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[英]Retrieving database data through PHP, based on the Javascript function value
[英]Value is not entered in mysql database through Javascript function
在下面給出的代碼中,javascript功能在phpmyadmin中不起作用-僅輸入包,但未在mysql數據庫中輸入$ clothes和$ totamt。 為什么會這樣呢?
<?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 $totamt = 2000; ?>
}
});
});
</script>
</head>
<body>
<form>
<table><tr>
<td>Name</td>
<td><input type="text" name="name" required="required"></td>
</tr>
<tr>
<td>Packagaes</td>
<td>
<select name="packages" onChange="slab(this)">
<option value="aslab" />Slab A 15 Clothes for 7500/-
<option value="bslab" />Slab B 10 Clothes for 6000/-
<option value="cslab" />Slab C 5 Clothes for 4000/-
<option value="dslab" />Slab D 2 Clothes for 2000/-
</select>
</td>
</tr>
<tr>
<td colspan="2"><center><input type="submit" name="submit"value="Submit"></center></td>
</tr>
</table>
</form>
</body>
</html>
當我查看數據庫值時,根據salb正確輸入了包文件,但$ totamt和$ clothes保留為空白。 無法理解為什么會這樣?
您用javascript代碼編寫的條件沒有用。 他們沒有道理。 將它們放入php中的后處理程序代碼中,並在那里設置變量。
喜歡
if(isset($_REQUEST['submit']))
{
$name=$_REQUEST['name'];
$packages=$_REQUEST['packages'];
if($packages == "aslab"){
$clothes = 15;
$totamt = 7500;
}
else if($packages == "bslab"){
$clothes = 10;
$totamt = 6000;
}
else if($packages == "cslab"){
$clothes = 5;
$totamt = 4000;
}
else if($packages == "dslab"){
$clothes = 2;
$totamt = 2000;
}
$abc=mysql_query("insert into cusrec(name,packages,totamt,clthpackage)values('$name','$packages','$totamt','$clothes')");
}
而且您可以完全刪除javascript代碼,因為通過讀取所選選項在JS內設置PHP變量沒有任何意義。
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