Código PHP:
<?php
//-----------------------DO NOT EDIT BELOW THIS LINE!----------------------------------
$db_host = $_POST['hostDB'];
$db_user = $_POST['userDB'];
$db_pass = $_POST['passDB'];
$db_db = $_POST['DB'];
$currentid = 0;
$finishid = $_POST['finishID'];
$connection = mysql_connect($db_host,$db_user,$db_pass);
$Tr =
////////////////////////////////////////////////////////////////////////////////////////
$currentid = $currentid + 1;
if ($currentid == $finishid + 1)
{
die('Traducción completada con éxito');
}
else
{
$query = 'SELECT entry, name, description FROM eses_item';
if ($r = mysql_query($query,$connection))
{
while ($row = mysql_fetch_array($r))
{if ($currentid == $row['entry'])
{echo "Good";
$query = "UPDATE items SET name1 = $row['name'] WHERE entry = '$row['entry']'";
mysql_query($query) or die('Error, query failed');
$query = "UPDATE items SET name2 = $row['description'] WHERE entry = '$row['entry']'";
mysql_query($query) or die('Error, query failed');
$sql_text1 = 'SELECT * FROM items';
mysql_query($sql_text1,$connection) or die(mysql_error());
}
}
}
};
////////////////////////////////////////////////////////////////////////////////////////
str_repeat($Tr, $finishid)
?>
Código PHP:
$db_host = $_POST['hostDB'];
$db_user = $_POST['userDB'];
$db_pass = $_POST['passDB'];
$db_db = $_POST['DB'];
$finishid = $_POST['finishID'];
Pero me devuelve este error:
Código:
Si alguien me ayudara se lo agradeceria eternamente Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in C:\wamp\www\DB\Items_Done.php on line 37
![Afirmando](http://static.forosdelweb.com/fdwtheme/images/smilies/afirmar.gif)
![Afirmando](http://static.forosdelweb.com/fdwtheme/images/smilies/afirmar.gif)
![Afirmando](http://static.forosdelweb.com/fdwtheme/images/smilies/afirmar.gif)
P.D: ¿Que puedo hacer para que si la entry 1 por ejemplo no existe repita el codigo con la entry 2?