How to filter by paramter

Support
1 
query  queryRecipesContents($slug: String!) {
  queryRecipesContents(filter: "data/slug/iv eq '$slug'") {...}

It responds here

{
  "errors": [
    {
      "message": "Variable '$slug' is never used in operation '$queryRecipesContents'.",
      "locations": [
        {
          "line": 1,
          "column": 29
        }
      ],
      "extensions": {
        "code": "NO_UNUSED_VARIABLES",
        "codes": [
          "NO_UNUSED_VARIABLES"
        ],
        "number": "5.8.4"
      }
    }
  ],
}

I also try to use “data/slug/iv eq \”$slug\“”, but the problem is still here.

How do you use the variable in the GraphiQL?

2

Please use code blocks to format your code. Not quotes.

3

I found a way to filter the data.

query QueryRecipe($filter: String!) {
        queryRecipesContents(filter: $filter) {...}

Variables:
{"filter": "data/slug/iv eq 'baked-lemon-herb-salmon'"}

We can pass the entire eq expression as a parameter.

Inspired xetera.dev/src/scripts/me.ts at 6521238915fee28505b4fe600b700d45bc8fc46e · Xetera/xetera.dev · GitHub

1 Like
4

Great,

I actually wanted to implement a filter language in GraphQL, where you have intellisense, but this would create so much types. Basically you need one field for every schema field and operator, e.g. something like

iv_eq: "11"

or you have complex objects like

iv: { eq: "11" }
5

Yep,
I remember the origin GraphQL support eq key-value filter.
So far, we can use this way as a temporary solution.

6

This is usually built in the application code. Some libraries also support GraphQL filters, but the problem is the amount of additional types, you could easily create 100.000+ per schema and they are expensive to create and you have to keep them in memory.