Javascript 获取组合框更改的用户详细信息
当我在组合框中选择pname时,我需要获取患者地址和pnum的代码 我该怎么做Javascript 获取组合框更改的用户详细信息,javascript,php,mysql,combobox,Javascript,Php,Mysql,Combobox,当我在组合框中选择pname时,我需要获取患者地址和pnum的代码 我该怎么做 <script> function getVal() { document.getElementById("text2").value = document.getElementById("model").value; } </script> <body> //code in opening and getting my addr and pnum in dbas
<script>
function getVal() {
document.getElementById("text2").value = document.getElementById("model").value;
}
</script>
<body>
//code in opening and getting my addr and pnum in dbase
<?php
include('connect.php');
$pname=$_GET['tpname'];
$result = mysql_query("SELECT * FROM tblnpatient where pname='$pname'");
while($row = mysql_fetch_array($result))
{
$pnum=$row['pnum'];
$addr=$row['addr'];
}
?>
//code for choosing pname
<tr><td>Patient Name:
<div id="ma">
<select name="pname" class="textfields" id="model" style="width:180px;" onchange="getVal();">
<option id="0" >--Select Patient Name--</option>
<?php
$con=mysqli_connect("localhost","root","","dbnpatient");
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$pnum=$_GET['pnum'];
$query = mysqli_query($con, "SELECT * FROM tblnpatient");
while($row = mysqli_fetch_array($query)){
$pnum = $row['pnum'];
$pname = $row['pname'];
?>
<option id="<?php echo $pnum; ?>" value="<?php echo $pname; ?>"><?php echo $pname; ?></option>
<?php } ?>
</select>
//code for getting pname and addr
Address:<input type="text" name="ename" size="20" id="ma" value="<?php echo $addr ?>"/>
Name:<input type="text" name="ename" size="20" id="ma" value="<?php echo $pname ?>"/>
</body>
像这样试试
function getVal()
{
var pname=$('#model option:selected').val();
var pnum=$('#textboxidofpnum').val();
var addr=$("#ma").val();
console.log(pname,pnum,addr);
}
但是我的pnum也是一个textbox然后更改var pnum=$'textboxid'。这就是我打开连接的方式..并在数据库中获取值..我是否正确?我将pname的值放在tpname中