2017-04-19 117 views
-1

我在为一个项目编写一个简单的PHP验证代码,我正在验证它的名字,姓氏和电子邮件。我得到这个错误,并且我查看了它至少45分钟,并且无法弄清楚它是什么。解析错误:语法错误,第44行出现意外的'{'

<?php 
$fnameErr = $lnameErr = $emailErr = ""; 
$firstname = $lastname = $email = ""; 

if ($_SERVER["REQUEST_METHOD"] == "POST") { 
if (empty($_POST["firstname"])) { 
$fnameErr = "First name is required"; 
} 

这是线38-46

+0

检查代码仔细,在这个错误说你的代码有一个不需要的{符号。 –

回答

1

看起来像括号中的一个还没有被关闭。

<?php 
    $fnameErr = $lnameErr = $emailErr = ""; 
    $firstname = $lastname = $email = ""; 

    if ($_SERVER["REQUEST_METHOD"] == "POST") { 
     if (empty($_POST["firstname"])) { 
      $fnameErr = "First name is required"; 
     } 
    } // Add an ending brace 
0

使用IDE来帮助您解决语法错误。我认为你是在学习阶段,所以我建议你使用一个IDE(Eclipse或Netbeans的)

缺少括号:

$fnameErr = $lnameErr = $emailErr = ""; 
$firstname = $lastname = $email = ""; 

if ($_SERVER["REQUEST_METHOD"] == "POST") { 
if (empty($_POST["firstname"])) { 
$fnameErr = "First name is required"; 
} 
}//Here 
1

你已经错过了结束第二个,如果条件。代码应该是这样的。

if ($_SERVER["REQUEST_METHOD"] == "POST") { 
    if (empty($_POST["firstname"])) { 
     $fnameErr = "First name is required"; 
    } 
}//this } symbol you have missed. 
1

你错过了你的第二个条件支架,你可以更新您的代码将更具可读性

<?php 
// this is error variable 
$fnameErr = ""; 
$lnameErr = ""; 
$emailErr = ""; 

// this is post initial variable define 
$firstname = ""; 
$lastname = ""; 
$email = ""; 

if ($_SERVER["REQUEST_METHOD"] == "POST") { 
    if (empty($_POST["firstname"])) { 
    $fnameErr = "First name is required"; 
    } 
} 
?> 

了解更多信息

PHP Parse/Syntax Errors; and How to solve them?

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