Question

我有一个PHP脚本，根据用户输入的表单数据向用户发送电子邮件，但是当我在字符串中包含变量时，所有的电子邮件都是“0”

<?php
    $u_name = $_POST['u_name'];
    $email = $_POST['email'];
    $password = $_POST['password'];
    $to = $_POST['email'];
    $subject = "Site Activation";
    $message = "Hello " + $u_name + ",";
    $from = "admin@mytest4389.freeiz.com";
    $headers = "From:" . $from;
    mail($to,$subject,$message,$headers);
    echo "Mail Sent.";
?>

Answer 1

php不添加带“+”的字符串使用“。”对于concat字符串。

这是固定代码：

<?php
    $u_name = $_POST['u_name'];
    $email = $_POST['email'];
    $password = $_POST['password'];
    $to = $_POST['email'];
    $subject = "Site Activation";
    $message = "Hello " . $u_name . ",";
    $from = "admin@mytest4389.freeiz.com";
    $headers = "From:" . $from;
    mail($to,$subject,$message,$headers);
    echo "Mail Sent.";
?>

Answer 2

就像@MahdiParsa所说的那样，在PHP中，连接是用.进行的，但你也可以去做（取决于每个人的口味）：

$firstVariable = "Hey";
$secondVariable = "How are you?";

$mixedVariable1 = $firstVariable . ' Alex! ' . $secondVariable;
$mixedVariable2 = "$firstVariable Alex! $secondVariable";
$mixedVariable3 = "{$firstVariable} Alex! {$secondVariable}";
  // $mixedVariable3 is the same as $mixedVariable2,
  // only a different type of Variable Substitution 
  // Take a look at:
    // http://framework.zend.com/manual/1.12/en/coding-standard.coding-style.html

http://codepad.org/1aYvxjiC

PHP邮件功能无法正常工作

2 个答案: