通过example.com时parse_url()返回错误 [英] parse_url() returns an error when example.com is passed
问题描述
根据以下代码，如果$host_name是类似example.com的PHP，则会返回一个通知:Message: Undefined index: host，但是在完整的URL(例如http://example.com)上，PHP返回example.com.我试过if语句为FALSE和NULL，但没有用.
According to following code if $host_name is something like example.com PHP returns a notice: Message: Undefined index: host but on full URLs like http://example.com PHP returns example.com. I tried if statements with FALSE and NULL but didn't work.
$host_name = $this->input->post('host_name');
$parse = parse_url($host_name);
$parse_url = $parse['host'];
如何修改脚本以接受example.com并将其返回?
How can I modify the script to accept example.com and return it?
推荐答案
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升级您的php.
5.4.7 Fixed host recognition when scheme is ommitted and a leading component separator is present.
Upgrade your php.
5.4.7 Fixed host recognition when scheme is ommitted and a leading component separator is present.
手动添加方案:if(mb_substr($host_name, 0, 4) !== 'http') $host_name = 'http://' . $host_name;
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