未通过Javascript函数在mysql数据库中输入值
在下面给出的代码中,javascript函数不起作用,在phpmyadmin中,只输入了包,但mysql数据库中没有输入$costs和$totam。为什么会这样未通过Javascript函数在mysql数据库中输入值,javascript,php,mysql,Javascript,Php,Mysql,在下面给出的代码中,javascript函数不起作用,在phpmyadmin中,只输入了包,但mysql数据库中没有输入$costs和$totam。为什么会这样 <?php mysql_connect("localhost","root",""); mysql_select_db('test'); if(isset($_REQUEST['submit'])) { $name=$_REQUEST['name']; @$packages=$_REQUEST['packages']; $
<?php
mysql_connect("localhost","root","");
mysql_select_db('test');
if(isset($_REQUEST['submit']))
{
$name=$_REQUEST['name'];
@$packages=$_REQUEST['packages'];
$abc=mysql_query("insert into cusrec(name,packages,totamt,clthpackage)values('$name','$packages','$totamt','$clothes')");
}
?>
<html>
<head>
<script src="js/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('input[name=packages]').function(slab(abc){
var slabOption=slab.options[slab.selectedIndex];
if(slabOption == "aslab"){
<?php $clothes = 15; ?>
<?php $totamt = 7500; ?>
}
else if(slabOption == "bslab"){
<?php $clothes = 10; ?>
<?php $totamt = 6000; ?>
}
else if(slabOption == "cslab"){
<?php $clothes = 5; ?>
<?php $totamt = 4000; ?>
}
else if(slabOption == "dslab"){
<?php $clothes = 2; ?>
<?php $totamt = 2000; ?>
}
});
});
</script>
</head>
<body>
<form>
<table><tr>
<td>Name</td>
<td><input type="text" name="name" required="required"></td>
</tr>
<tr>
<td>Packagaes</td>
<td>
<select name="packages" onChange="slab(this)">
<option value="aslab" />Slab A 15 Clothes for 7500/-
<option value="bslab" />Slab B 10 Clothes for 6000/-
<option value="cslab" />Slab C 5 Clothes for 4000/-
<option value="dslab" />Slab D 2 Clothes for 2000/-
</select>
</td>
</tr>
<tr>
<td colspan="2"><center><input type="submit" name="submit"value="Submit"></center></td>
</tr>
</table>
</form>
</body>
</html>
当我查看我的数据库值时,根据salb正确输入了包的字段,但$totamt和$COUNTS仍然为空。无法理解为什么会这样?用javascript代码编写的条件没有任何用处。它们毫无意义。将它们放入php中的post处理程序代码中,并在其中设置变量
<?php
mysql_connect("localhost","root","");
mysql_select_db('test');
if(isset($_REQUEST['submit']))
{
$name=$_REQUEST['name'];
@$packages=$_REQUEST['packages'];
$abc=mysql_query("insert into cusrec(name,packages,totamt,clthpackage)values('$name','$packages','$totamt','$clothes')");
}
?>
<html>
<head>
<script src="js/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('input[name=packages]').function(slab(abc){
var slabOption=slab.options[slab.selectedIndex];
if(slabOption == "aslab"){
<?php $clothes = 15; ?>
<?php $totamt = 7500; ?>
}
else if(slabOption == "bslab"){
<?php $clothes = 10; ?>
<?php $totamt = 6000; ?>
}
else if(slabOption == "cslab"){
<?php $clothes = 5; ?>
<?php $totamt = 4000; ?>
}
else if(slabOption == "dslab"){
<?php $clothes = 2; ?>
<?php $totamt = 2000; ?>
}
});
});
</script>
</head>
<body>
<form>
<table><tr>
<td>Name</td>
<td><input type="text" name="name" required="required"></td>
</tr>
<tr>
<td>Packagaes</td>
<td>
<select name="packages" onChange="slab(this)">
<option value="aslab" />Slab A 15 Clothes for 7500/-
<option value="bslab" />Slab B 10 Clothes for 6000/-
<option value="cslab" />Slab C 5 Clothes for 4000/-
<option value="dslab" />Slab D 2 Clothes for 2000/-
</select>
</td>
</tr>
<tr>
<td colspan="2"><center><input type="submit" name="submit"value="Submit"></center></td>
</tr>
</table>
</form>
</body>
</html>
像
您可以完全删除javascript代码,因为通过读取所选选项在JS中设置PHP变量毫无意义。您的查询在javascript代码之前运行。您需要在javascript代码之后添加insert查询。如果是这样,那么为什么它接受并在数据库中输入$packages。?请澄清这一点confusion@Pawan您需要在查询之前定义$packages,以便正确插入其包。因此,在运行查询之前,您还需要定义$Cloth和$totamt。您是正确的,它工作正常。谢谢你,伙计。