If there is a way to do this, it should be a fairly straightforward question. Using the twitter search api, is there a way to return results that contain a part of the word (searching "#dog" would return "#dogs" as-well)?
Thanks
No. What you can do is search for "#dog OR #dogs" and that will return results for both.
Related
I want to use the Deezer API to find a good match for a track using its title & artist, so I tried to use the advanced search described in http://developers.deezer.com/api/search#infos
But this api returns a lot of erratic results. For example:
https://api.deezer.com/search?q=artist:%27jeff%20mills%27 //Completely à côté de la plaque
https://api.deezer.com/search?q=jeff%20mills //Nice
It's pretty weird that when using the advanced search the result is less accurate than without it...
So is there a way to receive an accurate result for a track by specifying the artist and the title?
Thanks
Actually, it is not simple quotes that should be used but double quotes, and it works. We will update the documentation about this point, thank you!
I have some itens available from structured-data testing-tool, like:
pmr-recipe-recipetype-bolos
pmr-recipe-totaltime-30
pmr-recipe-name-chocolate
How can i use this values to filter the results from CSE?
Like, show pages with pmr-recipe-recipetype-bolos.
try your search term this way
more:pagemap:recipe-name:chocolate
more:pagemap:recipe-totaltime:30
more:pagemap:recipe:type:bolos
It should return the search result you look for.
Hope this help :)
Many of the Wiktionary pages for Chinese Characters (Hanzi) include links at the top of the page to other similar-looking characters. I'd like to use the Wiktionary API to send a single character in the query and receive a list of similar characters as the response. Unfortunately, I can't seem to find any query that includes the "See Also" field. Is this kind of query possible?
The “see also” field is just a line of wiki code in the page source, and there is no way for the API to know that it's different from any other piece of text on the page.
If you are happy with using only the English version of Wiktionary, you can fetch the wikicode: index.php?title=太&action=raw, and then parse the result for the template also. In this case, the line you are looking for is {{also|大|犬}}.
To check if the template is used on the page at all, query the API for titles=太&prop=templates&tltemplates=Template:also
Similar templates are avilable in more language editions of Wiktionary, in case you want to use other sources than the English one. The current list is:
br:Patrom:gwelet
ca:Plantilla:vegeu
cs:Šablona:Viz
de:Vorlage:Siehe auch
el:Πρότυπο:δείτε
es:Plantilla:desambiguación
eu:Txantiloi:Esanahi desberdina
fi:Malline:katso
fr:Modèle:voir
gl:Modelo:homo
id:Templat:lihat
is:Snið:sjá einnig
it:Template:Vedi
ja:テンプレート:see
no:Mal:se også
oc:Modèl:veire
pl:Szablon:podobne
pt:Predefinição:ver também
ru:Шаблон:Cf
sk:Šablóna:See
sv:Mall:se även
It has been suggested that the WikiData project be expanded to cover Wiktionary. If and when that happens, you might be able to query theWikiData API for that kind of stuff!
Google Analytics API: how to extract pageviews for a specific page?
I tried using something like
ga:pagePath=~page.php%3fid%3d44 (page.php?id=44)
but it doesn't seem to work... I get "no results found" where I have 20 pageviews for sure
UPDATE
I think I found the solution
ga:pagePath==/website/page.php?id=44
for some reason I had to include the complete path and ==
To use a partial path to match for a page in filters you should use
ga:pagePath=#page.php?id=44
=# tells ga to match a substring.
What you were originally using was incorrect for this.
I think your problem is that you put the hex version of the ? and = characters into your query, which doesn't match how Analytics stores the page paths. If you change these to the normal characters it should work:
ga:pagePath=~page.php?id=44
Your other solution should work as well but is a bit more inflexible in case you wanted to tweak the query to return other pages.
I am making some PHP code which takes a given search phrase and url and searches through the google search results until it finds the url (only first 100 results). My problem is, this is only working for the US. I have tried adding the "&cr=" option, but it still on returns US results.
The full URL I am using for the request is:
https://www.googleapis.com/customsearch/v1?key=API_KEY&cx=CX_VALUE&q=KEYWORD&cr=COUNTRY&alt=JSON
Does anyone have any experience with this? I want to be able to see UK results. Tried inserting &cr=countryUK , but still only does US results.
Thanks :)
Regards,
Stian
Use the gl=<country code> param to limit it to your country of choice (so gl=gb for the uk).
More info here:
http://googleajaxsearchapi.blogspot.com/2009/10/web-search-in-your-country.html