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ถ้าเราเรียก json แบบมีข้อมูลเป็นชุดๆ หลายชุดอ่ะค่ะ แล้วต้องมี key ไว้ให้ฝั่ง app ใช้เรียกต้องทำไงบ้างคะ |
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จาก โค้ดนี้
Code (PHP)
<?
$objConnect = mysql_connect("mysql4.000webhost.com","a7173290_sitny66","218820sinE");
mysql_query("SET NAMES UTF8");
$objDB = mysql_select_db("a7173290_AR");
$strSQL = "SELECT * FROM building_food WHERE 1 ";
$objQuery = mysql_query($strSQL);
$intNumField = mysql_num_fields($objQuery);
$resultArray = array();
while($obResult = mysql_fetch_array($objQuery))
{
$arrCol = array();
for($i=0;$i<$intNumField;$i++)
{
$arrCol[mysql_field_name($objQuery,$i)] = $obResult[$i];
}
array_push($resultArray,$arrCol);
}
mysql_close($objConnect);
$json = json_encode($resultArray);
echo substr($json, 1, strlen($json)-2);
?>
ผลลัพธ์ json ก็จะออกมาประมาณนี้
{"building_id":"1","building_name":"true coffee (\u0e15\u0e36\u0e01\u0e20\u0e32\u0e04\u0e27\u0e34\u0e0a\u0e32\u0e27\u0e34\u0e28\u0e27\u0e01\u0e23\u0e23\u0e21\u0e44\u0e1f\u0e1f\u0e49\u0e32)","latitude":"13.7359396","longitude":"100.5320388","picture":""},{"building_id":"2","building_name":"true coffee (\u0e15\u0e36\u0e01 3)","latitude":"13.7368873","longitude":"100.5326165","picture":""},{"building_id":"3","building_name":"i-canteen","latitude":"13.7366251","longitude":"100.5339695","picture":""},{"building_id":"4","building_name":"\u0e2d\u0e34\u0e19\u0e17\u0e19\u0e34\u0e25","latitude":"13.7356474","longitude":"100.5340211","picture":""}
แต่ทีนี้เราอยากให้สามารถเรียกเป็นแบบนี้อ่ะค่ะ
{result:[{building:"1",.....}]}
เราจะสร้าง key result ยังไงบ้างคะ
Tag : PHP, MySQL, iOS
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| Date :
2014-02-02 20:16:00 |
By :
sitny66 |
View :
655 |
Reply :
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Code (PHP)
echo '{result:',json_encode($resultArray),'}';
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| Date :
2014-02-02 20:35:59 |
By :
itpcc |
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