Question

我有一个验证检查器，检查数据库以查看＆＃34;电子邮件和用户名＆＃34;正在使用，如果是，那么回复电子邮件和用户名为＆＃34; false＆＃34;。如果它不在数据库中，则回显到＆＃34; true＆＃34;。我将返回用户电子邮件和用户名在sql服务器中的结果。所以我想追加＆＃34; $ json1或$ json2＆＃34;到第一个或第二个＆＃34; $像这样

 {"Email":false, "Username":true};

PHP代码：

<?php

if(isset($_GET['submit'])){

    //Connect to database
    $link = mysqli_connect("localhost", "root", "");
    mysqli_select_db($link, "magicsever");

    if(mysqli_connect_error()){

        die ("Database connection error");
    }

    //Find the email in the database
    $query = "SELECT * FROM app_signup WHERE email = '".mysqli_real_escape_string($link, $_GET['email'])."'";
    $query2 = "SELECT * FROM app_signup WHERE username = '".mysqli_real_escape_string($link, $_GET['username'])."'";

    $result = mysqli_query($link, $query);
    $result2 = mysqli_query($link, $query2);

    $email_checker1 = json_encode(array("Email"=>false));
    $email_checker2 = json_encode(array("Email"=>true));

    $username_checker1 = json_encode(array("Username"=>false));
    $username_checker2 = json_encode(array("Username"=>true));

    if(mysqli_num_rows($result)>0){

        echo $email_checker1;


    }

}
?>

Answer 1

为电子邮件和用户名结果制作新数组

//Find the email in the database
$query = "SELECT * FROM app_signup WHERE email = '".mysqli_real_escape_string($link, $_GET['email'])."'";
$query2 = "SELECT * FROM app_signup WHERE username = '".mysqli_real_escape_string($link, $_GET['username'])."'";

$result = mysqli_query($link, $query);
$result2 = mysqli_query($link, $query2);

$dataResult = array();
$dataResult['Email'] = (mysqli_num_rows($result)>0)? false: true;
$dataResult['Username'] = (mysqli_num_rows($result2)>0)? false: true;

echo json_encode($dataResult );

Answer 2

JSON只是一种数据格式，是一种呈现数据结构的方式。在您拥有可以呈现的数据结构之前，您不希望将其转换为JSON。

您应该做的是构建一个包含验证错误的数组：

// I'm assuming that the true/false means that there was an error with validating that value
// You can initialize the validation array with false values for all fields and change them to true in case of validation error
$errors = ['Email' => false, 'Username' => false];

// Then, change the values in case of an error
if (... /* invalid email */) {
    $errors[] = ['Email' => true];
}
if (... /* invalid username */) {
    $errors[] = ['Username' => true];
}

然后仅在显示数组时将数组转换为JSON：

echo json_encode($errors);

Answer 3

为什么不在数组附加后执行json_encode。

$first = json_encode(array("Email"=>false, "Username"=>false));

加入。

$data['Email'] = false;
if('''')
{
  $data['Username'] = false;
}
$result = json_encode($data);

Answer 4

 $first = json_encode(array("Email"=>false));
    $second = json_encode(array("Email"=>true));

    $json1 = json_encode(array("Username"=>false));
    $json2 = json_encode(array("Username"=>true));

像这样使用

$first = json_encode(array("Email"=>false)+array("Username"=>true));
$second = json_encode(array("Email"=>true)+array("Username"=>false));

Answer 5

如果您真的希望将它们合并在一起，请阅读此参考资料

好的方法就像

$resData1['email']=false;
$resData2['email']=true;

$resData1['Username']=false;
$resData2['Username']=true;

$json1 = json_encode($resData1);
$json2 = json_encode($resData2);

如何附加到json_encode数组

5 个答案: