問題描述

我有一個 mysqli 查詢，我需要將其格式化為移動應用程序的 JSON.

我已經設法為查詢結果生成了一個 XML 文檔，但是我正在尋找更輕量級的東西.(請參閱下面我當前的 XML 代碼)

$mysql = new mysqli(DB_SERVER,DB_USER,DB_PASSWORD,DB_NAME) or die('連接數據庫時出現問題');$stmt = $mysql->prepare('SELECT DISTINCT title FROM Section ORDER BY title ASC');$stmt->execute();$stmt->bind_result($title);//創建xml格式$doc = new DomDocument('1.0');//創建根節點$root = $doc->createElement('xml');$root = $doc->appendChild($root);//為每一行添加節點while($row = $stmt->fetch()) :$occ = $doc->createElement('data');$occ = $root->appendChild($occ);$child = $doc->createElement('section');$child = $occ->appendChild($child);$value = $doc->createTextNode($title);$value = $child->appendChild($value);終了；$xml_string = $doc->saveXML();header('Content-Type: application/xml; charset=ISO-8859-1');//輸出 xml jQuery 就緒回聲 $xml_string;

$mysqli = new mysqli('localhost','user','password','myDatabaseName');$myArray = array();if ($result = $mysqli->query("SELECT * FROM phase1")) {while($row = $result->fetch_array(MYSQLI_ASSOC)) {$myArray[] = $row;}回聲 json_encode($myArray);}$result->close();$mysqli->close();

1. $row = $result->fetch_array(MYSQLI_ASSOC)
2. $myArray[] = $row

輸出如下:

如果你想要另一種風格，你可以試試這個:

1. $row = $result->fetch_row()
2. $myArray[] = $row

輸出如下:

I have a mysqli query which I need to format as JSON for a mobile application.

I have managed to produce an XML document for the query results, however I am looking for something more lightweight. (See below for my current XML code)

$mysql = new mysqli(DB_SERVER,DB_USER,DB_PASSWORD,DB_NAME) or die('There was a problem connecting to the database');

$stmt = $mysql->prepare('SELECT DISTINCT title FROM sections ORDER BY title ASC');
$stmt->execute();
$stmt->bind_result($title);

// create xml format
$doc = new DomDocument('1.0');

// create root node
$root = $doc->createElement('xml');
$root = $doc->appendChild($root);

// add node for each row
while($row = $stmt->fetch()) : 

    $occ = $doc->createElement('data');  
    $occ = $root->appendChild($occ);  

    $child = $doc->createElement('section');  
    $child = $occ->appendChild($child);  
    $value = $doc->createTextNode($title);  
    $value = $child->appendChild($value);  

endwhile;

$xml_string = $doc->saveXML();  

header('Content-Type: application/xml; charset=ISO-8859-1');

// output xml jQuery ready

echo $xml_string;

$mysqli = new mysqli('localhost','user','password','myDatabaseName');
$myArray = array();
if ($result = $mysqli->query("SELECT * FROM phase1")) {

    while($row = $result->fetch_array(MYSQLI_ASSOC)) {
            $myArray[] = $row;
    }
    echo json_encode($myArray);
}

$result->close();
$mysqli->close();

1. $row = $result->fetch_array(MYSQLI_ASSOC)
2. $myArray[] = $row

output like this:

[
    {"id":"31","name":"pruduct_name1","price":"98"},
    {"id":"30","name":"pruduct_name2","price":"23"}
]

If you want another style, you can try this:

1. $row = $result->fetch_row()
2. $myArray[] = $row

output will like this:

[
    ["31","pruduct_name1","98"],
    ["30","pruduct_name2","23"]
]

這篇關于如何將 mysqli 結果轉換為 JSON?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！