如何将mysqli结果转换为JSON？

Question

我有一个mysqli查询，我需要将其格式化为移动应用程序的JSON。

我已经设法为查询结果生成一个XML文档，但是我正在寻找更轻量的东西。（有关我当前的XML代码，请参见下文）

$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;

Answer 1

$mysqli = new mysqli('localhost','user','password','myDatabaseName');$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();$row = $result->fetch_array(MYSQLI_ASSOC)$myArray[] = $row这样的输出：[    {"id":"31","name":"pruduct_name1","price":"98"},    {"id":"30","name":"pruduct_name2","price":"23"}]如果您想要其他样式，可以尝试以下方法：$row = $result->fetch_row()$myArray[] = $row输出将是这样的：[    ["31","pruduct_name1","98"],    ["30","pruduct_name2","23"]]

Answer 2

这是我制作JSON Feed的方法：$mysqli = new mysqli('localhost', 'user', 'password', 'myDatabaseName');$myArray = array();if ($result = $mysqli->query("SELECT * FROM phase1")) {    $tempArray = array();    while ($row = $result->fetch_object()) {        $tempArray = $row;        array_push($myArray, $tempArray);    }    echo json_encode($myArray);}$result->close();$mysqli->close();

Answer 3

如前所述，json_encode将为您提供帮助。最简单的方法是在获取结果的同时获取结果，并建立可以传递给的数组json_encode。例：$json = array();while($row = $stmt->fetch()){  $json[]['foo'] = "your content  here";  $json[]['bar'] = "more database results";}echo json_encode($json);您$json将是一个常规数组，其中每个元素都位于其自己的索引中。上面的代码应该几乎没有什么变化，或者，因为大多数代码是相同的，所以您可以返回XML和JSON。