我一直试图从API检索屏幕截图，但当我解码并保存图像时，我得到了一个损坏的图像。下面是我正在使用的代码。如果你想测试的话，我为一个包含谷歌响应的样本文件创建了一个tinyurl
$name = 'test';
$result = file_get_contents('http://tinyurl.com/q4smyod');
$result = json_decode($result, true);
$decoded=base64_decode($result['screenshot']['data']);
file_put_contents('img/'.$name.'.jpg',$decoded);
正如我在评论中提到的，这个问题是由使用phpapi时谷歌加密的错误引起的。如果您遇到此问题，只需使用以下替换函数来修复编码。
$data = str_replace('_','/',$result['screenshot']['data']);
$data = str_replace('-','+',$data);
$decoded = base64_decode($data);
这将帮助您更接近目标。您没有定义名称，并且您的json_edding有点奇怪：
<?php
error_reporting(-1);
ini_set('display_errors', 'On');
$result = file_get_contents('http://tinyurl.com/q4smyod');
$name = 'test2';
$result = json_decode($result, true);
$decoded = base64_encode($result['screenshot']['data']);
file_put_contents($name.'.jpg',$decoded);
?>
目前我正在使用Google Page Speed API通过Web URL捕获图像并将其保存到我指定的位置。这样做没有任何问题。请看一看。评论易于理解。
import urllib2
import json
import base64
import sys
import requests
import os
import errno
# The website's URL as an Input
site = #specify the URL here
imagePath = #specify your path to save the image
# The Google API. Remove "&strategy=mobile" for a desktop screenshot
api = "https://www.googleapis.com/pagespeedonline/v1/runPagespeed?screenshot=true&strategy=mobile&url=" + urllib2.quote(site)
# Get the results from Google
try:
site_data = json.load(urllib2.urlopen(api))
except urllib2.URLError:
print "Unable to retreive data"
sys.exit()
try:
screenshot_encoded = site_data['screenshot']['data']
except ValueError:
print "Invalid JSON encountered."
sys.exit()
# Google has a weird way of encoding the Base64 data
screenshot_encoded = screenshot_encoded.replace("_", "/")
screenshot_encoded = screenshot_encoded.replace("-", "+")
# Decode the Base64 data
screenshot_decoded = base64.b64decode(screenshot_encoded)
if not os.path.exists(os.path.dirname(impagepath)):
try:
os.makedirs(os.path.dirname(impagepath))
except OSError as exc:
if exc.errno != errno.EEXIST:
raise
# Save the file
with open(imagePath, 'w') as file_:
file_.write(screenshot_decoded)