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| | Добрый день!
Использую Google Ajax Search API по web поиску, вот собственно Пример результата, возвращаемого API Google
foo('bar',{
"results": [
{
"GsearchResultClass": "GwebSearch",
"unescapedUrl": "http://en.wikipedia.org/wiki/Paris_Hilton",
"url": "http://en.wikipedia.org/wiki/Paris_Hilton",
"visibleUrl": "en.wikipedia.org",
"cacheUrl": "http://www.google.com/search?q\u003dcache:TwrPfhd22hYJ:en.wikipedia.org",
"title": "\u003cb\u003eParis Hilton\u003c/b\u003e - Wikipedia, the free encyclopedia",
"titleNoFormatting": "Paris Hilton - Wikipedia, the free encyclopedia",
"content": "In 2006, she released her debut album \u003cb\u003eParis\u003c/b\u003e..."
},
{
"GsearchResultClass": "GwebSearch",
"unescapedUrl": "http://www.imdb.com/name/nm0385296/",
"url": "http://www.imdb.com/name/nm0385296/",
"visibleUrl": "www.imdb.com",
"cacheUrl": "http://www.google.com/search?q\u003dcache:1i34KkqnsooJ:www.imdb.com",
"title": "\u003cb\u003eParis Hilton\u003c/b\u003e",
"titleNoFormatting": "Paris Hilton",
"content": "Self: Zoolander. Socialite \u003cb\u003eParis Hilton\u003c/b\u003e was..."
},
...
],
"cursor": {
"pages": [
{ "start": "0", "label": 1 },
{ "start": "4", "label": 2 },
{ "start": "8", "label": 3 },
{ "start": "12","label": 4 }
],
"estimatedResultCount": "59600000",
"currentPageIndex": 0,
"moreResultsUrl": "http://www.google.com/search?oe\u003dutf8..."
}
}
, 200, null)
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Вот ссылка на документацию:
https://developers.google.com/web-search/docs/reference?hl=ru#_fonje_response
Вопрос, как получить данные с "estimatedResultCount" ? и показать данные этого класса, в <div> | |
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