我是 PHP 编程的初学者。我有一个代码，我需要获取第一个表中的前 3 个数据和下一个表中的所有数据。数据来自 db 作为 json。
以下是我的php文件
<?php
$host = "localhost";
$user = "root";
$pass = "";
$database = "TMS_Sample";
$linkID = mysql_connect($host, $user, $pass) or die("Could not connect to host.");
mysql_select_db($database, $linkID) or die("Could not find database.");
$result = mysql_query("SELECT * FROM Completed_Training");
while ($row = mysql_fetch_assoc($result)) {
$array[] = $row;
}
echo json_encode($array);
?>
结果我得到以下json：
[{"Date":"2012-12- 04","Topic":"Collections","Trainer":"Prabhakaran.G","Status":"Invitation Sent"},{"Date":"2012-12-12","Topic":"Collections","Trainer":"Prabhakaran.G","Status":"Invitation Sent"},{"Date":"2012-12-07","Topic":"ffb","Trainer":"vcvxcv","Status":"cvxcv"},{"Date":"2012-12-08","Topic":"xcv","Trainer":"cvxcv","Status":"vxcv"},{"Date":"2012-12-09","Topic":"cvxcv","Trainer":"cvxcv","Status":"cvxcvxc"},{"Date":"2012-12-10","Topic":"xcv","Trainer":"vxcvxc","Status":"vxcv"},{"Date":"2012-12-11","Topic":"vv","Trainer":"vv","Status":"vxcv"},{"Date":"2012-12-12","Topic":"vv","Trainer":"vcv","Status":"cvxcv"},{"Date":"2012-12-13","Topic":"vv","Trainer":"cvxcv","Status":"cvv"},{"Date":"2012-12-14","Topic":"vxcv","Trainer":"vccv","Status":"xcvcv"},{"Date":"2012-12-14","Topic":"vcxvxc","Trainer":"cvxcv","Status":"cvxcv"},{"Date":"2012-12-15","Topic":"cvxcv","Trainer":"xcvxcv","Status":"xcvxcv"},{"Date":"2012-12-16","Topic":"sdasd","Trainer":"sdasd","Status":"dscxzc"},{"Date":"2012-12-16","Topic":"scxzcxzc","Trainer":"sdfscxzc","Status":"sadffcvzxc"},{"Date":"2012-12-17","Topic":"fxzcvxzc","Trainer":"zcvxzcz","Status":"xzcxzcxzcxz"},{"Date":"2012-12-18","Topic":"xzceafsdfv","Trainer":"vxvxv","Status":"xgsdfgvsd"},{"Date":"2012-12-12","Topic":"xcdvxvxcv","Trainer":"vxzdgvgSv","Status":"gbvsgv"},{"Date":"2012-12-27","Topic":"SDgvsdv","Trainer":"sdvsdv","Status":"sdgvsdv"},{"Date":"2012-12-11","Topic":"sdvsvd","Trainer":"sdvsdv","Status":"vdv"},{"Date":"2012-12-22","Topic":"dvsdv","Trainer":"vssdv","Status":"vsdvV"}]
现在我需要在 html 表中显示这个 json 数据
我已经在 1 个表格中显示了所有数据。但我需要的是获取第一个表中的前 3 个数据和下一个表中的所有数据。
这是我的html代码：
<!DOCTYPE HTML>
<html>
<head>
<script type="text/javascript" src="jquery182.js"></script>
<script type="text/javascript" language="javascript">
$(document).ready(function () {
$.getJSON('CompletedTraining.php', null, function (data) {
var Completed_counter = 1;
if (data) {
if (Completed_counter <= 3) {
var table = '<table border="1">';
$.each(data, function (i, element) {
table += '<tr>';
table += '<td>' + element.Date + '</td>';
table += '<td>' + element.Topic + '</td>';
table += '<td>' + element.Trainer + '</td>';
table += '<td>' + element.Status + '</td>';
table += '</tr>';
});
table += '</table>';
$('#getname').html(table);
}
var table1 = '<table border="1">';
$.each(data, function (j, element1) {
table1 += '<tr>';
table1 += '<td>' + element1.Date + '</td>';
table1 += '<td>' + element1.Topic + '</td>';
table1 += '<td>' + element1.Trainer + '</td>';
table1 += '<td>' + element1.Status + '</td>';
table1 += '</tr>';
});
table1 += '</table>';
$('#getname1').html(table1);
Completed_counter = Completed_counter + 1;
} else {
alert("error");
}
});
});
</script>
</head>
<body>
<form name="index">
<table id="getname" ></table>
<table id="getname1" ></table>
</form>
</body>
</html>