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 rowwhile($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 readyecho $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。