Question

现在我的代码：

$sql = "SELECT * FROM schedule_feednow WHERE IDMac = '".$_GET["IDMac"]."'";
$result = mysql_query($sql) or die(mysql_error());
$json_data = array();

while($rec = mysql_fetch_assoc($result)){
	$Subjson = array();
	$Subjson['IDMac'] = $rec['IDMac'];
	$Subjson['DATETIME'] = $rec['DATETIME'];
	$Subjson['Weight'] = $rec['Weight'];
	
	array_push($json_data,$Subjson);
	

}

echo json_encode ($json_data);

和结果：

[{"IDMac":"C-01","DATETIME":"12:05:19 AM on June 23, 2017","Weight":"50"},{"IDMac":"C-01","DATETIME":"12:05:55 AM on June 23, 2017","Weight":"50"},{"IDMac":"C-01","DATETIME":"02:02:20 PM on June 23, 2017","Weight":"50"}]

但我不想要这个结果

我想要这个结果 - ＆gt;

["Schedule":{"IDMac":"C-01","DATETIME":"12:05:19 AM on June 23, 2017","Weight":"50"},{"IDMac":"C-01","DATETIME":"12:05:55 AM on June 23, 2017","Weight":"50"},{"IDMac":"C-01","DATETIME":"02:02:20 PM on June 23, 2017","Weight":"50"}]

请帮我生成代码或训练我。非常感谢你。

Answer 1

只需将当前结果放入带有“Schedule”键的数组中：

echo json_encode(array('Schedule' => $json_data));

Answer 2

只需创建另一个数组并将您的json数据放入：

$newJsonData = ['Schedule' => $json_data];
echo json_encode ($newJsonData);

Answer 3

替换

echo json_encode ($json_data);

与

$myResult = ['Schedule' => $json_data];
echo json_encode ($myResult);

Answer 4

＆＃13;

$sql = "SELECT * FROM schedule_feednow WHERE IDMac = '".$_GET["IDMac"]."'";
$result = mysql_query($sql) or die(mysql_error());
$json_data = array();

while($rec = mysql_fetch_array($result)){
	$json_data[] = $rec;
}

echo json_encode ('Schedule' => $json_data);

＆＃13;

使用php json编码选择mysql

4 个答案: