代码:
如果url中包含类似于 的字符串
当我输出$college_name时,它只显示(Agra Public Institute of Technology),而不是完整的名称,即(Agra Public Institute of Technology & Computer Education, Artoni),而$match[1]包含(药房),但在打印$match[1]时没有显示,而是显示'All'。那么,我该如何解决这个问题?
谢谢。
<?php
include('conn.php');
$student_id = $_SESSION['student_id'];
$college_name = $_GET['college_name'];
$college_name22 = explode('(', $college_name);
if(preg_match('#\((.*?)\)#', $college_name, $match))
{
$match[1] = lcfirst($match[1]);
}
else
{
$match[1] = 'All';
}
如果url中包含类似于 的字符串
cstest/view.php?college_name=Agra%20Public%20Institute%20of%20Technology%20&%20Computer%20Education,%20Artoni%20(Pharmacy)
我想要拆分以下字符串,即:
Agra%20Public%20Institute%20of%20Technology%20&%20Computer%20Education,%20Artoni%20(Pharmacy)
当我输出$college_name时,它只显示(Agra Public Institute of Technology),而不是完整的名称,即(Agra Public Institute of Technology & Computer Education, Artoni),而$match[1]包含(药房),但在打印$match[1]时没有显示,而是显示'All'。那么,我该如何解决这个问题?
谢谢。
var_dump($_GET)
看起来像什么?首先,它不会包含URL编码的字符串。如果您访问带有,
和(
的URL,则您的浏览器会出现问题,它们也将被编码为%2C和%28。 - miken32