php - 如何取出下面字符串中的多个src路径

Question

取出image/之后的20170626/1498445461112119.png（日期/文件名）路径也可以，前面的我可以手动拼。。。

很抱歉不能采纳多个，我是取出来后才登陆segmentfault的，非常感谢各位。

附正则：$preg='//d/d.[jpg|jpeg|png|bmp|gif]*/i';

Answer 1

<?php
$str = '<p>图1<img src="/public/upload/ueditor/image/20170626/1234567890.png" title="1234567890.png" alt="QQ图片1234567890.png"></p><p>图1<img src="/public/upload/ueditor/image/20170626/0987654321.png" title="0987654321.png" alt="QQ图片0987654321.png"></p>';

if (preg_match_all('/src="(.+?)"/', $str, $matches)) {
    print_r($matches);
} else {
    echo '没有匹配';
}

输出:

(
    [0] => Array
        (
            [0] => src="/public/upload/ueditor/image/20170626/1234567890.png"
            [1] => src="/public/upload/ueditor/image/20170626/0987654321.png"
        )

    [1] => Array
        (
            [0] => /public/upload/ueditor/image/20170626/1234567890.png
            [1] => /public/upload/ueditor/image/20170626/0987654321.png
        )

)

感兴趣欢迎看我的正则表达式入门讲堂: 一个小时学会使用正则表达式

Answer 2

正则表达式比较方便

Answer 3

正则匹配下，然后要取的那块用()当子集取出来就可以了

Answer 4

雷雷

Answer 5

<?php
define('CLI_SCRIPT', true);

$tt = '<p><img class="icon" alt="141220154.zip" title="141220154.zip" src="http://11111/theme/image.php/ gourmet/core/1491882323/f/archive" /> </p><p><img class="icon" alt="141220154.zip" title="141220154.zip" src="http://3333/theme/image.php/gourmet/core/         1491882323/f/archive" /></p>';

preg_match_all('/<img [^>]*src=\"(.+?)\"/', $tt, $matches);                                                                                                                 
var_dump($matches);    // 需要的字符串在$matches[1]中，具体信息可以打印了看

参考link: http://php.net/manual/zh/func...