有个疑问,在数据提交时,有按钮判断其状态,然后跳转到新的页面。如何根据单选按钮的选择,在跳转页面进行相应的数据库操作?
例子如下:<form name="submit1" method="post" action="submit.php?id=<?php echo $id; ?>">
<?php
$sql = "SELECT `username`,`lotno`,`barcode`,`date` FROM `barcode_2d` WHERE `ID`=".$id;
echo displayitem($sql, 1);
echo '<br />';
echo '<font color="blue"><b>是否为合格品</b></font>';
echo '<br />';
echo '<br />';
echo '<input type="radio" name="status" value="yes" /> 是';
echo '<input type="radio" name="status" value="no" /> 否';
echo '<br />';
echo '<br />';
?>
<input name="btnEditOK" type="button" title="Submit" class="btnbig" value="提交" onclick="javascript:add();"
onmouseover="this.className='btnbig_over'" onMouseOut="this.className='btnbig'" onMouseDown="this.className='btnbig_down'"
onMouseUp="this.className='btnbig'" />
</form>
<script>
function add(){
if(confirm('请再次确认是否为合格品!'))
{
var url = 'submit.php?id=<?php echo $id; ?>';
window.location.href = url;
}
}
</script>我如何根据单选按钮的值,在submit.php页面进行相应的数据库操作:insert or delete?
例子如下:<form name="submit1" method="post" action="submit.php?id=<?php echo $id; ?>">
<?php
$sql = "SELECT `username`,`lotno`,`barcode`,`date` FROM `barcode_2d` WHERE `ID`=".$id;
echo displayitem($sql, 1);
echo '<br />';
echo '<font color="blue"><b>是否为合格品</b></font>';
echo '<br />';
echo '<br />';
echo '<input type="radio" name="status" value="yes" /> 是';
echo '<input type="radio" name="status" value="no" /> 否';
echo '<br />';
echo '<br />';
?>
<input name="btnEditOK" type="button" title="Submit" class="btnbig" value="提交" onclick="javascript:add();"
onmouseover="this.className='btnbig_over'" onMouseOut="this.className='btnbig'" onMouseDown="this.className='btnbig_down'"
onMouseUp="this.className='btnbig'" />
</form>
<script>
function add(){
if(confirm('请再次确认是否为合格品!'))
{
var url = 'submit.php?id=<?php echo $id; ?>';
window.location.href = url;
}
}
</script>我如何根据单选按钮的值,在submit.php页面进行相应的数据库操作:insert or delete?
// 合格时的操作
}else {
// 不合格时的操作
}
老大,不对啊。我echo $_POST['status'].'<br />';后没有结果出来。
function add(){
if(confirm('请再次确认是否为合格品!'))
{
var url = '?id=32';
window.location.href = url;
}
}
</script>再把提交按钮改为
<input name="btnEditOK" type="Submit" title="Submit" class="btnbig" value="提交" onclick="return confirm('请再次确认是否为合格品!')"
onmouseover="this.className='btnbig_over'" onMouseOut="this.className='btnbig'" onMouseDown="this.className='btnbig_down'"
onMouseUp="this.className='btnbig'" />即可