Question

我使用stream_socket_client()创建一个套接字，并将它们放在名为$sockets的数组中。当我运行代码时，会出现错误：

严格标准：只应通过引用传递变量第63行的D：\ www \ study \ cl \ s.php

代码：

while (count($sockets)) {
    $read = $write = $sockets;
    stream_select($read, $write, $e = null, $timeout); //line 63
    if (count($read)) {
        foreach ($read as $r) {
            $id = array_search($r, $sockets);
            $data = fgets($r, 1024);
            echo $data;
            if (strlen($data) == 0) {
                if ($status[$id] == "in progress") {
                    $status[$id] = "failed to connect";
                }
                fclose($r);
                unset($sockets[$id]);
            } else {
                $status[$id] .= $data;
            }
        }
        foreach ($write as $w) {
            $id = array_search($w, $sockets);
            fwrite($w, "HEAD / HTTP/1.0rnHost: " . $host . "rnrn");
            $status[$id] = "waiting for response";
        }
    } else {
        foreach ($sockets as $id => $s) {
            $status[$id] = "timed out " . $status[$id];
        }
        break;
    }
}

Answer 1

这一行：

stream_select($read, $write, $e = null, $timeout);

应该是：

$e=null;
stream_select($read, $write, $e, $timeout);

虽然第一个代码可以工作，但PHP会抛出E_STRICT消息。我不确定为什么PHP开发人员会这么做。

但是，您应该检查stream_select()的返回类型：

$ret = stream_select($read, $write, $e, $timeout);
if($ret === FALSE) {
    die('Error: stream_select');
}

检查返回类型后，您还可以使用静默运算符@：

$ret = @stream_select($read, $write, $e = NULL, $timeout);
if($ret === FALSE) {
    die('Error: stream_select');
}

stream_select错误：严格标准

1 个答案: