]
Once in your GCP project Console, access the Storage bucket with the same name as the one you have in your Firebase project. They are the same bucket. Then make the bucket public by following these steps. After that, you will be able to construct your links as mentioned above and they will be accessible with no token and no alt=media param. If you do not want to make the public to everyone, you will be able to play around with the permissions there as you wish.
You could split the url string into two halves by using String.componentsSeparatedByString(_ separator:)
Storage.storage().reference().child(filePath).downloadURL(completion: { (url, error) in
let urlString = url.absoluteString
let urlStringWithoutQueryString = urlString.componentsSeparatedByString("?").first!
})
Calling .downloadURL on a StorageReference will return you that URL, but this method can be used to remove the query string from any URL. String.componentsSeparatedByString(_ separator:) breaks a String into an array of Strings, splitting the string by any occurrence of a given separator, in this case ?.
NOTE this method assumes that ? occurs only once within the url string, which I believe is the case for all Firebase Storage urls.
You should treat the download URL that you get back from Firebase as an opaque string. There's no way to strip the parameters from a download URL without breaking that download URL.
If you want to allow public access to the files in your bucket with simpler URLs, consider making the object in your (or even your entire) bucket public.
I've been trying to get file uploads to work, following the instructions for both Dropbox and S3 but each time I just get this message:
File Upload URL not provided
It doesn't seem to be making any calls to the server. I've found this mention of a bug around file uploads:
https://github.com/formio/ngFormio/issues/322
But I suspect that applies if you're hosting it yourself. I'm using the cloud version.
I've configured it with e.g. the S3 bucket's URL, authentication etc.
What does this error actually mean?
Update: here's the syntax I'm using:
<formio form="https://formview.io/#/xxxxxxxxxxxxxxxxxxx/applicationform" url="'https://formview.io/#/xxxxxxxxxxxxxxxxxxx/applicationform'"></formio>
Thanks
In order to make the uploads work, you need to provide the URL of your form, which is used to generate the upload token to upload the files to the 3rd party providers. This can be done in one of two ways.
<formio src="'https://examples.form.io/example'"></formio>
You would use above if you wish to render the form from the JSON REST API of the form. In many cases, you may wish to provide the actual form object (which I suspect is what you are doing) like so.
<formio form="{...}"></formio>
This works fine for rendering the form, but it does not provide the URL context for file uploads. For this reason, we have the url parameter which you can include along with your form object for file uploads to work.
<formio form="{...}" url="'https://examples.form.io/example'"></formio>
Providing the url this way is passive. The form will not try to submit to that url, but rather just use it as the url configuration for file uploads.
In golang, is there a way to pipe a variable to part of a web form?
For example, sending "123 Random St." to the Street address part of https://www.dominos.com/en/pages/order/#/locations/search/ and so on? I found pizza_party*, but the GUI used is no longer available, I have also found pizzadash**, but this uses a credit card where I want to use cash. I even found a list of golang ones, but the links that they use doesn't work anymore.***
Therefore, my goal is so: order a pizza in golang through the dominos website API!
NOTE: Please suggest a package or function with example!
NOTE: I do not want to make a web scraper/data getter.
NOTE: Your answer must work on at least one box of my linked website.
NOTE: I want to fill out links similar to the provided link from the linux command line.
*https://github.com/coryarcangel/Pizza-Party-0.1.b
**https://github.com/bhberson/pizzadash
***https://golanglibs.com/top?q=pizza
This is how you post any form values onto an online form. Provided you know the POST endpoint of the service.
func main():
resp, err := http.PostForm(targetPostUrlHere,
url.Values{"Service_Type": {"Delivery"},
"Address_Type_Select": {"House"},
"Street": {"123 E 24th St"},
"Address_Line_2": {"4D"},
"City": {"New York"},
"Region": {"NY"},
"Postal_Code": {"10027"}})
}
**Note: The field keys and values are guesstimates. You must inspect the actual key names expected in the form.
In your case, https://www.dominos.com/en/pages/order/ is an endpoint for the form page. Once the form is filled and submitted, the information is submitted using POST method akin to the code afore-mentioned to a dedicated CREATE endpoint (C in the CRUD), which normally can be found in the <form> html tag.
<form action="posttargetendpoint" method="POST">...</form>
Once the POST operation is successful, usually a web service would redirect you to another page. In your case, it is https://www.dominos.com/en/pages/order/#/section/Food/category/AllEntrees/
However, any good web service wouldn't expose the POST endpoint in the clear since it is the vulnerable point of attack. You're welcome to find out by inspect he Domino's page source and adjust the field values in the Go code accordingly.
Now to make a command line prompt to wrap around the PostForm code, I suggest you look into https://github.com/codegangsta/cli which is a very nice package for creating quick command line app.
I assume you mean pipe information originating from your backend to another site on behalf of a user?
The standard way of passing information between domains is via HTTP params, usually via a GET request, but this capability would need to be supported by established protocols the remote site. You can also use an iframe to embed the page of another site onto your page, however, you wouldn't be able to remotely interact, call JS code, or even query the page at all. Cross-domain security safeguards justifiably prohibit such capability, and generally speaking, interacting on behalf of the user via their browser is also restricted for security reasons.
However, if you're looking to emulate user behavior such as with a bot or web scraper from your own host or browser then that's a different story. There are tons of frameworks provide rich capability for interacting with a page. I'd recommend checking out Selenium, which acts as a virtual browser. There are also tons of libraries in Python for processing data from HTML and structured data. You might want to check out Beatiful Soup and Scrapy.
Hope this helps.
I'm trying to integrate my rails app with SugarCRM. Is it possible to fetch the Contact picture from SugarCRM using REST API? Please let me know.
To get the profile image for a user do the following:
Call the login method through REST
Call the get_entry_list method through REST, with the following parameters:
Module: Users
Query: users.user_name = 'xxxx'
Select_fields: picture
The response contains the filename for the profile image, which is stored in /uploads.
However, it is not possible to view the image in that folder due to .htaccess restrictions for security reasons, but other options exist:
Extend the REST API with a method to serve profile images (similar to get_document_revision)
Login on the server from your rails app and get the image
Create a simple entrypoint+module in SugarCRM, which can show the picture
Remove the .htaccess restiction for images (if it doesn't create a security risk in your setup)
In such scenario, I faced a problem where the upload folder stores file with name of the record id, i.e the GUID without file extension.
So to cop up with this I did write a function to copy the file at same hierarchy but with its extension.
Example:
A png extension file at upload folder with name say, '32sdft-tg35f-Tuhis-675rtyf-77666-46dgc' will end up as, '32sdft-tg35f-Tuhis-675rtyf-77666-46dgc.png'
Now only the path will be require to render the image.
Rest all things as applicable as suggested by our friend, Kare !!