[英]JSON jQuery display to table
如何在表中呈現JSON數據。 我嘗試了以下代碼:
HTML :
<table>
<thead>
<tr>
<th>id</th>
<th>Name</th>
<th>Password</th>
</tr>
</thead>
<tbody id="tbody"></tbody>
</table>
<button id="get">Get</button>
<script src="script/ajax.googleapis.com_ajax_libs_jquery_1.10.1_jquery.min.js" type="text/javascript"></script>
<!--<script src="script/myScript.js" type="text/javascript"></script>-->
<script src="script/json_array.js" type="text/javascript"></script>
腳本jQuery
$(document).ready( function() {
done();
});
function done() {
setTimeout( function() {
updates();
done();
}, 200);
}
function updates() {
$.getJSON("fetch.php", function(data) {
$.each(data, function (index, item) {
var eachrow = "<tr>"
+ "<td>" + item[1] + "</td>"
+ "<td>" + item[2] + "</td>"
+ "<td>" + item[3] + "</td>"
+ "<td>" + item[4] + "</td>"
+ "</tr>";
$('#tbody').append(eachrow);
});
});
}
PHP腳本
<?php
include "./pdoConn.php";
$output = array();
$query = "select * from wishers";
$stmt = $pdo->query($query);
$stmt->execute();
$name = $stmt->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($name);
?>
json的輸出 :
[
{"id":"1","name":"Tom","password":"098f6bcd4621d373cade4e832627b4f6"},
{"id":"2","name":"Jerry","password":"098f6bcd4621d373cade4e832627b4f6"},
{"id":"3","name":"Kate","password":"098f6bcd4621d373cade4e832627b4f6"},
{"id":"4","name":"Joan","password":"098f6bcd4621d373cade4e832627b4f6"},
{"id":"5","name":"Cyril","password":"098f6bcd4621d373cade4e832627b4f6"},
{"id":"8","name":"Ama","password":"ama"},
{"id":"7","name":"Akusika","password":"mummy"},
{"id":"9","name":"Abetiafa","password":"joko"}
]
當使用表和AJAX時,我建議使用jQuery Datatables插件。 它為您完成了很多工作,如果您使用的是Laravel之類的框架,則有很多軟件包
不要使用item [1]使用item [id],這是json,不僅是數組,甚至可以使用item.id。
將json數據視為對象(此處為對象數組),並且此對象具有帶有名稱(而不是數字)的屬性(除非您將其設置為名稱^^)
它應該是
var eachrow = "<tr>"
+ "<td>" + item[index].id + "</td>"
+ "<td>" + item[index].name + "</td>"
+ "<td>" + item[index].password + "</td>"
+ "</tr>";
因為您具有對象數組,所以必須使用index
訪問元素,然后才能訪問其屬性。
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