我有这个URL-
http://localhost/app_demo/sample.php?jsonRequest={"GenInfo":{"type":"Request","appname":"XXX","appversion":"1.0.0"},"searchDish":{"userId":"295","dishName":"","est":"Pizza & Wings","location":"","type":"","priceRange":"","deviceos":"value","deviceId":"<UDID>","deviceType":"value","pageNo":"1"}}
当我点击这个URL并打印
print_r($_REQUEST['jsonRequest']);
字符串只打印到
{"GenInfo":{"type":"Request","appname":"XXX","appversion":"1.0.0"},"searchDish":{"userId":"295","dishName":"","est":"2 Pizza
我上网搜索，但没有得到答案。这个问题的解决方案是什么?请帮忙,谢谢。
查询字符串通常由键/值对组成，查询字符串的开头是问号(?)，然后所有的键/值对以&分隔。在你的值中加入&号就像开始一个新参数。
然而，这不是正确的方法。你不应该把JSON放到查询字符串中。
如果确实必须在查询字符串中使用&符号，则使用%26而不是&。%26，这是与符号的十六进制值。
您应该发出POST请求而不是GET请求:
- <
- 编码碰撞/gh>
- URI长度限制
字符"&"是个问题，因为它是保留的。(是查询字符串参数分隔符)在GET请求中使用字符串之前，必须对其进行"urlencode"。所以像&被转换。但正如jValdron所指出的，你不应该在查询字符串中放入JSON，但你可以这样做。
所以你要对字符串进行urlencode:
$url = 'http://localhost/app_demo/sample.php?jsonRequest=';
$jsonRequest = urlencode('{"GenInfo":{"type":"Request","appname":"XXX","appversion":"1.0.0"},"searchDish":{"userId":"295","dishName":"","est":"Pizza & Wings","location":"","type":"","priceRange":"","deviceos":"value","deviceId":"<UDID>","deviceType":"value","pageNo":"1"}}');
$url .= $jsonRequest;
然后是urldecode
print_r(urldecode($_REQUEST['jsonRequest']));
同样，您不应该在查询字符串中放入JSON