我已经在使用商店收入的每日报告,但我需要我的报告执行以下操作:我希望查看特定日期,周或月的收入报告....我可以选择我想看的日期或天数,以便我可以打印我的选择。
这里是inc.php
<?
$dbtype = "mysql";
$dbhost = "localhost";
$dbname = "anillos";
$dbuser = "rubi";
$dbpass = "----";
$conn = new PDO("mysql:host=$dbhost;dbname=$dbname",$dbuser,$dbpass);
$conn->exec("set names utf8");
$conn->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
?>
这是我的代码:
<table class="table-bordered table-striped table-condensed">
<thead>
<tr>
<th>Fecha</th>
<th>Total Anillos Vendidos</th>
<th>Total ganado del día</th>
</tr>
</thead>
<tbody id="result_fechas">
<tr>
<?
include 'inc.php';
$sql = $conn->prepare("SELECT DATE(start) AS date, COUNT(id_anillos) AS total_anillos, SUM(ventas) AS total_diario_ganado
FROM PRODUCTOS WHERE start >= CURDATE() AND start < CURDATE() + INTERVAL 1 DAY ORDER BY start ASC");
$sql->execute();
while($row = $sql->fetch(PDO::FETCH_ASSOC)) {
?>
<td class="center"><? echo $row['date']; ?></td>
<td class="center"><? echo $row['total_anillos']; ?></td>
<td class="center"><? echo $row['total_diario_ganado']; ?></td>
</tr> <? } ?>
</tbody>
你能帮我解决这个问题......我真的没有线索......
---- ----- UPDATE
@ Chris78我做了这个表格:
<form name="fechas" id="fechas" method="post" >
<fieldset>
<legend>Reporte desde : </legend>
<input type="text" value="" placeholder="YYYY-MM-DD" name="f_desde" id="datepicker"/>
<legend>Reporte hasta : </legend>
<input type="text" value="" placeholder="YYYY-MM-DD" name="f_hasta" id="datepicker2" />
<button class="btn btn-inverse" type="submit" name="enviar" >
<i class="icon icon-print icon-white"></i>
Ver Reporte </button>
</fieldset>
</form>
新的SELECT CODE:
<?
$sql = $conn->prepare("SELECT DATE(start) AS date, COUNT(id_anillos) AS total_anillos, SUM(ventas) AS total_ganado FROM PRODUCTOS WHERE start BETWEEN 'f_desde' AND 'f_hasta' ORDER BY start ASC");
$sql->execute();
while($row = $sql->fetch(PDO::FETCH_ASSOC)) {
?>
表:
<table class="table-bordered table-striped table-condensed">
<thead>
<tr>
<th>rango de fechas seleccionado</th>
<th>Total Anillos Vendidos</th>
<th>Total ganado</th>
</tr>
</thead>
<tbody id="result_fechas">
<tr>
<?
////////SELECT CODE HERE
?>
<td class="center"><? echo $row['date']; ?></td>
<td class="center"><? echo $row['total_anillos']; ?></td>
<td class="center"><? echo $row['total_ganado']; ?></td>
</tr> <? } ?>
</tbody>
我尝试在同一页面中使用ajax调用显示结果
<script type="text/javascript">
$(function(){
$("#fechas").submit(function(){
$.ajax({
type:"POST",
url:".reportes.php?ts=" + new Date().getTime(),
dataType:"text",
data:$(this).serialize(),
beforeSend:function(){
$("#loading").show();
},
success:function(response){
$("#result_fechas").append(response);
$("#loading").hide();
}
})
return false;
});
});
});
</script>
reportes.php是同一个页面,其中是表单和等待结果的表....但我不知道错误在哪里,因为没有捕获ajax数据,当我的页面刷新按钮...你可以帮我解决这个问题。
答案 0 :(得分:0)
您创建了一个表格,其中包含您所谈到的那两个输入,当您检索输入值时,您给予处理“date('Y-m-d',inputvalue)”(如果尚未处理)。然后在sql语句中替换已处理的输入值的“CURDATE()”。
日期应始终以“yyyy-mm-dd”格式提供给MySql。
来源:http://dev.mysql.com/doc/refman/5.5/en/date-and-time-types.html
答案 1 :(得分:0)
如果您想要使用日期范围运行query
,可以使用BETWEEN
声明。
$sql = $conn->prepare("SELECT DATE(start) AS date, COUNT(id_anillos) AS total_anillos, SUM(ventas) AS total_diario_ganado
FROM PRODUCTOS WHERE start BETWEEN 'Older Date Here' AND 'Newer Date Here' ORDER BY start ASC");