【发布时间】：2013-09-12 11:01:11
【问题描述】：
我正在使用以下脚本检索特定月/年的所有发票。 php 从数据库中提取所有年份和日期，将它们组合在一起并将它们放入选择菜单中，结果类似于：
June - 2013
July - 2013
August - 2013
September - 2013
这是 selectsummary.php ：
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Sales Summary</title>
<script type="text/javascript" src="../js/jquery/jquery.js"></script>
<script type="text/javascript" src="../js/jqueryui/js/jquery-ui.js"></script>
<link href="../js/select2/select2.css" rel="stylesheet"/>
<script type="text/javascript" src="../js/select2/select2.js"></script>
<script type="text/javascript">
$(document).ready(function() { $("select").select2(); });
</script>
<?php
include '../connectmysqli.php';
include '../menu.php';
echo '<link rel="stylesheet" href="../css/template/template.css" />';
$salesID = rand().rand();
$today = date("Y-m-d");
?>
<script type="text/javascript" charset="utf-8">
$(document).ready(function(){
$('#selectmonth').on('change', function (){
// THIS IS WHERE I AM TRYING TO WORK OUT HOW TO RETRIEVE THE JSON AND THEN PLACE THE RESULTS INSIDE THE "summarycontent" DIV.
$.getJSON('select.php', {monthyear: $(this).val()}, function(data){
var invoicerow = '';
for (var x = 0; x < data.length; x++) {
invoicerow += '<p>' + data[x]['invoiceID'] + '">' + data[x]['date'] + ' - ' + data[x]['grandtotal'] + ' - ' + data[x]['customerID'] + '</p>';
}
$('#summarycontent').html(invoicerow);
$("select").select2();
});
});
});
</script>
</head>
<body>
<form method="post" action="addsalesubmit.php">
<p>
<select id="selectmonth">
<option>Please Select A Monthly Summary To View</option>
<?php
$sql = <<<SQL
SELECT YEAR(date) AS 'year', MONTHNAME(date) AS 'month'
FROM `sales`
GROUP BY YEAR(date), MONTHNAME(date) DESC
SQL;
if(!$result = $db->query($sql)){ die('There was an error running the query [' . $db->error . ']');}
while($row = $result->fetch_assoc()){
echo '<option value="'.$row['month'].'-'.$row['year'].'">'.$row['month'].' - '.$row['year'].'</option>';
}
echo '</select>';
?>
<br />
<br />
</form>
<div id="summarycontent"></div>
</body>
</html>
这是 ajax 用来查找结果并将其发送回主脚本的 select.php：
<?php include '../connectmysqli.php'; ?>
<?php
$monthyear = strtotime($_GET['monthyear']);
$sql = 'SELECT * FROM sales WHERE date = ' . (int)$monthyear;
$result = $db->query($sql);
$json = array();
while ($row = $result->fetch_assoc()) {
$json[] = array(
'invoiceID' => $row['invoiceID'],
'date' => $row['date'],
'grandtotal' => $row['grandtotal'],
'customerID' => $row['customerID']
);
}
echo json_encode($json);
?>
我遇到的问题是我不确定如何将诸如“2013 年 7 月”之类的文本转换为 select.php 可以使用的内容。目前，如果我使用 chrome 开发人员工具查看发生了什么，我会得到以下信息：
select.php?monthyear=July-2013
/manda/salessummary
所以日期没问题，但我不知道如何在另一端使用它来选择那个月/年的日期，因为它的格式错误。
数据库如下：
id invoiceID salesID customerID vehicleID date comments subtotal vat grandtotal description1 qty1T linetotal1T stock1T stock2T description2 qty2T linetotal2T stock3T description3 qty3T linetotal3T stock4T description4 qty4T linetotal4T stock5T description5 qty5T linetotal5T discount
68 1 1512428605 82428579 134722464 2013-07-08 22.48 4.50 26.98 Bridestone Pt34 - 175/55/18/W/63 - (99 In Stock) -... 1 22.48 711022407
编辑>>>>
select.php 现在看起来像这样：
<?php include '../connectmysqli.php';
$date_convert = date('Y-m-d', strtotime($_POST['monthyear']));
// 或者，使用 LIST，这将允许您稍后使用两个日期部分。
list($month, $year) = explode('-', $_POST['monthyear']);
$SQL = "SELECT * FROM sales WHERE date = :date"; $STH = $db->prepare($SQL); $STH->bindParam(':date', $date_convert);
$json = array();
while ($row = $result->fetch_assoc()) {
$json[] = array(
'invoiceID' => $row['invoiceID'],
'date' => $row['date'],
'grandtotal' => $row['grandtotal'],
'customerID' => $row['customerID']
);
}
echo json_encode($json);
?>
【问题讨论】：
-
请看我的回答。
标签： php jquery mysql ajax json