为什么strpos PHP
不能使用fsockopen响应?
加载此代码时。此代码将请求sdgsgsdgsfsdfsd.ca
到whois.cira.ca
服务器,并找到strpos PHP
文本Domain status: available
,如果找到,它将返回
{"domain":"sdgsgsdgsfsdfsdca","availability":"available"}
,但如果没有找到文本。它将是echo
{"domain":"sdgsgsdgsfsdfsdca","availability":"TAKEN"}
本例中为found text
,但仍为echo
{"domain":"sdgsgsdgsfsdfsdca","availability":"TAKEN"}
我该怎么办?
<?php
$server = "whois.cira.ca";
$response = "Domain status: available";
showDomainResult(sdgsgsdgsfsdfsd.ca,$server,$response);
function checkDomain($domain_check,$server,$findText)
{
$con = fsockopen($server, 43);
if (!$con) return false;
fputs($con, $domain_check."'r'n");
$response = ' :';
while(!feof($con))
{
$response .= fgets($con,128);
}
echo $response."<BR><BR><BR><BR><BR>";
fclose($con);
if (strpos($response, $findText))
{
return true;
}
else
{
return false;
}
}
function showDomainResult($domain_check,$server,$findText)
{
if (checkDomain($domain_check,$server,$findText))
{
class Emp
{
public $domain = "";
public $availability = "";
}
$e = new Emp();
$e->domain = $domain_check;
$e->availability = "available";
echo json_encode($e);
}
else
{
class Emp
{
public $domain = "";
public $availability = "";
}
$e = new Emp();
$e->domain = $domain_check;
$e->availability = "TAKEN";
echo json_encode($e);
}
}
?>
你使用strpos错误,如果字符串与你正在搜索的内容开始,它将返回int(0),这是"有点假" PHP的定义。显式检查false
,像这样
return false!==strpos($response, $findText);
!==
而不是!=
和作为经验法则,永远不要在PHP中使用松散的比较运算符,如果你可以避免它,可能会发生滑稽的错误:https://3v4l.org/tT4l8