MySQL重新格式化mysql结果来满足json hansontable [英] MySQL reformat mysql result to meet json hansontable

问题描述

根据我以前解决的问题：重新排列mysql结果为hansontable的预期格式。
$ b

[Superior]，[Deluxe - City View]，[豪华 - 阳台]，[Junior
Suite]，[Andaman Studio] >

  [Superior，Deluxe  -  City View，Deluxe  -  Balcony，Junior Suite，Andaman Studio] 
  

从这些代码： $ sql_rName =从room_db选择标题where hotel ='1'by id asc;
$ result_rName = mysql_db_query（$ dbname，$ sql_rName）;
while（$ rec_rName = mysql_fetch_array（$ result_rName））{
$ _rName = $ rec_rName ['title'];
$ _array [] = $ _ rName;

echo{\data \：.json_encode（$ _ array）。};


mysql表：room_db

ps。感谢Olaf Dietsche提供的所有帮助。

解决方案

为了制作正确的JSON，请尝试：
`

  $ result =从room_db中选择标题where hotel ='1'by id asc; 
 $ messages = array（）; 
 while（$ message_data = mysql_fetch_assoc（$ result））{
 $ message = array（
'id'=> $ message_data ['userid']，
'title' => $ message_data ['title'] 
）; 
 $ messages [] = $ message; 
} 
 echo json_encode（$ messages）; 
 
 
 $ / code $ / pre 
 
< p>在关闭方这样做：
 `
 data1 = $。parseJSON（data）; 
 
 if（data1.length === 0）{
 
 $（'＃table> #table_body'）。append（'< tr>< td colspan =4align =centerstyle =color：red> NO matching data< / td>< / tr>'）; 
} 
 else {
 for（var i = 0; i< data1.length; i ++）
 {
 $（'＃table> #table_body'） .append（'< tr id ='+ data1 [i] ['id'] +'>< td id ='+ data1 [i] ['id'] +'align =center < td>'+ data1 [i] ['title'] +'< / td>< / tr>'）; 
 
 
 $（'＃table'）。append（'< / tbody>'）; 
 
`
  
 
According to my previous solved issue : Re-arrange mysql result in an expected format for hansontable. I'm going to re-format mysql result from 

  
["Superior"],["Deluxe - City View"],["Deluxe - Balcony"],["Junior
  Suite"],["Andaman Studio"]
into
["Superior","Deluxe - City View","Deluxe - Balcony","Junior Suite","Andaman Studio"]
From these codes:
$sql_rName="select title from room_db where hotel='1' order by id asc";
$result_rName=mysql_db_query($dbname,$sql_rName);
while($rec_rName=mysql_fetch_array($result_rName)){
    $_rName=$rec_rName['title'];
    $_array[]=$_rName;
}
echo "{\"data\": ".json_encode($_array)."}";
mysql Table : room_db


Please suggest.

ps. Thanks to Olaf Dietsche for all of these help.
解决方案
To make a correct JSON, Try :
`
$result="select title from room_db where hotel='1' order by id asc";    
$messages = array();
            while($message_data = mysql_fetch_assoc($result)) {
                $message = array(
                'id' => $message_data['userid'],
                'title' => $message_data['title']
                );
                $messages[] = $message;
                }
                echo json_encode($messages);
            }
`
and on the reseiver side do this :
`
data1=$.parseJSON(data);

            if(data1.length===0){

                $('#table > #table_body').append('<tr><td colspan="4" align="center" style="color:red">NO matching data </td></tr>');
                }
        else{
            for(var i=0;i<data1.length;i++)
            {
                $('#table > #table_body').append('<tr id="' + data1[i]['id'] +'"> <td id="' + data1[i]['id'] +'" align="center" <td>'+data1[i]['title']+'</td> </tr>');
            }
            }
            $('#table').append('</tbody>');

    `


                        
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