Metadata-Version: 1.2
Name: google-search-results
Version: 1.2.0
Summary: this pip package is meant to scrape and parse Google results using SERP API. Feel free to fork this repository to add more backends.
Home-page: https://github.com/serpapi/google-search-results-python
Author: lf2225
Author-email: lf2225@gmail.com
License: UNKNOWN
Description-Content-Type: UNKNOWN
Description: # Google Search Results in Python
        
        This Python package is meant to scrape and parse Google results using [SERP API](https://serpapi.com). Feel free to fork this repository to add more backends.
        
        ## Simple Example
        ```python
        from lib.google_search_results import GoogleSearchResults
        query = GoogleSearchResults({"q": "coffee"})
        html_results = query.get_html()
        ```
        
        ## Set SERP API key
        
        ```python
        GoogleSearchResults.SERP_API_KEY = "Your Private Key"
        ```
        Or
        ```python
        query = GoogleSearchResults({"q": "coffee", "serp_api_key": "Your Private Key"})
        ```
        
        ## Example with all params and all outputs
        
        ```python
        query_params = {
          "q": "query",
          "google_domain": "Google Domain",
          "location": "Location Requested",
          "device": device,
          "hl": "Google UI Language",
          "gl": "Google Country",
          "safe": "Safe Search Flag",
          "num": "Number of Results",
          "start": "Pagination Offset",
          "serp_api_key": "Your SERP API Key"
        }
        
        query = GoogleSearchResults[query_params]
        query.params_dict["location"] = "Portland"
        
        html_results = query.get_html()
        json_results = query.get_json()
        json_results_with_images = query.get_json_with_images()
        ```
        
        ## Example of Python Dictionary Output (GoogleSearchResults#get_dictionary)
        
        dictionary_results = query.get_dictionary()
        dictionary_results_with_images = query.get_dictionary_with_images()
        
Platform: UNKNOWN
Classifier: License :: OSI Approved :: MIT License
Classifier: Programming Language :: Python :: 2.7
Requires-Python: >=2.6, !=3.0.*, !=3.1.*, !=3.2.*
