将mysqli结果转换为json

Question

我有一个mysqli查询,我需要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级的东西.(请参阅下面的我当前的xml代码)

任何帮助或信息都非常感谢人们!

$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();

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

输出如下:

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

如果你想要另一种风格,你可以试试这个:

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

输出将是这样的:

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

Answer 2

这是我如何制作我的json feed:

$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

如上所述,json_encode会对你有所帮助.最简单的方法是获取您已经完成的结果,并构建一个可以传递给的数组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将成为一个常规数组,其中每个元素都有自己的索引.

在上面的代码中应该没什么变化,或者,你可以返回XML和JSON,因为大多数代码是相同的.

Answer 4

JSON 有一件重要的事情——数据必须是 UTF-8 编码的。因此，必须为数据库连接设置正确的编码。

其余的与任何其他数据库操作一样愚蠢

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$db = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME);
$db->set_charset('utf8mb4');

$sql = 'SELECT DISTINCT title FROM sections ORDER BY title ASC';
$data = $db->query($sql)->fetch_all(MYSQLI_ASSOC);
echo json_encode($data);