Question

希望这是一个非常简单的解决方案，我是PHP的新手，所以我可能会遗漏一些明显的东西。我正在使用ScraperWiki构建一个刮刀（虽然这是PHP的一个问题，与SW无关）。代码如下：

<?php
require 'scraperwiki/simple_html_dom.php';

$allLinks = array();

function nextPage($nextUrl, $y)
{
    getLinks($nextUrl, $y);    
}

function getLinks($url) // gets links from product list page   
{
    global $allLinks;
    $html_content = scraperwiki::scrape($url);
    $html         = str_get_html($html_content);

    if (isset($y)) {
        $x = $y;
    } else {
        $x = 0;
    }

    foreach ($html->find("div.views-row a.imagecache-product_list") as $el) {
        $url          = $el->href . "\n";
        $allLinks[$x] = 'http://www.foo.com';
        $allLinks[$x] .= $url;
        $x++;
    }

    $next = $html->find("li.pager-next a", 0)->href . "\n";
    print_r("Printing $next:");
    print_r($next);

    if (isset($next)) {
        $nextUrl = 'http://www.foo.com';
        $nextUrl .= $next;
        print_r($nextUrl);
        $y = $x;
        print_r("Printing X:");
        print_r($x);
        print_r("Printing Y:");
        print_r($y);

        nextPage($nextUrl, $y);
    } else {
        return;
    }

}

getLinks("http://www.foo.com/department/accessories");

print_r($allLinks);

?>

预期输出：脚本应该从第一页抓取所有链接，找到“下一页”按钮，从其URL抓取链接，从该URL找到“下一页”，等等等等。当没有剩下“下一页”链接时，它应该停止。

CURRENT OUTPUT ：代码运行正常，但它应该停止运行。这是关键路线：

$next = $html->find("li.pager-next a", 0)->href . "\n";
if (isset($next)) { }

如果页面上存在li.pager-next a，我只想运行“nextPage（）”函数。以下是控制台的输出：

     http://www.foo.com/department/accessories?page=1
        http://www.foo.com/department/accessories?page=2
        http://www.foo.com/department/accessories?page=3
        http://www.foo.com/department/accessories?page=4
        http://www.foo.com/department/accessories?page=5
        http://www.foo.com/department/accessories?page=6
        http://www.foo.com/department/accessories?page=7
        http://www.foo.com/department/accessories?page=8
        http://www.foo.com/department/accessories?page=9
        http://www.foo.com/department/accessories?page=10

    PHP Notice:  Trying to get property of non-object in /home/scriptrunner/script.php on line 31
 // THE LOOP SHOULD BREAK HERE BUT DOESN'T

        http://www.foo.com
        http://www.foo.com/home?page=1
        http://www.foo.com/home?page=2
        http://www.foo.com/home?page=3
        http://www.foo.com/home?page=4
        http://www.foo.com/home?page=5
        http://www.foo.com/home?page=6
        http://www.foo.com/home?page=7

Answer 1

这个怎么样：

$next = $html->find("li.pager-next a", 0);

if (isset($next)) {
    $nextUrl = 'http://www.foo.com';
    $nextUrl .= $next->href; // move ->href here
    print_r($nextUrl . "\n"); // put \n here since we don't actually want that char in the url
    $y = $x;
    print_r("Printing X:");
    print_r($x);
    print_r("Printing Y:");
    print_r($y);

    nextPage($nextUrl, $y);
} else {
    return;
}

Answer 2

返回的是什么值

$next = $html->find("li.pager-next a", 0)->href . "\n";

在向isset($next)附加"\n"时，永远不会导致$nextElement = $html->find("li.pager-next a", 0); if(isset($nextElement)) { $nextUrl = 'http://www.foo.com' . $nextElement->href . PHP_EOL; print_r($nextUrl); $y = $x; print_r("Printing X:"); print_r($x); print_r("Printing Y:"); print_r($y); nextPage($nextUrl, $y); }返回false。

使用类似的东西：

{{1}}

Answer 3

只需删除isset（）

    if($next){
    }

当未设置变量时，PHP ISSET函数仍在运行

3 个答案: