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使用php从url(get方法)获取值_Php_Html_Get - Fatal编程技术网

使用php从url(get方法)获取值

使用php从url(get方法)获取值,php,html,get,Php,Html,Get,我从url中获取值,url如下所示 http://*****/test.php?register=true$gcm_regid=Amit&name=amit.sid@gmail.com&email=asdxcvxcvxzcvxc 我用下面的代码打印这个值 $gcm_regid = $_GET["gcm_regid"]; $name = $_GET["name"] $email = $_GET["email"] echo $name; ======>this is lin

我从url中获取值,url如下所示

 http://*****/test.php?register=true$gcm_regid=Amit&name=amit.sid@gmail.com&email=asdxcvxcvxzcvxc
我用下面的代码打印这个值

$gcm_regid = $_GET["gcm_regid"];
$name = $_GET["name"]
$email = $_GET["email"]
echo $name; ======>this is line 14-<==========

很简单,您在
$name
$email
行后缺少分号。

您缺少分号
$\u GET[“name”]
$\u GET[“email]

$gcm_regid = $_GET["gcm_regid"];
$name = $_GET["name"];
$email = $_GET["email"];
echo $name;
您缺少的分号。 及

在此链接中,您使用的是
$
符号,而不是
&

$gcm_regid
更改为
&gcm_regid
简单地说,有两个问题, 第一个,你缺少分号

$name = $_GET["name"]
$email = $_GET["email"]
&第二个问题是url传递元素

http://*****/test.php?register=true$gcm_regid=Amit&name=amit.sid@gmail.com&email=asdxcvxcvxzcvxc
把这行改成这行

http://*****/test.php?register=true&gcm_regid=Amit&name=amit.sid@gmail.com&email=asdxcvxcvxzcvxc
错误是“true$gcm_regid”,它将以“true&gcm_regid”的形式出现

http://*****/test.php?register=true$gcm_regid=Amit&name=amit.sid@gmail.com&email=asdxcvxcvxzcvxc
http://*****/test.php?register=true&gcm_regid=Amit&name=amit.sid@gmail.com&email=asdxcvxcvxzcvxc