我被賦予了使用datepicker，ajax，php和mysql在兩個給定日期之間生成報告的任務。以下是我的html：

日期明智的報告

From date: To date:

User informathions will be listed here.

腳本：

$(function() {

$( "#fromdate" ).datepicker();

$( "#todate" ).datepicker();

});

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();

}

這是應該生成報告的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 "

iddatestartnameemailphonecommentsapproved

while($row = mysql_fetch_array($result))

{

echo "

";

echo "

" . $row['id'] . "";

echo "

" . $row['date'] . "";

echo "

" . $row['start'] . "";

echo "

" . $row['name'] . "";

echo "

" . $row['email'] . "";

echo "

" . $row['phone'] . "";

echo "

" . $row['comments'] . "";

echo "

" . $row['approved'] . "";

echo "

";

}

echo "

";

問題是當我同時選擇兩個日期時什么也沒發生。在這種情況下，請幫我。簡單的例子將不勝感激。謝謝。

總結

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