我有一个php页面,我有一些javascript代码,可以运行一些字段。我很好地复制了我的工作测试,并修改了一些适合的代码。好吧它不起作用,我似乎无法工作。有什么东西显然我不知道还是有其他原因导致它没有运行?
<?php
//server connection info
?>
<html>
<head>
<title>Survey</title>
<link rel="stylesheet" type="text/css" href="styles.css" />
<script type="text/javascript">
function Total()
{
var a=document.getElementById("a").value;
var b=document.getElementById("b").value;
var c=document.getElementById("c").value;
var d=document.getElementById("d").value;
a=parseInt(a);
b=parseInt(b);
c=parseInt(c);
var total=a+b+c;
document.getElementById("total").value=total;
}
</script>
</head>
<body>
<h1>QUALITY OF LABOR SURVEY</h1>
<p />
<h2>ABOUT YOUR COMPANY</h2>
<div class="Wrapper">
<form id="Main" method="post" action="Process.php">
<div class="Question">
1. In what state and county is your business located? (click below)
</div>
<div class='answer'>
<?php
$tsql = "select
StateCountyID,
State,
County
from dbo.StateCounty
where State='MO'
and Active='True'
order by State";
$tsql2 = "select
StateCountyID,
State,
County
from dbo.StateCounty
where State='IL'
and Active='True'
order by State";
/* Execute the query. */
$stmt = sqlsrv_query( $conn, $tsql);
if ( $stmt )
{
echo "<span><select name='ListMO'>";
echo "<option value='0'>MO-County</option>";
while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_NUMERIC))
{
echo "<option value='".$row[0]."'>";
echo $row[2];
echo "</option>";
}
echo "</select></span>";
}
else
{
echo "Error in statement execution.\n";
die( print_r( sqlsrv_errors(), true));
}
$stmt2 = sqlsrv_query( $conn, $tsql2);
if ( $stmt2 )
{
echo "<span><select name='ListIL'>";
echo "<option value='0'>IL-County</option>";
while( $row = sqlsrv_fetch_array( $stmt2, SQLSRV_FETCH_NUMERIC))
{
echo "<option value='".$row[0]."'>";
echo $row[2];
echo "</option>";
}
echo "</select></span>";
}
else
{
echo "Error in statement execution.\n";
die( print_r( sqlsrv_errors(), true));
}
?>
</div>
<table width="700px">
<tr>
<td style="font-size: 1.2em; font-weight: bold;">
ABOUT YOUR EMPLOYMENT
</td>
<td style="font-weight: bold;">
(Exclude Temporary Employees Throughout Survey)
</td>
</tr>
</table>
<p />
<b>Please estimate the following:</b>
<p />
<div class="Question">
4. Number of Full-Time Hourly Employees (Eligible for Full-Time Benefits)
</div>
<div class="Answer">
<input type="text" id="a" value="0" onchange="Total();" />
</div>
<div class="Question">
5. Number of Part-Time Hourly Employees(Not Eligible for Full-Time Benefits)
</div>
<div class="Answer">
<input type="text" id="b" value="0" onchange="Total();" />
</div>
<div class="Question">
6. Salaried Employees
</div>
<div class="Answer">
<input type="text" id="c" value="0" onchange="Total();" />
</div>
<div class="Question">
7. Is this your current number of employees? If not, change responses to 4, 5, and 6.
</div>
<div class="Answer">
<input type="text" id="total" value="0" onchange="Total();" />
</div>
答案 0 :(得分:8)
var d=document.getElementById("d").value;
产生错误,因为您似乎没有id="d"
答案 1 :(得分:4)
我还想提供一些改进代码的建议。
https://addons.mozilla.org/en-US/firefox/addon/1843
http://jquery.com/
http://developer.yahoo.com/yui/
http://toys.lerdorf.com/archives/38-The-no-framework-PHP-MVC-framework.html/
答案 2 :(得分:1)
要检查的一些事项。
您还应该在Firefox中使用调试器,甚至可以为Firefox找到Firebug之类的工具。帮助找到这样的问题。
答案 3 :(得分:0)
答案 4 :(得分:0)
HTML中没有元素“d”。如果你注释掉这一行:
var d=document.getElementById("d").value;
它应该可以正常工作