Php 基于复选框的下拉列表
我有几个名为HR、visitor、gaurd的复选框,现在我想根据属于该团队的员工的姓名(无论是HR、警卫还是访客)选择哪个复选框,并显示在下拉列表中Php 基于复选框的下拉列表,php,html,Php,Html,我有几个名为HR、visitor、gaurd的复选框,现在我想根据属于该团队的员工的姓名(无论是HR、警卫还是访客)选择哪个复选框,并显示在下拉列表中 <select name=cmbname id="cmbname" width='50%'> 全部 ` $objDB->SetQuery($sql); $res=$objDB->GetQueryReference(); 如果(!$res) 退出(“SQL中的错误:$SQL”); 如果($objDB->GetNumRows($re
<select name=cmbname id="cmbname" width='50%'>
全部
`
$objDB->SetQuery($sql);
$res=$objDB->GetQueryReference();
如果(!$res)
退出(“SQL中的错误:$SQL”);
如果($objDB->GetNumRows($res)>0)
{
while($row=mysql\u fetch\u row($res))
{
打印(“
{$row[0]}”);
}
}
mysql_免费_结果($res);
“Himani
。希望它对你有用。您可以使用下拉列表代替文本区域。试试这种方法,它会为您提供解决方案
<script type="text/javascript">
//javascript
function clicked_checkbox()
{
document.form.submit();
}
</script>
<?php
$hr = isset($_REQUEST['HR'])?$_REQUEST['HR']:false;
$guest = isset($_REQUEST['guest'])?$_REQUEST['guest']:false;
$visiter = isset($_REQUEST['visiter'])?$_REQUEST['visiter']:false;
//Prepare query with retrieved value and put value in Dropdown
$sql = 'select * from table ';
if($hr) { $sql .= "where user = '$hr'" };
if($guest) { $sql .= "where user = '$guest'" };
if($visiter) { $sql .= "where user = '$hr'" };
$objDB->SetQuery($sql);
$res = $objDB->GetQueryReference();
if(!$res)
exit("Error in SQL : $sql");
if($objDB->GetNumRows($res) > 0)
{
while($row = mysql_fetch_row($res))
{
print("<option value='{$row[0]}'>{$row[0]}</option>");
}
}
mysql_free_result($res);
?>
//on change of checkbox we'll call above function and set data to dropdown
<form name='form' method='get' action='#'>
<input type="checkbox" id="checkbox_HR" name="HR" value="true" <?php if($hr) echo "checked='checked'"; ?> onchange="return clicked_checkbox();">
<input type="checkbox" id="checkbox_guest" name="guest" value="true" <?php if($guest) echo "checked='checked'"; ?> onchange="return clicked_checkbox();">
<input type="checkbox" id="checkbox_user" name="visiter" value="true" <?php if($visiter) echo "checked='checked'"; ?> onchange="return clicked_checkbox();">
</form>
//javascript
函数已单击\u复选框()
{
document.form.submit();
}
请显示相应的代码。您尝试了什么??请出示ITOLL,第一个答案似乎总是与jQuery相关粘贴一些您尝试过的代码…@JonTaylor,因为此解决方案最好使用jQuery.thanx完成,但我使用了下拉列表希望基于复选框填充选项标记尝试使用此选项,但它显示了整个列表,即下拉列表不依赖于复选框这对我来说很有用!!您必须检查可能有问题的查询或向我显示您的查询
<script type="text/javascript">
//javascript
function clicked_checkbox()
{
document.form.submit();
}
</script>
<?php
$hr = isset($_REQUEST['HR'])?$_REQUEST['HR']:false;
$guest = isset($_REQUEST['guest'])?$_REQUEST['guest']:false;
$visiter = isset($_REQUEST['visiter'])?$_REQUEST['visiter']:false;
//Prepare query with retrieved value and put value in Dropdown
$sql = 'select * from table ';
if($hr) { $sql .= "where user = '$hr'" };
if($guest) { $sql .= "where user = '$guest'" };
if($visiter) { $sql .= "where user = '$hr'" };
$objDB->SetQuery($sql);
$res = $objDB->GetQueryReference();
if(!$res)
exit("Error in SQL : $sql");
if($objDB->GetNumRows($res) > 0)
{
while($row = mysql_fetch_row($res))
{
print("<option value='{$row[0]}'>{$row[0]}</option>");
}
}
mysql_free_result($res);
?>
//on change of checkbox we'll call above function and set data to dropdown
<form name='form' method='get' action='#'>
<input type="checkbox" id="checkbox_HR" name="HR" value="true" <?php if($hr) echo "checked='checked'"; ?> onchange="return clicked_checkbox();">
<input type="checkbox" id="checkbox_guest" name="guest" value="true" <?php if($guest) echo "checked='checked'"; ?> onchange="return clicked_checkbox();">
<input type="checkbox" id="checkbox_user" name="visiter" value="true" <?php if($visiter) echo "checked='checked'"; ?> onchange="return clicked_checkbox();">
</form>