[英]PHP img url with preg_match
想要只獲得img網址
"//lh5.googleusercontent.com/-0qGW989h9DM/AAAAAAAAAAI/AAAAAAAAG3g/aXkL8mbboKY/photo.jpg"
從
<meta itemprop="name" content="Saurabh Sharma">
<meta itemprop="description" content="Works at Google as Product Manager. Lives in San Francisco. Lived in New Delhi - Mumbai - Canonsburg - Pittsburgh - Austin - San Francisco.">
<meta itemprop="image" content="//lh5.googleusercontent.com/-0qGW989h9DM/AAAAAAAAAAI/AAAAAAAAG3g/aXkL8mbboKY/photo.jpg">
<meta itemprop="url" content="https://plus.google.com/+SaurabhSharma">
</head>
<body class="Td lj">
<input type="text" name="hist_state" id="hist_state" style="display:none;">
如何用preg_match做到這一點?
要獲取圖像內容,請使用以下內容
$ data = '<meta itemprop="name" content="Saurabh Sharma"><meta itemprop="description" content="Works at Google as Product Manager. Lives in San Francisco. Lived in New Delhi - Mumbai - Canonsburg - Pittsburgh - Austin - San Francisco."><meta itemprop="image" content="//lh5.googleusercontent.com/-0qGW989h9DM/AAAAAAAAAAI/AAAAAAAAG3g/aXkL8mbboKY/photo.jpg"><meta itemprop="url" content="https://plus.google.com/+SaurabhSharma"></head><body class="Td lj"><input type="text" name="hist_state" id="hist_state" style="display:none;">';
preg_match('/<meta itemprop=\"image\" content=\"(.*?)\">/i', $data, $matches);
$ matches將返回一個數組
Array
(
[0] =>
[1] => //lh5.googleusercontent.com/-0qGW989h9DM/AAAAAAAAAAI/AAAAAAAAG3g/aXkL8mbboKY/photo.jpg
)
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