# Python (Django)

In a Django application, you might want to create a view that makes the API call and returns the result to the client. Here's how you can do it:

First, in your views.py file, add the following code:

```python
from django.http import JsonResponse
import requests
import os
import json

def book_search(request):
    query = request.GET.get('query', '')
    url = "https://api.promptjoy.com/api/mVMCpq"
    headers = {
        'Content-Type': 'application/json',
        'x-api-key': os.getenv('PROMPTJOY_API_KEY')
    }
    data = {
        'query': query
    }
    response = requests.post(url, headers=headers, data=json.dumps(data))

    if response.status_code == 200:
        return JsonResponse(response.json(), safe=False)
    else:
        return JsonResponse({'error': f"Request failed with status {response.status_code}"}, status=response.status_code)

```

In this example, the search term is expected to be passed as a query parameter named `query` in the request to the `book_search` view. The search results are then returned as a JSON response.

Then, in your urls.py file, add a URL pattern for this view:

```python
from django.urls import path
from . import views

urlpatterns = [
    path('book_search/', views.book_search, name='book_search'),
]

```

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This example uses the `requests` library, which is a popular choice for making HTTP requests in Python. If you haven't installed it yet, you can do so with `pip install requests`.

Remember to handle exceptions and errors as needed in your actual application code.

Note: In this example, the API key is retrieved from environment variables for security reasons. Ensure that you've set the `PROMPTJOY_API_KEY` environment variable in your environment where this code will be executed.

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