The idea is as follows: for a keyword which I will give, curl should find pictures in Google Images, then I want to find the right div in which all these pictures lie and through PHPQuery get links to all these pictures. In the code I prescribe what I need to look for.

$ch = curl_init('https://www.google.com/search?q=Steve&tbm=isch'); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true); curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false); curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 2); curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/56.0.2924.87 Safari/537.36'); $out = curl_exec($ch); curl_close($ch); $document = phpQuery::newDocument($out);

In case I enter echo $ document, then everything comes off the page. But when I do this:

 $imgs = $document->find('img');

and bring this whole thing out, then it turns out like this. Where do src tags go? Why pictures are not displayed? What should be done? I fight for a long time and really in a stupor.

it turns out like this

  • the <img> tag in src does not have a normal url, there is a line of the form data:image/jpeg;base64,/9j/4AAQSkZJRgABAQAAAQA..<и далее набор символов, описывающих превью изображения в base64> . It seems to me that you are not looking for your image for your images = / but the parent <a> tag in href already contains something similar to the image link, although starting with "imgurl =" and then the url is a little encoded - MrModest
  • Indeed, thank you. And how to be in this case? Updated: read, thank you - idriverx

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