Capture page screenshot by URL PHP script

• 23 Mar 2020
• |
2 comments

There are many third-party API available to capture screenshot of a page or website. For this I will used simple API `Google PageSpeed Insights API`, which is used to measure the performance of website. It will create a website screenshot when user try to check performance of any website. I will used file_get_contents() function in php to capture website screenshot. Website or page URL is required to pass in this function as a parameter. Use below code to take a website or page screenshot. Replace $URL variable value with your website link which you want to capture.

<?php

//page or website URL which you wants to capture
$siteURL = "phpesperto.com";
$jsonData = file_get_contents("https://www.googleapis.com/pagespeedonline/v2/runPagespeed?url=$siteURL&screenshot=true");

//decode $jsonData json data
$jsonData = json_decode($jsonData, true);

//Website data in array $jsonData
// Image data key is in $jsonData['screenshot']['data'];
$screenshot = $jsonData['screenshot']['data'];

//change some characters
$screenshot = str_replace(array('_','-'),array('/','+'),$screenshot); 

//pass image data in img src
echo "<img src=\"data:image/jpeg;base64,".$screenshot."\" />";
?>

<?php

//page or website URL which you wants to capture
$siteURL = "phpesperto.com";
$jsonData = file_get_contents("https://www.googleapis.com/pagespeedonline/v2/runPagespeed?url=$siteURL&screenshot=true");

//decode $jsonData json data
$jsonData = json_decode($jsonData, true);

//Website data in array $jsonData
// Image data key is in $jsonData['screenshot']['data'];
$screenshot = $jsonData['screenshot']['data'];

//change some characters
$screenshot = str_replace(array('_','-'),array('/','+'),$screenshot); 

//pass image data in img src
echo "<img src=\"data:image/jpeg;base64,".$screenshot."\" />";
?>