Php preg_match从html中查找json数据

文本html

,["pid","VJHggI_XrKLWQ"]
,["image_full","1080|https://pic.website.com/bdnsns,800|https://pic2.website.com/bdnsns"]

如何获取字符串

https://pic.website.com/bdnsns

我尝试了以下代码：

$str = file_get_contents($str);

if(preg_match('/"image_full","1080|(.*?)"/', $str, $m)){
  echo  $m[1] . "\n";
}

---

试试这个，我希望它能对你有所帮助。如果url在两个delimeters之间，那么这个解决方案是完美的

$str = file_get_contents("test.html");
print_r(extract_unit($str,'|',','));
function extract_unit($string, $start, $end)
 {
  $pos = stripos($string, $start);

  $str = substr($string, $pos);

  $str_two = substr($str, strlen($start));

  $second_pos = stripos($str_two, $end);

  $str_three = substr($str_two, 0, $second_pos);

  $unit = trim($str_three); // remove whitespaces

  return $unit;
 }

---

实际上，解析这种输出不是正确的做法。您需要以JSON格式正确获取数据。如果你需要它，那么这里是你可以做的

<?php

$str = ',["pid","VJHggI_XrKLWQ"],["image_full","1080|https://pic.website.com/bdnsns,800|http://pic2.website.com/bdnsns"],["image_full","1080|https://pic.website.com/bdnsns,800|https://pic2.website.com/bdnsns"]';

$m = array();

if(preg_match_all('/\d+\|http(s)?:\/\/.+(\,|\")/U', $str, $m)){
    foreach($m[0] as $each_match){
        $each_match = trim(trim($each_match,","),'"');
        $new_matches = array(); 
        if(preg_match('/http(s)?:\/\/.+/',$each_match,$new_matches)){
            echo $new_matches[0],"\n";
        }
    }
}

---

如果输入的JSON格式正确，为什么不正确解析JSON数据？没有JSON数据源，它只是来自文件的html\u get\u content如果是正确的html，为什么不解析它？如何解析？我使用json_decode（$str，true），但是空白的
json_decode
只能用于有效的json结构。如果您的源数据是HTML格式，则类似于
DOMDocument
的内容将有所帮助。但是给出的示例看起来不像是有效的JSON或HTML。。。