How to do web-scraping of contents in TripAdvisor using YQL -


i quite new web scraping question might little simple bothers me lot. want scrap contents tripadvisor, when run following command in yql, returns nothing.

select * html url="http://www.tripadvisor.com/search?q=sunny+relax&geo=191#&ssrc=a&o=0.html"

can tells me why? there wrong commands?

thank in advance kind help.

it because "/search" page disallowed in http://www.tripadvisor.com/robots.txt , yql checks in robots.txt.

you can try page , use xpath select nodes, e.g.:

select * html xpath = '//div[@class="listing_title"]/a' , url = 'http://www.tripadvisor.com/hotels-g45963-las_vegas_nevada-hotels.html'

Comments

Post a Comment

Popular posts from this blog

routing - AngularJS State management ->load multiple states in one page -

Image
i developing site in angularjs ui-router. have situation on button click need change , load 3 state page , 3 looks image .please me how can load multiple pages. in 1 page you need configure router config this: $stateprovider .state("state1", { url: "/url1", controller: "firstcontroller", views: { view1: { templateurl: "templates/template1.html" }, view2: { templateurl: "templates/template2.html" }, view3: { templateurl: "templates/template3.html" } } }).state("state2", { url: "/url2", controller: "secondcontroller", views: { view1: { templateurl: "templates/template4.html" }, view2: { templateurl: "templates/template5.html" }, view3: { templateurl: "templates/template6.html" } }).state("state3", { ...
Read more

python - GRASS parser() error -

i'm trying use grass on python 2.7, i'm having problems @ setting script on idle, i'm getting error @ parser() function: here script: import os import sys gisbase = os.environ['gisbase'] = 'c:\program files (x86)\grass gis 7.0.1rc1' gisrc = 'c:\grassdata' gisdbase = 'c:\grassdata' location = 'newlocation' mapset = 'tc' ld_library_path = 'c:\program files (x86)\grass gis 7.0.1rc1\lib' path = 'c:\program files (x86)\grass gis 7.0.1rc1\etc';'c:\program files (x86)\grass gis 7.0.1rc1\etc\python';'c:\program files (x86)\grass gis 7.0.1rc1\lib';'c:\program files (x86)\grass gis 7.0.1rc1\bin';'c:\python27';'c:\program files (x86)\grass gis 7.0.1rc1\python27';'c:\program files (x86)\grass gis 7.0.1rc1\msys' pythonlib = 'c:\python27' pythonpath = 'c:\program files (x86)\grass gis 7.0.1rc1\etc\python' sys.path.append(os.path.join(os.environ['...
Read more

post - imageshack API cURL -

i credentials correct. when send image through form this:(action="http://api.imageshack.com/v2/images/"). working fine when tried send $ch = curl_init($url); failed. code: enter code here <form action="" method="post" enctype="multipart/form-data"> select image upload: <input type="file" name="file" id="file" multiple> <input type="text" name="auth_token" id="auth_token" value="xxxxxxxxxxxxxxxx"> <input type="text" name="api_key" id="api_key" value="xxxxxxxxxxxxxxx"> <input type="submit" value="upload image" > </form> $url = ' http://api.imageshack.com/v2/images/ '; $temp = $_files["file"]['tmp_name']; echo $temp."<br>" ; $ch = curl_init($url); curl_setopt($ch, curlopt_post, true); curl...
Read more