如何在 Swift 中从 PHP 返回 JSON？

Question

我正在尝试从 PHP/PDO 返回 json，但在 Swift 中出现此错误。

错误域=NSCocoaErrorDomain 代码=3840 “垃圾结束。” UserInfo={NSDebugDescription=垃圾结束。}

这是 PHP 文件。

//*FUNCTION TO GET CARD FROM SEARCH WORD CALLED FROM GetCards.php   
public function getAllCards($word) {

//Connect to db using the PDO not PHP
$db = new PDO('mysql:host=localhost;dbname=xxxx', 'xxxx', 'xxxxx');

//Here we prepare the SELECT statement from the search word place holder :word
$sql = $db->prepare('SELECT * FROM carddbtable WHERE businessNameDB=:word OR lastNameDB=:word OR firstKeywordDB=:word OR    secondKeywordDB=:word OR thirdKeywordDB=:word OR fourthKeywordDB=:word OR fithKeywordDB=:word');

//We execute the $sql with the search word variable"$word"
$sql->execute([':word' => $word]);

//Looping through the results
while ($row = $sql->fetch(PDO::FETCH_ASSOC)) {

//Print to screen
 //  echo json_encode($row). "<br>"."<br>";

//Store all return rows in $returnArray
$returnArray[] = $row;
}

//Feedback results
return $returnArray;

}  

这是 Swift。

    //Search and retrieve card / users
func doSearch (word : String) {

    //Search word from searchKeyWordVar
    let word = "TODAY"

    // URL path to GetCards.php
    let url = NSURL(string: "http://www.xxxxxxx.com/xxx/xx/GetCards.php")

    //Create URL request
    let request = NSMutableURLRequest(url: url! as URL)

    //Method to pass info to GetCards.php
    request.httpMethod = "POST"

    //body that passing info to php
//        let body = "word=\(word)"
    let body = "TODAY" //This is hard coded for testing

    //convert string to utf8 for all languages
    request.httpBody = body.data(using: String.Encoding.utf8)


    //Launch session
    URLSession.shared.dataTask(with: request as URLRequest) { (Data, response, error) in


        //Get main Queue
        DispatchQueue.main.async(execute: {

            if error == nil {

                do {
                    // declare json var to store $returnArray inf we got from GetCards.php
                    let json = try JSONSerialization.jsonObject(with: Data!, options: .mutableContainers) as? NSDictionary


                    // delcare new secure var to store json
                    guard let parseJSON = json else {
                        print("Error while parsing")
                        return
                    }

                    // declare new secure var to store $returnArray["users"]
                    guard let parseUSERS = parseJSON["users"] else {
                        print(parseJSON["message"] ?? [NSDictionary]())
                        return
                    }

                } catch {
                    print(error)
                }


                } else {
                    print(error as Any)
                }


    })

}.resume()


}

快速错误是 错误域 = NSCocoaErrorDomain 代码 = 3840 “垃圾结束。” UserInfo={NSDebugDescription=垃圾结束。}

我只是没看到。当我从网页运行测试时，我得到的 json 看起来像这样。

{"users":[{"idDB":"383","addressNotsDB":"\n","alternateNameDB":"","alternateNumberDB":"","businessMainCategoryDB":"News"," businessNameDB":"TODAY"}]}

应用调用 GetCards.php 这调用 DBopperation.php 这有一个名为 getAllCards($word) 的 public_function

这是GetCards.php

//STEP: 1 Make connection to DB
//Including the db operation file for connection to DB
$cardConnect = require_once 'DbOperation.php';

//Checking if there is a connection to DB
if ($cardConnect) {
$returnArray1['Connected to DB'] = '200';
} else {
$returnArray1['Did not connect ot DB'] = '400';
}
//echo json_encode($returnArray1). "<br>"."<br>";


//STEP: 2 Connecting to Public Function
//Connecting the DbOperation.php file public fuction getAllCards to the       variable $card
$card = new DbOperation();

//If connected  
if ($card) {

//Checking connection to the DbOperation.php 
    $returnArray2['Connected to GetCards.php'] = '200';
} else {
    $returnArray2['Could not connect to GetCards'] = '400';
}
//echo json_encode($returnArray2). "<br>"."<br>";


//STEP: 3 Running the search
//Creating a varable to hold the search word and setting it to null
$word = null;

//Getting to search word from the app
if (!empty($_REQUEST["word"])) {
    $word = htmlentities($_REQUEST["word"]);
}
// STEP 4. Access searching func and retrieve data from server
$users = $card->getAllCards($word);

if (!empty($users)) {
    $returnArray3["users"] = $users;
} else {
    $returnArray3["message"] = 'Could not find records in GetCards';
}


// STEP 4. Close connection
$card->disconnect();

// STEP 5. Pass information back as json to user
echo json_encode($returnArray);

Answer 1

首先，您的问题在于服务器发送的数据：

echo json_encode($row). "<br>"."<br>";

为什么最后是垃圾？你应该发回 json，你为什么要添加错误的 html 标签？

echo json_encode($row);

其次，为什么要从函数中打印？

你定义你的函数，对于它获取的每一行，将打印一个 JSON 对象。在发布的示例中，您似乎只有一行，多于一行这将失败，因为{}{} 不是有效的 JSON 对象。

改成这样：

public function getAllCards($word) {

    ...

    //Looping through the results
    while ($row = $sql->fetch(PDO::FETCH_ASSOC)) {

        //Store all return rows in $returnArray
        $returnArray[] = $row;
    }

    //Feedback results
    return $returnArray;
}  

那么你的调用者应该打印它：

$arrayOfResults = getAllCards();
echo json_encode( $arrayOfResults );

Answer 2

代码在每一行的末尾回显"<br>"."<br>"，使 JSON 结构无效。你看不到这一点，因为这个 PHP 输出了一个 HTML 响应，在这种情况下，"<br>"."<br>" 只是添加了一些行......但是对于解析 JSON，它们是无效的。

我认为最正确的是将数组输出为 JSON 并告诉 HTTP 客户端这是一个 JSON 输出，如下所示：

//*FUNCTION TO GET CARD FROM SEARCH WORD CALLED FROM GetCards.php   
public function getAllCards($word) {

//Connect to db using the PDO not PHP
$db = new PDO('mysql:host=localhost;dbname=xxxx', 'xxxx', 'xxxxx');

//Here we prepare the SELECT statement from the search word place holder :word
$sql = $db->prepare('SELECT * FROM carddbtable WHERE businessNameDB=:word OR lastNameDB=:word OR firstKeywordDB=:word OR    secondKeywordDB=:word OR thirdKeywordDB=:word OR fourthKeywordDB=:word OR fithKeywordDB=:word');

//We execute the $sql with the search word variable"$word"
$sql->execute([':word' => $word]);

//Empty the returnArray
$returnArray = array();

//Looping through the results
while ($row = $sql->fetch(PDO::FETCH_ASSOC)) {

//Store all return rows in $returnArray
$returnArray[] = $row;
}

// Tell that is a JSON output
header('Content-Type: application/json');

//Feedback results
return json_encode($returnArray);

}  

Answer 3

如果您的 JSON 输出是真正合法的 JSON，请尝试将您的 DispatchQueue 替换为：

    DispatchQueue.main.async(execute: {

        if error == nil {

            do {

                guard let jsonData = Data? else{
                    return
                }

                let json = try? JSONSerialization.jsonObject(with: jsonData)

                guard let parseJSONDict = json as? [String : Any] else{
                    print("Error while parsing")
                    return
                }

                guard let parseUSERS = parseJSONDict["users"] else{
                    return
                }


            } catch {
                print(error)
            }


            } else {
                print(error as Any)
            }


})