djax | Dynamic pjax , responds to layout changes

by beezee JavaScript Version: Current License: No License

X-Ray Key Features Code Snippets Community Discussions(1)Vulnerabilities Install Support

kandi X-RAY | djax Summary

djax is a JavaScript library. djax has no bugs, it has no vulnerabilities and it has low support. You can download it from GitHub.

djax is very quick to set up, with a few markup requirements to let it work smoothly.

Support

Quality

Security

License

Reuse

Support

djax has a low active ecosystem.

It has 175 star(s) with 29 fork(s). There are 16 watchers for this library.

It had no major release in the last 6 months.

There are 21 open issues and 13 have been closed. On average issues are closed in 47 days. There are 6 open pull requests and 0 closed requests.

It has a neutral sentiment in the developer community.

The latest version of djax is current.

Quality

djax has 0 bugs and 0 code smells.

Security

djax has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

djax code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

djax does not have a standard license declared.

Check the repository for any license declaration and review the terms closely.

Without a license, all rights are reserved, and you cannot use the library in your applications.

Reuse

djax releases are not available. You will need to build from source code and install.

Installation instructions are not available. Examples and code snippets are available.

djax saves you 237 person hours of effort in developing the same functionality from scratch.

It has 579 lines of code, 0 functions and 11 files.

It has low code complexity. Code complexity directly impacts maintainability of the code.

Top functions reviewed by kandi - BETA

kandi has reviewed djax and discovered the below as its top functions. This is intended to give you an instant insight into djax implemented functionality, and help decide if they suit your requirements.

Similar to G .
if we check a group
wrap XMLHttpRequest
step 2D .
generate the c c header
Check if an object is dirty
test 1 .
Check if there is a negative number
Returns true if element is not
diff between a and b

Get all kandi verified functions for this library.

djax Key Features

No Key Features are available at this moment for djax.

djax Examples and Code Snippets

No Code Snippets are available at this moment for djax.

Community Discussions

Trending Discussions on djax

AppSync, pass array as an input argument with graphql-tag

QUESTION

AppSync, pass array as an input argument with graphql-tag

Asked 2018-Sep-12 at 22:35

I'm using appSync with react-native, I have a mutation defined in my schema like this:

Mutation Definition

And it works fine when I test the mutation in the query section

Test mutation

The problem is that I'm not sure how to represent that with graphql-tag, I cannot figure it out how to pass the argument.

This is the representation I got:

graphql-tag representation

This is how I call the mutation:

Call Mutaiton

And this is the query definition, where vitals is and array

Query definition

But the mutation does not execute, it rolls back. I only can see this:

Mutation execution

Anyone can help me out??

Thanks in advance

...

ANSWER

Answered 2018-Sep-12 at 22:35

You have declared a variable $vitals but are not providing that key in variables. You are setting variables with variables: initValues.vitals where initValues.vitals is the array of inputs that you want. You need to change it to variables: initValues so vitals remains a top level key that GraphQL can then find and replace the $vitals variable with.

Let me know if this fixes your issue.

Source https://stackoverflow.com/questions/52301379

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install djax

You can download it from GitHub.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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