从GET传递值到URL

问题描述 投票:0回答:2

我如何通过从变量$名1的值(比方说,它的bobsmith)取代janedoe

<?php

$name1 = $_GET["name"];

$curl = curl_init();
$user = "user1";
$password = "pass1";
curl_setopt_array($curl, array(
CURLOPT_URL =>'https://betawebservices.whatever.com/login/service/v4/SchemaData/INDIVIDUAL- 
ACTIVITIES-University/USERNAME:janedoe/',

我希望我能说

<?php

$name1 = $_GET["name"];

$curl = curl_init();
$user = "user1";
$password = "pass1";
curl_setopt_array($curl, array(
CURLOPT_URL=>'https://betawebservices.whatever.com/login/service/v4/SchemaData/INDIVIDUAL- ACTIVITIES-University/USERNAME:$name1/',

......但是,这并不工作。谢谢

string variables
2个回答
1
投票

如果你把CURLOPT_URL的内容在双引号代替单引号应该能正常运行。


0
投票

这将工作..

CURLOPT_URL=>'https://betawebservices.whatever.com/login/service/v4/SchemaData/INDIVIDUAL- ACTIVITIES-University/USERNAME:'.$name1.'/'
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