Question

由于某种原因，我得到未定义的变量和字符串到数组转换我不明白为什么会发生这些事情

Notice:  Undefined variable: body in C:\Users\New\Desktop\xampp\htdocs\yakov\sendemail.php</b> on line <b>20

Notice:  Array to string conversion in C:\Users\New\Desktop\xampp\htdocs\yakov\sendemail.php on line 33
"Array'services.html:  ''services.html'\n\n'new york:  ''new york'\n\n'new york:  ''new york'\n\n'round_trip:  ''round_trip'\n\n'2016-09-16:  ''2016-09-16'\n\n'2016-09-23:  ''2016-09-23'\n\n'nonstop:  ''nonstop'\n\n'flexible:  ''flexible'\n\n'Business:  ''Business'\n\n'1 Adult:  ''1 Adult'\n\n'some:  ''some'\n\n'one:  ''one'\n\n'someonesemail@gmail.com:  ''someonesemail@gmail.com'\n\n'new york:  ''new york'\n\n'dsfa\n:  ''dsfa\n'\n\n'4127117117:  ''4127117117'\n\n'me:  ''me'\n\n;"

这是我的代码导致我尝试使用它的问题

<?php
    header('Content-type: application/json');
    $status = array(
        'type'=>'success',
        'message'=>'Thank you for contacting us. We will contact you as early as possible.'
    );
    //print phpinfo(); 
    error_reporting(-1);
ini_set('display_errors', 'On');
//set_error_handler("var_dump");
$body;
$email;
$subject;
$email_from;
$email_to = 'Sales@ElyonTravel.com';
if (!empty($_REQUEST)) {
    $body;
    foreach($_REQUEST as $key => $val) {
        if (isset($_REQUEST[$key])) {
            $body .= "'". $_REQUEST[$key] .":  '" . $val . "\n\n";
        }

    }
$email = isset($_REQUEST['email']) ? trim(stripslashes($_REQUEST['email'])) : "NA";
$subject = isset($_REQUEST['subject']) ? trim(stripslashes($_REQUEST['subject'])) : "NA";
    $body .= ";";
    $email_from = $email;
    //$email_to = 'Sales@Travel.com';// your email
    $body;
}
$success = mail($email_to, $subject, $body, 'From: <'.$email_from.'>');

    echo json_encode($status .$body);
//}
    die;

任何其他建议都会受到赞赏，因为我不熟悉后端

Answer 1

我不确定你对此做了什么：

$body;

我在您的代码中看到三个流浪$body;。让那些人离开那里。正如评论中所提到的，只需将其定义为顶部的字符串：

$body = "";

然后你可以将所有其他字符串连接到它。

数组到字符串错误可能是因为您正在尝试将字符串与数组连接：

echo json_encode($status .$body); 
// ^-- this won't work. $status is an array. $body is a string.

如果你只是为了好玩而回应json，你总是可以将你的身体字符串添加到那个状态数组中，然后将其回显：

$status['body'] = $body;
echo json_encode($status);

PHP中未定义的变量变量声明和数组到字符串的转换

1 个答案: