解析 JSON 网页中的值

Parsing a value from a JSON webpage

本文关键字:网页 JSON 解析      更新时间:2023-09-26

使用javascript或php,如何从网页中获取值?这是我尝试解析的 JSON:

{"错误":["],"模板Html":", "_visitor_conversationsUnread":"0","_visitor_alertsUnread":"0"}

我正在尝试获取"_visitor_alertsUnread"的值。我将如何做到这一点?

谢谢!

您可以使用正则表达式、json 解码或简单索引来解析它。 但是,在这三者中,json 是最干净、最正确的方法。

1) JSON解码:

$page = file_get_contents($url);
$json_arr = json_decode($string,true);
return $json_arr['_visitor_alertsUnread'];

2)正则表达式:

$page = file_get_contents($url);
$pattern = ".*?_visitor_alertsUnread'''":'''"(''d)'''"";
preg_match($pattern, $page, $matches);
return $matches[1];

3) 索引:

$page = file_get_contents($url);
$needle = "_visitor_alertsUnread";
$startpos = strrpos($page, $needle) + strlen($needle) + 3;
$endpos = strrpos($page, "'"", $startpos);
return substr($page, $startpos, $endpos);

响应采用 JSON 格式。

如果你只是想得到_visitor_alertsUnread的值,那么你可以做这样的事情。

var JSONObj = {"error":["Security error occurred. Please press back, refresh the page, and try again."],"templateHtml":"'n'n'n'n't</a>'n't'n't'tThe following error occurred:</h2>'n't't'n't't'n't't'n't't'tSecurity error occurred. Please press back, refresh the page, and try again.</label>'n't't'n't't</div>'n't'n</div>","_visitor_conversationsUnread":"0","_visitor_alertsUnread":"0"};
alert(JSONObj._visitor_alertsUnread);