我需要一个PHP脚本,它接受一个网页的URL,然后回传一个单词被提到多少次。
示例
这是一个通用的HTML页面:
<html>
<body>
<h1> This is the title </h1>
<p> some description text here, <b>this</b> is a word. </p>
</body>
</html>
--这将是PHP脚本:
<?php
htmlurl="generichtml.com";
the scri
如何从两个表中检索和查询执行绑定形成json数组传递
cont_search.php
<?php
include "connection.php";
$keyword=$_GET["context"];
//Checking if any error occured while connecting
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error(
我使用jQuery getJson方法从json中得到值.下面是我的代码。
//var data_url="message.json";
var data_url="php/server_processing.php";
var count = 0;//will be computed(no of entries in json file)
var paginationCount=20;
var i=0,j=0;
var symflag=0;
这是我的密码:
代码
<?php
if(isset($_POST['Submit'])){
$title ='myPost.php';
echo $title;
//the data
$data = "Hey I am Aidan\n";
//open the file and choose the mode
$fh = fopen($title, "a");
fwrite($fh, $data);
//close the file
f
得到验证错误:Error: an attribute specification must start with a name or name token,但我不能批评它,我认为它需要新的视角!
该脚本位于PHP页面上的<script>标记中,因此php执行得很好。
var Target = ' target="_blank"';
var twtTitle = "I am going to <?php echo $fb_title;?>";
var tinyUrl = "<?php