PHP从图像扩展名中删除参数 [英] PHP Remove parameters from image extension

问题描述

我从一个函数获取图像URL.我需要找到该图像的扩展名.有时，图片网址带有类似" http://slimages.macys.com/is/image/MCY/products/4/optimized/1776484_fpx.tif ?$ filterlrg $& wid = 370'.因此，文件扩展名类似于"tif?$ filterlrg $& wid = 370".如何获得确切的扩展名.

I am getting the image url from one function. I need to find the extension for the image. Sometimes the image url comes with parameters like 'http://slimages.macys.com/is/image/MCY/products/4/optimized/1776484_fpx.tif?$filterlrg$&wid=370'. So, the file extension comes like 'tif?$filterlrg$&wid=370'. How can I get the exact extension.

下面是我的代码

<?php
    $srcimg = 'http://slimages.macys.com/is/image/MCY/products/4/optimized/1776484_fpx.tif?$filterlrg$&wid=370';
    $fullpath = basename($srcimg);
    $userImageDetails = pathinfo($fullpath);
    $extension = strtolower($userImageDetails['extension']);
    echo $extension;
?>

推荐答案

您可以使用此答案

$link = 'http://slimages.macys.com/is/image/MCY/products/4/optimized/1776484_fpx.tif?$filterlrg$&wid=370';

if ($url = parse_url($link)) { 
   echo pathinfo($url['path'], PATHINFO_EXTENSION);
}

输出

tif

没有第二个参数，parse_url返回一个关联数组，但是如果只需要路径，则可以传递第二个参数parse_url($link, PHP_URL_PATH)，它将返回一个单独的变量.

with no second argument, parse_url returns an associative array but if you only want the path, you can pass a second argument parse_url($link, PHP_URL_PATH) and it will return a single variable instead.

if ($path = parse_url($link, PHP_URL_PATH)) { 
   echo pathinfo($path, PATHINFO_EXTENSION);
}

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