Question

我有一个像这样的JSON数组：

[
    {
        "title": " Bones of the Hills <a href="/series/44108-conqueror" class="greyText">(Conqueror #3) </a>",
        "author": "Conn Iggulden",
        "format": "Hardcover",
        "pages": "518",
        "rating": "4.29",
        "image": "http://d.gr-assets.com/books/1347616868l/3276637.jpg"
    }
]

有时，（并非总是）标题字符串中包含链接。如何删除链接及其中的所有内容？我甚至不希望锚内的文本。我想最终以JUST“The Bones of the Hills”为标题。我怎么能在PHP中这样做？我用google搜索负载，但似乎没什么用。

Answer 1

你可以在标题上使用php内置函数strip_tags（），它将从标题中删除所有html元素，包括链接

Answer 2

您可以使用strip_tag（）函数来避免此错误

 $str='[{"title": " Bones of the Hills <a href="/series/44108-conqueror" class="greyText">(Conqueror #3) </a>","author": "Conn Iggulden","format": "Hardcover",';
        $str.='"pages": "518","rating": "4.29","image": "http://d.gr-assets.com/books/1347616868l/3276637.jpg"}]';
  echo      $str2=strip_tags($str);
  $array=json_decode($str2);
print_r($array);

Answer 3

首先你的json数组不正确，请检查

[
    {
        "title": " "title": " Bones of the Hills <a href=\\"/series/44108-conqueror\\" class=\\"test\\">(Conqueror #3) </a>",
        "author": "Conn Iggulden",
        "format": "Hardcover",
        "pages": "518",
        "rating": "4.29",
        "image": "http://d.gr-assets.com/books/1347616868l/3276637.jpg"
    }
]

使用"title": " Bones of the Hills <a href=\\"/series/44108-conqueror\\" class=\\"test\\">(Conqueror #3) </a>"

而不是

<a href="/series/44108-conqueror" class="greyText"

然后

$arr = json_decode($json,true);
$arr['title'] = preg_replace("/<a.+?href.+?>.+?<\/a>/is","",$arr['title']);

var_dump($arr['title']);

现在你的头衔将是

string(20) " Bones of the Hills "

删除不确定的字符串块（URL）

3 个答案: