Php 如何将json数据从js显示或附加到html
如何将JSON数据从js显示为html 这是我尝试过的代码Php 如何将json数据从js显示或附加到html,php,jquery,arrays,ajax,json,Php,Jquery,Arrays,Ajax,Json,如何将JSON数据从js显示为html 这是我尝试过的代码 <script> $.ajax({ url:'load_categories.php', type:'get', data:{'from':loaded,'to':loadmore}, success: function (res) { var categories = $.parseJSON(res); var i=0; for (var x in c
<script>
$.ajax({
url:'load_categories.php',
type:'get',
data:{'from':loaded,'to':loadmore},
success: function (res) {
var categories = $.parseJSON(res);
var i=0;
for (var x in categories){
alert(categories.date_+i);
$('#categories').append('<div>'+(categories.date_+i+'</div>'); //not displaying on html
i++;
}
$('#loadmore').attr('num_loaded',(loaded+10));
}
});
</script>
<div id="categories"></div>
<?php
//load_categories.php
$res = mysql_query("SELECT * FROM impressions LIMIT $from,$to");
//echo "SELECT * FROM impressions LIMIT $from,$to";
$arr = array();
$i= 0;
while($row = mysql_fetch_array($res)) {
$arr['rec_id_'.$i] = $row['rec_id'];
$arr['date_'.$i] = $row['date'];
$i++;
}
echo json_encode($arr);
$.ajax({
url:'load_categories.php',
类型:'get',
数据:{'from':loaded,'to':loadmore},
成功:功能(res){
var categories=$.parseJSON(res);
var i=0;
对于(类别中的var x){
警报(类别、日期+i);
$(“#categories”).append(“”+(categories.date#+i+“”);//不在html上显示
i++;
}
$('loadmore').attr('num#u loaded',(loaded+10));
}
});
我已经修正了你的密码,请再试一次
正确的代码:
<script>
$.ajax({
url:'load_categories.php',
type:'get',
data:{'from':loaded,'to':loadmore},
success: function (res) {
var categories = $.parseJSON(res);
for(var i=0;i< categories.length;i++)
{
var row = categories[i];
alert(row.date);
$('#categories').append('<div>'+row.date+'</div>'); //not displaying on html
}
$('#loadmore').attr('num_loaded',(loaded+10));
}
});
</script>
<div id="categories"></div>
<?php
//load_categories.php
$res = mysql_query("SELECT * FROM impressions LIMIT $from,$to");
//echo "SELECT * FROM impressions LIMIT $from,$to";
$arr = array();
while($row = mysql_fetch_array($res)) {
$arr[] = row;
}
echo json_encode($arr);
$.ajax({
url:'load_categories.php',
类型:'get',
数据:{'from':loaded,'to':loadmore},
成功:功能(res){
var categories=$.parseJSON(res);
对于(变量i=0;i警报(categories.date+i)
告诉您有一个错误的(
:$(“#categories”)。附加(“”+(categories.date+i+);
检查我的答案,我已经更新了,您可以使用javascript json中数据库中的相同字段名