Php 为什么这个问题会破裂？

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"}}

字符串打印仅限于

{"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。有什么解决办法？ 请帮忙，

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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（？），然后所有对都用符号（&）分隔。在您的值中有一个与符号就像启动一个新参数

然而，这不是正确的方法。您不应该将JSON放在查询字符串中

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如果确实在查询字符串中必须有一个与，请使用%26而不是&%26，它是符号AND的十六进制值。

您应该发出POST请求，而不是GET请求：

• 编码冲突
• 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，必须将其“”为“”。所以像&这样的字符被转换。但正如jValdron所指出的，不应该将JSON放入查询字符串中，但可以这样做

因此，您需要对字符串进行URL编码：

$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;

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同样，您不应该将JSON放在查询字符串中。

不要使用

$\u REQUEST

指定

$\u POST

或

$\u GET

$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;

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