Question

我有一个用PHP编写的渲染页面输出这段代码：

    <html>
    <body>
   random code 

    <u>Hello everyone</u>

   random code    
    </body>
    </html>

现在我需要检查页面是否包含字符串'Hello everyone'以及找到时的内容，或者在找不到时不执行任何操作。呈现的页面位于同一服务器中。感谢。

Answer 1

尝试使用jQuery

<html>
<head>
</head>
<body>
random code 
<u>Hello everyone</u>
random code 
</body>
</html>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script>
var url="filecont.php"//THE PHP FILE LOCATION
$.post(url, function(data) {
  var count = data.match(/Hello everyone/g); 
  if(count.length<=1){//If its in the same file, because in .match(/Hello everyone/g); already exist
    alert("NOT FOUND");
  }
  else{
    alert("YEAH!, FOUND");
  }
});
</script>

Answer 2

使用strpos

$isexist = strpos($the_page_text, "Hello everyone");

Answer 3

您也可以尝试 -

$page_source=file_get_contents("page_url"); //or use curl

if(preg_match("/Hello everyone/i",$page_source)){
    //do something
}
else{
    //do nothing
}

使用PHP在HTML源呈现页面中搜索字符串

3 个答案: