PHP / jQuery AJAX JSON错误

Question

当我尝试使用jQuery Ajax从PHP页面接收代码时，我发现一个奇怪的错误：“未定义变量：错误”

<?php
$errors = array("already_signed" => "You are already signed in", "field_empty" => "All fields must be filled", "long_username" => "Username length must be less then 40 symbols", "incorrect_email" => "Your mail is incorrent", "user_exists" => "User with such username already exists", "account_not_found" => "Account not found", "passwords_arent_same" => "Passwords must be the same");
Function check_post() {
    $answer = array("ok" => "false", "answer" => $errors["field_empty"]);
    echo json_encode($answer);
}

check_post();
?>

如果我没有功能就回声-一切都会好的。 谢谢你的帮助

Answer 1

您似乎在其中至少缺少一个} 。 照原样，您的函数定义没有结束，因此它读为无限递归调用。

同样，您在函数之外定义了$errors 。 PHP不允许“较低”的代码作用域查看在较高作用域中定义的变量。 您需要在函数内将$ errors声明为全局变量：

<?php

$errors = array(....);
function check_post() {
   global $errors;
   $answer = ...
   ...
}

check_post();

Answer 2

您正在尝试从函数内部访问全局变量。 为了实现这一点，您需要使用“ global”关键字：

<?php
$errors = array("already_signed" => "You are already signed in", "field_empty" => "All fields must be filled", "long_username" => "Username length must be less then 40 symbols", "incorrect_email" => "Your mail is incorrent", "user_exists" => "User with such username already exists", "account_not_found" => "Account not found", "passwords_arent_same" => "Passwords must be the same");
function check_post() {
    global $errors;
    $answer = array("ok" => "false", "answer" => $errors["field_empty"]);
    echo json_encode($answer);
}
check_post();
?>