Pick site name by link

Question

//  IRIA PUXAR DO BANCO DE DADOS OS LINKS, MAS ASSIM JÁ DA UM EXEMPLO DE COMO IRIA FICAR.

$listalinks = "http://site1.com/embed1/CODIGO http://site2.com/embed2/CODIGO http://site3.com/embed3/CODIGO http://site4.com/embed4/CODIGO"; 

function pegaListadeUrlDaBase() {
 

   // Exemplo com links em uma única string da base, separados por ' ';
   $links_text = "$listalinks";
   return explode(" ", $links_text);
}

$urls = pegaListadeUrlDaBase();


foreach($urls as $url) {
   $nome = explode('.', parse_url($url, PHP_URL_HOST))[0];
   echo "<a href='$url'>$nome</a>";
}

I put it like this and it won’t catch.

Answer 1

I think this will do for you:

index php.

<form id="form" name="form" method="post" action="envia.php">
   Links: <label for="nome"></label>
          <textarea name="nome" id="nome" cols="20" rows="8"></textarea>
          <input type="submit" name="button" id="button" value="Enviar" />
</form>

<?php
include("conexao.php");

$sql = "SELECT seu_campo FROM sua_tabela";

$result = mysqli_query($conn, $sql);

while ($linha = mysqli_fetch_assoc($result)){

  echo "<a href=".$linha['nome'].">".$linha['nome']."</a><br>";
}
?>

In the index.php, you enter the values in the textarea and then they are shown as link below, after insertion.

php.

<?php
include_once("conexao.php");

$nomes = $_POST['nome'];
$separar = explode("\n", $nomes);

foreach($separar as $nome){
    $query = "INSERT INTO sua_tabela (seu_campo) VALUES ('".$nome."')";
    $resultado = mysqli_query($conn, $query);
}

echo "Seu cadastro foi realizado com sucesso!";

?>

This is where it does all the separating magic by line breaking using the explode and then you can enter it into the database.

php connection.

<?php
  $servidor = "localhost";
  $usuario = "root";
  $senha = "1234";
  $dbname = "teste";

  $conn = mysqli_connect($servidor, $usuario, $senha, $dbname);

?>

And this is the connection file I used, just exchange the value of the variables for your mysql data.

Finally, replace where it says "your table" and "your field" and I believe it will work smoothly.

Answer 2

You can use php’s native function parse_url to retrieve information from a URL. That way:

$listalinks = "http://site1.com/embed1/CODIGO http://site2.com/embed2/CODIGO http://site3.com/embed3/CODIGO http://site4.com/embed4/CODIGO";

$urls = explode(" ", $listalinks);


foreach($urls as $url) {
   $nome = explode('.', parse_url($url, PHP_URL_HOST))[0];
   echo "<a href='$url'>$nome</a>";
}

This example will print:

<a href='http://site1.com/embed1/CODIGO'>site1</a>
<a href='http://site2.com/embed2/CODIGO'>site2</a>
<a href='http://site3.com/embed3/CODIGO'>site3</a>
<a href='http://site4.com/embed4/CODIGO'>site4</a>

View documentation:

http://php.net/manual/en/function.parse-url.php

http://php.net/manual/en/function.explode.php

Answer 3

$ssite=('http://site.com'); 

$q=@file_get_contents($ssite); 

preg_match_all('#<title>([^<\/]{1,})<\/title>#i',$q,$match);

$otitle=($match[1][0]); 

echo $otitle; // Mostra frase do title