如何通过json从研究中恢复数据?我想恢复单独的项目作为结果显示在php页面创建新元素,例如恢复项目。f,项目。M创建新的元素,如item。f & lt;/div>
(错误显示在)
Php页面(执行查询)
if ($_GET['action'] == "chatheartbeat") { chatHeartbeat(); }
function chatHeartbeat() {
$sql = "select * from mensagens ";
$query = mysql_query($sql);
$items = '';
$chatBoxes = array();
while ($chat = mysql_fetch_array($query)) {
$items. = << < EOD
{
"s": "0",
"f": "{$chat['de']}",
"m": "{$chat['mensagem']}",
"i": "{$chat['img']}"
},
EOD;
}
}
索引(调用pair查询显示结果-这里有错误-这里需要帮助)
$.ajax({
url: "asd.php?action=chatheartbeat",
cache: false,
dataType: "json",
success: function(data) {
$.each(data.items, function(i, item) {
alert(item.f)
});
}
});
正如@jeroen所说,不要手动生成JSON字符串,使用json_encode()
从PHP数组或对象创建JSONString。
if ($_GET['action'] == "chatheartbeat") {
$sql = "select * from mensagens ";
$query = mysql_query($sql);
$chatBoxes = array();
while ($chat = mysql_fetch_array($query)) {
$t - new stdClass();
$t->s = "0";
$t->f = $chat['de'];
$t->m = $chat['mensagem'];
$t->i = $chat['img'];
$chatBoxes[] = $t;
}
$items = json_encode($chatBoxes);
// now just echo $items to return it to the javascript in the browser
// here or later in the code if there is more to your code than you showed us
echo $items;
}
引用:
json_encode ()
json_decode ()
感谢大家的帮助,我设法解决了如下问题
Php页面(执行查询)
$result = mysql_query("SELECT * FROM mensagens");
while( $array = mysql_fetch_assoc($result)){
$table[]= array("de"=>$array['de'],"mensagem"=>$array['mensagem']); }
echo json_encode($table);
指数 $.ajax({
url: 'asd.php',
dataType: 'json',
success: function(data)
{
for($i=0; $i < data.length; $i++){
$("#chatbox_"+chatboxtitle+" .chatboxcontent").append(
'<div class="message"><span class="from">'+data[$i].de+':
</span><span class=content">'+data[$i].mensagem+'</span></div>');
}}
});