【发布时间】:2014-05-14 19:57:18
【问题描述】:
您好,我尝试循环我的json_encode() 以获取所有基于我的查询相关的数据
脚本工作正常,但循环部分不行
这是我目前所拥有的:
$con3 = new PDO("mysql:host=". db_host .";dbname=db", db_username , db_password);
$con3->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
$con4 = new PDO("mysql:host=". db_host .";dbname=chat_db", db_username , db_password);
$con4->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
$sql5 = "SELECT * FROM users WHERE id = :rid LIMIT 1";
$stmt6=$con4->prepare($sql5);
$stmt6->bindValue( 'rid',$_GET['rid'], PDO::PARAM_STR);
$stmt6->execute();
foreach($stmt6->fetchAll()as $res)
{
$usern = $res['username'];
$user_lvl = $res['ulvl'];
}
$comb = $usern . $_GET['name'];
$sql6="SELECT msgid FROM thread WHERE combination1=:msgids OR combination2=:submsgids";
$msg_id = $con4->prepare($sql6);
$msg_id->bindParam(':msgids', $comb, PDO::PARAM_STR);
$msg_id->bindParam(':submsgids', $comb, PDO::PARAM_STR);
$msg_id->execute();
$msgd = $msg_id->fetchColumn();
$tbpre = $msgd;
$sql7 = "SELECT message_content, username , message_time, recipient FROM ".$tbpre."chat_conversation WHERE msgid=:chat";
$stmt7=$con3->prepare($sql7);
$stmt7->bindValue( 'chat', $msgd, PDO::PARAM_STR);
$stmt7->execute();
$message_query = $stmt7;
if(count($message_query) > 0) {
while($message_array = $stmt7->fetchAll(PDO::FETCH_ASSOC)) {
echo json_encode($message_array);
}
}
它只从我的数据库中返回一个数据..
这是我用于从我的 php 端检索的 javascript:
function AjaxRetrieve()
{
var rid = document.getElementById('trg').value,
data = {chat: uid, rid: rid, name: user};
$.get('includes/getChat.php', data, function (result) {
var res = $([]);
$.each(result[0], function(key, value) {
res = res.add($('<div />', {text : value}));
});
$("#clog").html(res);
}, 'json');
}
【问题讨论】:
-
&那个数据一定是最后一个?
-
$json_string += json_encode($message_array);
-
@DeepakMane 这不是正确的方式
-
是 $json_string_array[]= ($message_array);最后一行将是 echo json_encode($json_string_array)
-
@DeepakMane 你给的东西没有显示任何东西。这是为什么?
标签: javascript php json while-loop