Question

我有以下功能，在单词中应该获取用户凭据并验证它们。如果它们有效，则将它们插入到数据库中。插入后检查它们是否已成功插入。如果没有返回false。如果电子邮件首先无效，则返回false。 Eclipse + Aptana插件报告了最后一个else部分的语法错误。当我尝试合理化代码时，它似乎应该正常运行。我可以确认所有{方括号}都是正确的层次结构

以下是代码：

   public function storeUser($name, $email, $password) {
    $uuid = uniqid('', true);
    $hash = $this->hashSSHA($password);
    $encrypted_password = $hash["encrypted"]; // encrypted password
    $salt = $hash["salt"]; // salt
    if((filter_var($email, FILTER_VALIDATE_EMAIL)) && preg_match('/@.+\./', $email)); { //validate the email 
    $result = mysql_query("INSERT INTO users(unique_id, name, email, encrypted_password, salt, created_at) VALUES('$uuid', '$name', '$email', '$encrypted_password', '$salt', NOW())"); 

    // check for successful store
    if ($result) {
        // get user details 
        $uid = mysql_insert_id(); // last inserted id
        $result = mysql_query("SELECT * FROM users WHERE uid = $uid");
        // return user details
        return mysql_fetch_array($result);
    } else {
        return false; //unsuccessful store
    }
     else { //THIS IS LINE 45
        //not a valid email
        return false;
    }
    }
}

它返回以下错误：

05-23 11:07:10.758: E/JSON(1601): <b>Parse error</b>:  syntax error, unexpected T_ELSE in <b>/home/content/40/8529140/html/webapps/projects/functions/DB_Functions.php</b> on line <b>45</b><br />
05-23 11:07:10.758: E/JSON Parser(1601): Error parsing data org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject

Answer 1

if((filter_var($email, FILTER_VALIDATE_EMAIL)) 
      && preg_match('/@.+\./', $email)); { //<<<REMOVE THE ;

Answer 2

关于结束括号，您的else声明处于错误的位置。如果您在第二个}之前移动其中一个else，它将起作用。实际上也许它不会，你也有一个流浪;，但其他建议仍然存在。

如果您更加一致地缩进代码，您将不会被这类问题所困扰。

public function storeUser($name, $email, $password) {
    $uuid = uniqid('', true);
    $hash = $this->hashSSHA($password);
    $encrypted_password = $hash["encrypted"]; // encrypted password
    $salt = $hash["salt"]; // salt
    if ( //validate the email 
         filter_var($email, FILTER_VALIDATE_EMAIL) &&
         preg_match('/@.+\./', $email)
         ) {
        $result = mysql_query (
            "INSERT INTO users(unique_id, name, email, encrypted_password, salt, created_at) VALUES('$uuid', '$name', '$email', '$encrypted_password', '$salt', NOW())"
        );
        if ($result) { // check for successful store
            // get user details 
            $uid = mysql_insert_id(); // last inserted id
            $result = mysql_query("SELECT * FROM users WHERE uid = $uid");
            // return user details
            return mysql_fetch_array($result);
        } else {
            return false; //unsuccessful store
        }
    } else {
        //not a valid email
        return false;
    }
}

另外：不要使用mysql_*功能。学习使用PDO和参数化查询。

Answer 3

您没有正确关闭大括号，并且在外部if后面有一个额外的分号：

if((filter_var($email, FILTER_VALIDATE_EMAIL)) && preg_match('/@.+\./', $email)) { //validate the email 
   $result = mysql_query("INSERT INTO users(unique_id, name, email, encrypted_password, salt, created_at) VALUES('$uuid', '$name', '$email', '$encrypted_password', '$salt', NOW())"); 

   // check for successful store
   if ($result) {
      // get user details 
      $uid = mysql_insert_id(); // last inserted id
      $result = mysql_query("SELECT * FROM users WHERE uid = $uid");
      // return user details
      return mysql_fetch_array($result);
   } else {
     return false; //unsuccessful store
   }
} // close of outer if
else {
    //not a valid email
    return false;
}

Answer 4

你看到这个小分号吗？ if((filter_var($email, FILTER_VALIDATE_EMAIL)) && preg_match('/@.+\./', $email));

Answer 5

你错放了一个支架。试试这个

 public function storeUser($name, $email, $password) {
    $uuid = uniqid('', true);
    $hash = $this->hashSSHA($password);
    $encrypted_password = $hash["encrypted"]; // encrypted password
    $salt = $hash["salt"]; // salt
    if((filter_var($email, FILTER_VALIDATE_EMAIL)) && preg_match('/@.+\./', $email)) { //validate the email 
    $result = mysql_query("INSERT INTO users(unique_id, name, email, encrypted_password, salt, created_at) VALUES('$uuid', '$name', '$email', '$encrypted_password', '$salt', NOW())"); 

    // check for successful store
    if ($result) {
        // get user details 
        $uid = mysql_insert_id(); // last inserted id
        $result = mysql_query("SELECT * FROM users WHERE uid = $uid");
        // return user details
        return mysql_fetch_array($result);
    } else {
        return false; //unsuccessful store
    }
    }
     else {
        //not a valid email
        return false;
    }

}

Answer 6

为什么最后有两个else {}块...？你只能有一个。

如果Else语句语法错误

6 个答案: