php - 如何使用datepicker、ajax、php、mysql生成两个日期之间的报告？

Question

我接到一项任务，使用 datepicker、ajax、php 和 mysql 在两个给定日期之间生成报告。下面是我的 html:

按日期报告

        From date: <input type="text" id="fromdate" value="">   To date: <input type="text" id="todate" value="" onChange="showuser(document.getElementById('fromdate').value,document.getElementById('todate').value)">  
        <br>
        <div id="txtHint"><b>User informathions will be listed here.</b></div>

脚本:

<script>
  $(function() {
    $( "#fromdate" ).datepicker();
    $( "#todate" ).datepicker();
  });
  </script>

<script type="text/javascript">
function showUser(fromdate,todate)
{
  if (fromdate =="" && todate=="")
    {
     document.getElementById("txtHint").innerHTML="";
     return;
    } 

    if (window.XMLHttpRequest)
     {// code for IE7+, Firefox, Chrome, Opera, Safari
      xmlhttp=new XMLHttpRequest();
     }
  else
    {// code for IE6, IE5
     xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }

  xmlhttp.onreadystatechange=function()
    {
     if (xmlhttp.readyState==4 && xmlhttp.status==200)
      {
        document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
      }
    }
    xmlhttp.open("GET","bet_date.php?fromdate="+fromdate+"&todate="+todate,true);
    xmlhttp.send();
}
</script>

这是应该生成报告的 php 文件: bet_date.php

include("database.php"); 
$fromdate=$_GET["fromdate"];
$todate=$_GET["todate"];
 $sql = "SELECT * FROM bookings WHERE date between '$fromdate' and '$todate'";
 $result = mysql_query($sql);

 echo "<table border='1'>
<tr>
<th>id</th>
<th>date</th>
<th>start</th>
<th>name</th>
<th>email</th>
<th>phone</th>
<th>comments</th>
<th>approved</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['id'] . "</td>";
  echo "<td>" . $row['date'] . "</td>";
  echo "<td>" . $row['start'] . "</td>";
  echo "<td>" . $row['name'] . "</td>";
  echo "<td>" . $row['email'] . "</td>";
   echo "<td>" . $row['phone'] . "</td>";
    echo "<td>" . $row['comments'] . "</td>";
     echo "<td>" . $row['approved'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

问题是当我选择两个日期时什么也没有发生。 在这种情况下请帮助我。 简单的例子将不胜感激。 谢谢。

Answer 1

I have Created Whole Tutorial for Date Wise Report, so once try it

表结构

CREATE TABLE IF NOT EXISTS `bookings` (
  `id` int(12) NOT NULL AUTO_INCREMENT,
  `date` date NOT NULL,
  `start` varchar(255) NOT NULL,
  `name` varchar(255) NOT NULL,
  `email` varchar(255) NOT NULL,
  `phone` varchar(20) NOT NULL,
  `comments` longtext NOT NULL,
  `approved` varchar(10) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

INSERT INTO `bookings` (`id`, `date`, `start`, `name`, `email`, `phone`, `comments`, `approved`) VALUES
(1, '2014-12-17', 'yes', 'Mahendra', 'mahendra@XXXX', '89XXXXXXXX', 'nothing', 'yes'),
(2, '2014-12-18', 'no', 'Rahul', 'rahul@XXXXXX', '987XXXXXXX', 'very nice article', 'yes');

my_test.php

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
  $("#generate_report").click(function(){
	  var from_date=jQuery("#fromdate").val();
	  var to_date=jQuery("#todate").val();
	  var data =
	  {		
	  	from_date	 : from_date,
		to_date		 : to_date
	  }
	jQuery.ajax({	
					type: "POST",
					url: "test.php",
					data: data,
					success: function(responce){
						$("#txtHint").after(responce);
				  }	
			 });
  });
});
</script>

From date: <input type="text" id="fromdate" value="2014-12-17">   To date: <input type="text" id="todate" value="2014-12-18">  
<input type="button" name="generate_report" id="generate_report" value="Generate Report" />
<br>
<div id="txtHint"><b>User informations will be listed here.</b></div>

test.php(ajax调用文件)

<?php
error_reporting(0);
include_once("wp-config.php");
extract($_POST);
$sql1=mysql_query("Select * from bookings where date between '".$from_date."' AND '".$to_date."'");
echo "<table border='1'>
	  <tr>
		<th>id</th>
		<th>date</th>
		<th>start</th>
		<th>name</th>
		<th>email</th>
		<th>phone</th>
		<th>comments</th>
		<th>approved</th>
	 </tr>";
		while($row = mysql_fetch_array($sql1))
		{
		  echo "<tr>";
			  echo "<td>" . $row['id'] . "</td>";
			  echo "<td>" . $row['date'] . "</td>";
			  echo "<td>" . $row['start'] . "</td>";
			  echo "<td>" . $row['name'] . "</td>";
			  echo "<td>" . $row['email'] . "</td>";
			  echo "<td>" . $row['phone'] . "</td>";
			  echo "<td>" . $row['comments'] . "</td>";
			  echo "<td>" . $row['approved'] . "</td>";
		  echo "</tr>";
		}
echo "</table>";
?>

这个演示工作正常，我刚刚创建了，所以一旦检查它...

请任何人感兴趣，请正确回答并为此解决方案投票。 谢谢