php - 我只能在php中获取url参数的第一个数据

Question

我有这样的 php 代码

     print "<pre>";
    $string = http_build_query($data);
    echo "http://localhost:8888/moodle35/course/data.php?".$string;
    $ch = curl_init("http://localhost:8888/moodle35/course/data.php?".$string);
    curl_setopt($ch, CURLOPT_POST, true);
    curl_setopt($ch, CURLOPT_POSTFIELDS, $string);
    curl_setopt($ch, CURLOOPT_RETURNTRANSFER, ture);
    curl_exec($ch);
    curl_close($ch);
    print "</pre>";
    return;

我在 http_build_query 之后的网址是： http://localhost:8888/moodle35/course/data.php?returnto=catmanage&fullname=MadanGurung&shortname=keg&category=1&visible=1&startdate=1530140400&enddate=1561676400&idnumber=&mform_isexpanded_id_descriptionhdr=1&summary_editor%5Btext%5%5%5 5D=1&summary_editor%5Bitemid%5D=538633179&overviewfiles_filemanager=479174445&format=topics&numsections=4&hiddensections=0&coursedisplay=0&addcourseformatoptionshere=0&lang=&newsitems=5&showgrades=1&showreports=0&maxbytes=0&enablecompletion=1&groupmode=0&groupmodeforce=0&defaultgroupingid=0&role_1=&role_2=&role_3=&role_4=&role_5=&role_6= &role_7=&role_8=&saveandreturn=保存+and+return&id=0

我可以看到 returnto=catmanage&fullname=MadanGurung&shortname=keg 等等，但我无法获取所有

<?php
echo "something to print outside<\n>";
echo $_GET['returnto'];
echo $_GET['fullname'];
echo $_GET['shortname'];
?>

此代码仅打印返回值，但不打印全名和短名称。任何帮助将不胜感激。

标签： php