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发表于
2020-8-9 20:24:02
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来自 山东
<?php
header("Content-type:text/html; charset=utf-8");
// 获取form表单值
$username = $_POST['username'];
$password = $_POST['password'];
// 判断form表单中key
if(isset($_POST['username']) && isset($_POST['password'])){
// 判断username和password
if($username == "zhangwei" && $password == "123456"){
$result = array("success" => 1, "code" => 101, "data" => array("username" => $username, "password" => $password));
}else{
$result = array("success" => 0, "code" => 103, "data" => null);
}
}else{
$result = array("success" => 0, "code" => 100, "data" => null);
}
// 将错误信息(数组)转换成json类型,返回前端
echo(json_encode($result));
?> |
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