Question

我有这个网址 -

http://localhost/app_demo/sample.php?jsonRequest={"GenInfo":{"type":"Request","appname":"XXX","appversion":"1.0.0"},"searchDish":{"userId":"295","dishName":"","est":"Pizza &amp; Wings","location":"","type":"","priceRange":"","deviceos":"value","deviceId":"<UDID>","deviceType":"value","pageNo":"1"}}

当我点击此网址并打印

时

print_r($_REQUEST['jsonRequest']);

仅打印字符串

{"GenInfo":{"type":"Request","appname":"XXX","appversion":"1.0.0"},"searchDish":{"userId":"295","dishName":"","est":"2 Pizza

我在网上搜索但没有得到答案。这是什么解决方案？请帮忙，感谢。

Answer 1

查询字符串通常由键/值对组成，查询字符串的开头是问号（？），然后所有对都用＆符号（＆amp;）分隔。在您的值中使用＆符就像启动一个新参数。

但是，这不是正确的方法。您不应该将JSON放在查询字符串中。

如果真的在查询字符串中必须有＆符号，请使用％26 而不是＆amp; amp 。％26是＆符号的十六进制值。

Answer 2

您应该发出POST请求而不是GET请求：

编码冲突
URI长度限制

Answer 3

字符“＆amp;”是问题，因为它是保留的。（是查询字符串params分隔符）在您的GET请求中使用之前，您必须“urlencode”您的字符串。所以像＆amp;转换。但是当jValdron指出它时你不应该把JSON放在查询字符串中，但你可以这样做。

所以你urlencode字符串：

$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 &amp; Wings","location":"","type":"","priceRange":"","deviceos":"value","deviceId":"<UDID>","deviceType":"value","pageNo":"1"}}');
$url .= $jsonRequest;

然后你urldecode

print_r(urldecode($_REQUEST['jsonRequest']));

同样，您不应将JSON放在查询字符串中。

为什么此查询会中断？

3 个答案: