Question

我试图从PHP / PDO返回json但是我在Swift中得到了这个错误。

错误域= NSCocoaErrorDomain代码= 3840＆＃34;垃圾结束。＆＃34; UserInfo = {NSDebugDescription =结尾处的垃圾。}

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()


}

Agin迅速的错误是错误域= NSCocoaErrorDomain代码= 3840＆＃34;垃圾结束。＆＃34; UserInfo = {NSDebugDescription =结尾处的垃圾。}

我只是没有看到它。当我从网页上运行测试时，我得到的json看起来像这样。

{＆＃34;用户＆＃34;：[{＆＃34; IDDB＆＃34;：＆＃34; 383＆＃34;＆＃34; addressNotsDB＆＃34;：＆＃34; \ n＆＃34 ;，＆＃34; alternateNameDB＆＃34;：＆＃34;＆＃34;＆＃34; alternateNumberDB＆＃34;：＆＃34;＆＃34;＆＃34; businessMainCategoryDB＆＃34;：＆＃34 ;新闻＆＃34;＆＃34; businessNameDB＆＃34;：＆＃34; TODAY＆＃34;}]}

应用程序调用GetCards.php这会调用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

您的问题是服务器发送的数据，首先是：

echo json_encode($row);

为什么垃圾到底？您应该发回 json ，为什么要添加错误的 html 标签？

{}{}

其次，为什么要从功能中打印？

您定义了您的函数，对于它所获取的每一行，它将打印一个JSON对象。在发布的示例中，您似乎只有一个单行，多行一行会失败，因为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; }不是有效的JSON对象。

将其更改为：

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

然后你的来电者应该打印出来：

{{1}}

Answer 2

代码在每行的末尾回显"<br>"."<br>"，使JSON结构无效。您无法看到这一点，因为此PHP输出了一个HTML响应，在这种情况下"<br>"."<br>"只添加一些行...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客户端是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，请尝试将此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)
            }


})

如何在Swift中从PHP返回JSON？

3 个答案: