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;
答案 0 :(得分:64)
$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
输出如下:
[
{"id":"31","name":"pruduct_name1","price":"98"},
{"id":"30","name":"pruduct_name2","price":"23"}
]
如果您想要其他款式,可以试试这个:
$row = $result->fetch_row()
$myArray[] = $row
输出将是这样的:
[
["31","pruduct_name1","98"],
["30","pruduct_name2","23"]
]
答案 1 :(得分:16)
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();
答案 2 :(得分:13)
如上所述,json_encode
会对您有所帮助。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json_encode
的数组。
示例:
$json = array();
while($row = $stmt->fetch()){
$json[]['foo'] = "your content here";
$json[]['bar'] = "more database results";
}
echo json_encode($json);
您的$json
将是一个常规数组,其中每个元素都有自己的索引。
上面的代码应该没有什么变化,因此,你可以返回XML和JSON,因为大多数代码是相同的。
答案 3 :(得分:3)
我设法运行此代码:
<?php
//create an array
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
$emparray[] = $row;
}
return json_encode($emparray);
?>
答案 4 :(得分:0)
如果您在php中安装了mysqlnd扩展程序+,则可以使用:
$mysqli = new mysqli('localhost','user','password','myDatabaseName');
$result = $mysqli->query("SELECT * FROM phase1");
//Copy result into a associative array
$resultArray = $result->fetch_all(MYSQLI_ASSOC);
echo json_encode($resultArray);
mysqli :: fetch_all()需要先安装mysqlnd驱动程序 可以使用它。
答案 5 :(得分:0)
关于JSON有一件重要的事情-数据必须必须采用UTF-8编码。因此,必须为数据库连接设置正确的编码。
其余的与其他任何数据库操作一样愚蠢
$mysql = new mysqli(DB_SERVER,DB_USER,DB_PASSWORD,DB_NAME);
$mysql->set_charset('utf8mb4');
$sql = 'SELECT DISTINCT title FROM sections ORDER BY title ASC';
$data = $mysql->query(sql)->fetch_all(MYSQLI_ASSOC);
echo json_encode($data);