怎样的preg_match（）从

Question

我创建从一个URL与PHP获取数据非常基本的脚本（谷歌趋势）file_get_contents（）函数的变量：

$url = 'https://www.google.com/trends/fetchComponent?hl=en-US&q=battlefield%201&cid=TIMESERIES_GRAPH_0&export=5&w=500&h=300&gprop=youtube&date=today%201-m'; 
$url2 = file_get_contents($url); 

在源代码中有一个某些类型的数据我想提取

{"columns":[{"id":"d","label":"Date","type":"datetime"},{"role":"annotation","type":"string"},{"p":{"html":true},"role":"annotationText","type":"string"},{"id":"q0","label":"battlefield 1","type":"number"},{"role":"annotation","type":"string"},{"p":{"html":true},"role":"annotationText","type":"string"},{"role":"certainty","type":"boolean"}],"headlineDataPoints":[],"width":485,"axisAnnotations":[],"rows":[[{"v":new Date(2016, 8, 24, 12, 0),"f":"Saturday, September 24, 2016"},null,null,21,null,null,true],[{"v":new Date(2016, 8, 25, 12, 0),"f":"Sunday, September 25, 2016"},null,null,23,null,null,true],[{"v":new Date(2016, 8, 26, 12, 0),"f":"Monday, September 26, 2016"},null,null,19,null,null,true],[{"v":new Date(2016, 8, 27, 12, 0),"f":"Tuesday, September 27, 2016"},null,null,44,null,null,true],[{"v":new Date(2016, 8, 28, 12, 0),"f":"Wednesday, September 28, 2016"},null,null,54,null,null,true],[{"v":new Date(2016, 8, 29, 12, 0),"f":"Thursday, September 29, 2016"},null,null,39,null,null,true],[{"v":new Date(2016, 8, 30, 12, 0),"f":"Friday, September 30, 2016"},null,null,35,null,null,true],[{"v":new Date(2016, 9, 1, 12, 0),"f":"Saturday, October 1, 2016"},null,null,38,null,null,true],[{"v":new Date(2016, 9, 2, 12, 0),"f":"Sunday, October 2, 2016"},null,null,64,null,null,true],[{"v":new Date(2016, 9, 3, 12, 0),"f":"Monday, October 3, 2016"},null,null,46,null,null,true],[{"v":new Date(2016, 9, 4, 12, 0),"f":"Tuesday, October 4, 2016"},null,null,35,null,null,true],[{"v":new Date(2016, 9, 5, 12, 0),"f":"Wednesday, October 5, 2016"},null,null,34,null,null,true],[{"v":new Date(2016, 9, 6, 12, 0),"f":"Thursday, October 6, 2016"},null,null,34,null,null,true],[{"v":new Date(2016, 9, 7, 12, 0),"f":"Friday, October 7, 2016"},null,null,31,null,null,true],[{"v":new Date(2016, 9, 8, 12, 0),"f":"Saturday, October 8, 2016"},null,null,29,null,null,true],[{"v":new Date(2016, 9, 9, 12, 0),"f":"Sunday, October 9, 2016"},null,null,30,null,null,true],[{"v":new Date(2016, 9, 10, 12, 0),"f":"Monday, October 10, 2016"},null,null,31,null,null,true],[{"v":new Date(2016, 9, 11, 12, 0),"f":"Tuesday, October 11, 2016"},null,null,22,null,null,true],[{"v":new Date(2016, 9, 12, 12, 0),"f":"Wednesday, October 12, 2016"},null,null,40,null,null,true],[{"v":new Date(2016, 9, 13, 12, 0),"f":"Thursday, October 13, 2016"},null,null,63,null,null,true],[{"v":new Date(2016, 9, 14, 12, 0),"f":"Friday, October 14, 2016"},null,null,55,null,null,true],[{"v":new Date(2016, 9, 15, 12, 0),"f":"Saturday, October 15, 2016"},null,null,71,null,null,true],[{"v":new Date(2016, 9, 16, 12, 0),"f":"Sunday, October 16, 2016"},null,null,64,null,null,true],[{"v":new Date(2016, 9, 17, 12, 0),"f":"Monday, October 17, 2016"},null,null,84,null,null,true],[{"v":new Date(2016, 9, 18, 12, 0),"f":"Tuesday, October 18, 2016"},null,null,100,null,null,true],[{"v":new Date(2016, 9, 19, 12, 0),"f":"Wednesday, October 19, 2016"},null,null,null,null,null,true],[{"v":new Date(2016, 9, 20, 12, 0),"f":"Thursday, October 20, 2016"},null,null,null,null,null,true],[{"v":new Date(2016, 9, 21, 12, 0),"f":"Friday, October 21, 2016"},null,null,null,null,null,true],[{"v":new Date(2016, 9, 22, 12, 0),"f":"Saturday, October 22, 2016"},null,null,null,null,null,true],[{"v":new Date(2016, 9, 23, 12, 0),"f":"Sunday, October 23, 2016"},null,null,null,null,null,true]],"showHeadlines":false,"percentData":false,"colors":["#3f85f2"],"height":230} 

如何使用preg_match找到它？我试过

$data = preg_match('/^(var chartData = |; var htmlChart)\\/$/', $url2, $output_array); 
print_r($output_array); 

我使用的第一个和最后一个关键词，试图得到什么其间他们（即我想要的数据）

我使用phpliveregex试过这实际上不排序的回暖是什么我希望（找到VAR chartData字符串）http://www.phpliveregex.com/p/hCB然而当我尝试克隆它不工作。

我的问题的存在，如何从谷歌趋势使用的preg_match提取此JSON对象（chartData变量中），（），因为我已经试过一直没有工作。

Answer 1

这应该是有益的

preg_match('/chartData\s+=\s+\{(.+)\}/', $url2, $output_array); 
$charData = $output_array[0]; 

Answer 2

的问题似乎来自于“JSON字符串”几个值，看起来都像是new Date(2016, 9, 14, 12, 0)包含空格和引号之间的不封闭。您可以将解决他们身边的报价问题：

$str = preg_replace('~:(new Date\([^)]*\))~', ':"$1"', $str); 
$jsonArray = json_decode($str, true);