Question

我有以下代码来提取外部div的内容并显示在我的网页上：

首先在输出链接中使第一个字母大写的som样式：

<style>
h1:first-letter {text-transform: uppercase;} /* make first letter in link upper-case */
h1 {margin-bottom: -15px; font-size:1.2em;}
p {margin-bottom: 5px;}
a {text-decoration: none;}
</style>

然后是php代码（根据@ Bang的回答进行最终更改编辑）：

<?php   
    //The URL for the external content we want to pull
    $url = 'https://ekstern.vuq.visma.com/ra_recruitment/';
    $content = file_get_contents($url);

    //The div that includes the content '<div id="divid">'
    $first_step = explode( '<div id="ide">' , $content );
    $second_step = explode("</div>" , $first_step[1] );

    //Do some magic with the URL
    $url2 = $second_step[0];

    //What do you want to replace?
    $patterns = array(
        '#\./opening;jsessionid=.*\?#',
        '#<a href=#',
        '#span(.*?)>#'
    );

    //...with what?
    $replaces = array(
        'https://ekstern.vuq.visma.com/ra_recruitment/opening?',
        '<a target="_blank" href=',
        'h1>'
    );

    //Print the final output
    echo preg_replace($patterns, $replaces, $url2), $second_step[1], $second_step[2], $second_step[3], $second_step[4], $second_step[5], $second_step[6], '<hr>';
?>

此代码在最后一行的$second_step[0]中提取链接。此链接的格式如下：.opening\and_following_dynamic_link_text。

我尝试使用str_replace更改要移除.opening的网址，然后添加原始主机，后跟/opening。

我尝试在str_replace和$first_step之后插入$second_step字符串，正如您在$content之后的代码中看到的那样。但我无法让它发挥作用。

有人可以帮我解决这个问题吗？此外，最终输出包括一些显示为Ã¸的特殊字符。如何将其显示为原始角色？

以示例查看HERE。

编辑：感谢@ kadam-sunil和@bank，您的回答都帮助了我。我使用工作解决方案更新了代码！

我还设法删除了网址中的一些不需要的文字，将我的评论引用到@ bang的答案。

现在我正在尝试将target =“_ blank”插入到我提取的URL中。有人能指出我正确的方向吗？

EDIT2：以下是我要显示的原始数据

  <div id="ide">
        <div>
            <div class="openingTitle"><a href="./opening;jsessionid=E2A19018E967B4771224A9FA515AFBC0?0-1.ILinkListener-content-contentPanel-openings~view~container-openings~view-0-details"><span style="font-weight:bold;">fagarbeider</span></a></div>
            <div class="openingIngress"><p>Avdeling teknisk drift har ledig stilling som fagarbeider vann/avløp.<br/>Fast, 100 %, ledig snarest.</p></div>
            <div class="openingDetail"><i>Utlyst:&nbsp;<span>28.01.2015</span></i></div>
            <div class="openingDetail"><i>Søknadsfrist:&nbsp;<span style="color:red">01.03.2015</span></i></div>
            <div class="openingDetail"><i>Selskap:&nbsp;<span>Randaberg kommune</span></i></div>
            <div class="openingDetail"><i>Stillingstype:&nbsp;<span>Fast ansatt</span></i></div>
            <div class="openingDetail"><i>Lokasjon:&nbsp;<span>Avd. teknisk drift</span></i></div>
            <div class="openingDetail">

            </div>
        </div>

我还编辑了原始代码，以反映我所做的最新更改。

谢谢！

Answer 1

您应该影响str_replace：

$content = str_replace('./opening', 'https://ekstern.vuq.visma.com/ra_recruitment/opening', $content);

使用preg_replace方法：

$url = '<a href="./opening;jsessionid=E2A19018E967B4771224A9FA515AFBC0?0-1.ILinkListener-content-contentPanel-openings~view~container-openings~view-0-details"><span style="font-weight:bold;">fagarbeider</span></a>';

// What do you want to replace ?
$patterns = array(
    '#\./opening;jsessionid=.*\?#',
    '#<a href=#',
    '#span(.*?)>#'
);

// By What ?
$replaces = array(
    'https://ekstern.vuq.visma.com/ra_recruitment/opening?',
    '<a target="_blank" href=',
    'h1>'
);

echo preg_replace($patterns, $replaces, $url);

然后我得到<a target="_blank" href="https://ekstern.vuq.visma.com/ra_recruitment/opening?0-1.ILinkListener-content-contentPanel-openings~view~container-openings~view-0-details"><h1>fagarbeider</h1></a>

我保持代码简单而愚蠢。

Answer 2

<?php
$url = 'https://ekstern.vuq.visma.com/ra_recruitment/';
$content = file_get_contents($url);
$content1=str_replace('./opening', 'https://ekstern.vuq.visma.com/ra_recruitment/opening', $content);
$first_step = explode( '<div id="ide">' ,  $content1 );
$second_step = explode("</div>" , $first_step[1] );

echo $second_step[0], $second_step[1], $second_step[2], $second_step[3], $second_step[4], $second_step[5], $second_step[6];
?>

替换从外部div拉出的部分链接

2 个答案: