2016-02-26 181 views
-3

我是PHP的初學者,我無法解決此錯誤,感謝您的支持answers.It是一個簡單的連接到數據庫MySQL中的列:

Name 
Last Name 
Email 

解析錯誤:

syntax error, unexpected '""' (T_CONSTANT_ENCAPSED_STRING), expecting variable (T_VARIABLE) or '{' or '$' in C:\xampp\htdocs\Suscribe\registro.php on line 7

$db_host="localhost"; 
$db_user="root"; 
$db_password=""; 
$db_name="prueba"; 
$db_table_name="datos"; 
    $db_connection = mysql_connect("$localhost", "$root",""); 

if (!$db_connection) { 
    die('No se ha podido conectar a la base de datos'); 
} 
$Nombre = utf8_decode($_POST['nombre']); 
$Apellido = utf8_decode($_POST['apellido']); 
$Email = utf8_decode($_POST['email']); 
$resultado=mysql_query("SELECT * FROM ".$datos." WHERE Email = '".$Email."'", $db_connection); 

if (mysql_num_rows($resultado)>0){ 
    header('Location: Fail.html'); 
}else{ 
    $insert_value = 'INSERT INTO `' . $prueba. '`.`'.$datos.'` (`nombre` , `apellido` , `email`) VALUES 
    ("' . $Nombre . '", "' . $Apellido . '", "' . $Email . '")'; 
    mysql_select_db($prueba, $db_connection); 
    $retry_value = mysql_query($insert_value, $db_connection); 
    if (!$retry_value){ 
     die('Error: ' . mysql_error()); 
    } 
    header('Location: Success.html'); 
} 
mysql_close($db_connection); 
+1

檢查[此](http://stackoverflow.com/questions/18050071/php-parse-syntax-errors-and-how-to-solve-them)進行。對於初學者來說最好的書籤;) – giorgio

回答

0

mysql已被棄用,你應該使用mysqlipdo

變化

$db_connection = mysql_connect("$localhost", "$root",""); 

$db_connection = mysql_connect($localhost, $root,""); 

還有一件事你沒有連接DB

,你都一定MySQL後,寫

mysql_select_db($db_name); 

_connect

0

試試這個

<?php 

$db_host="localhost"; 
$db_user="root"; 
$db_password=""; 
$db_name="prueba"; 
$db_table_name="datos"; 
    $db_connection = mysql_connect($db_host, $db_user,$db_password); 

if (!$db_connection) { 
    die('No se ha podido conectar a la base de datos'); 
} 
$Nombre = utf8_decode($_POST['nombre']); 
$Apellido = utf8_decode($_POST['apellido']); 
$Email = utf8_decode($_POST['email']); 

$resultado=mysql_query("SELECT * FROM ".$db_table_name." WHERE Email = '".$Email."'", $db_connection); 

if (mysql_num_rows($resultado)>0) 
{ 

header('Location: Fail.html'); 

} else { 

    $insert_value = 'INSERT INTO `' . $prueba. '`.`'.$datos.'` (`nombre` , `apellido` , `email`) VALUES 
    ("' . $Nombre . '", "' . $Apellido . '", "' . $Email . '")'; 

mysql_select_db($db_name, $db_connection); 
$retry_value = mysql_query($insert_value, $db_connection); 

if (!$retry_value) { 
    die('Error: ' . mysql_error()); 
} 

header('Location: Success.html'); 

} 

mysql_close($db_connection); 

我敢變化 這$db_connection = mysql_connect("$localhost", "$root","");

$db_connection = mysql_connect($db_host, $db_user,$db_password);

$resultado=mysql_query("SELECT * FROM ".$datos." WHERE Email = '".$Email."'", $db_connection);

$resultado=mysql_query("SELECT * FROM ".$db_table_name." WHERE Email = '".$Email."'", $db_connection); 和 這

`mysql_select_db($prueba, $db_connection);` 

mysql_select_db($db_name, $db_connection);

+0

總是試圖指出錯誤,然後提供解決方案。不像寫整個劇本。所以最好指導他/她對他/她所犯的錯誤。 –

+0

@Maha Dev soryy –

+0

@naseebac:插入查詢是錯誤的,並有語法錯誤。選擇查詢應該是: - $ resultado = mysql_query(「SELECT * FROM $ db_table_name WHERE Email ='$ Email'」); –

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