Jump to content
Fórum Script Brasil
  • 0

Erro $end

Carlos Daniel



Estou com o seguinte erro: Parse error: syntax error, unexpected $end. Mas não consigo resolver. Colo o código abaixo. Agradeço a ajuda de todos. Abraços.






<title>Convite a Vida</title>

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">


<body bgcolor="#FFFFFF" text="#000000">

<form action"" method="post" name="formularioVideo">

<input type="hidden" name="h" value="ok" />

<label><span>URL do Video:</span>

<input type="text" name="url" id="urlvideo" size="40" />

<input type="submit" name="btnOK" id="" value="salvar" />



<h3>Convite a Vida</h3>

<div id="videos-youtube">


$sql = mysql_query("SELECT * FROM videosvy");

while($ln = mysql_fetch_array($sql)){


<img src="<?php echo $ln["imagem"]?>" width=120" height="90" style="clear:both; float:left; padding: 3px; border: 1px solid #999 margin-top:2px;" />

<?php echo $ln["titulo]?> / <?php echo $ln["descricao"]?>



<?php } ?>





if(isset($_POST["h"]) && $_POST["h"] == "ok") {

$url = $_POST["url"];

$conteudo = get_meta_tags($url);

$imagem = substr($url, 31, 11);

$urlimagem = "http://i2.ytimg.com/vi/$imagem/default.jpg";

$sql = mysql_query("INSERT INTO videosvy(titulo, descricao, urlvideo, imagem) values('".utf8_decode($conteudo["title"])."', '".utf8_decode($conteudo["description"])."', '$url', '$urlimagem')");

if($sql) {

header("Location: ./");



Link to comment
Share on other sites

1 answer to this question

Recommended Posts

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.


  • Forum Statistics

    • Total Topics
    • Total Posts
  • Create New...