I am creating a dropdown list in php. how can i put the selected item when someone selects an item.
my php code:
<form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
<select name="app" id="dropdown" value="" onchange="this.form.submit()" ><option>--select-app--</option>
<?php
$sql="select * from application";
$result=mysqli_query($con, $sql) or die("ereor selecting app ".mysqli_error($con));
while($row=mysqli_fetch_array($result))
{
$selected = $row['name'];
echo "<option id=". $row['id']."value = ".$row['id'].">".$row['name']."</option>";
}
echo "</select>";
?>
i want this: if I select an item it will show it as selected. how can I do this
You can do it in php like
<form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
<select name="app" id="dropdown" value="" onchange="this.form.submit()" >
<option>--select-app--</option>
<?php
$sql="select * from application";
$result=mysqli_query($con, $sql) or die("ereor selecting app ".mysqli_error($con));
$selected_val = $_POST['app']; //Should be $_GET, $_POST, $_SESSION whatever your selected value is
while($row=mysqli_fetch_array($result))
{
if(trim($row['id']) == trim($selected_val)) //<== Change this line
$selected = 'selected="selected"';
else
$selected = '';
echo '<option id="'. $row['id'].'" value="'.$row['id'].'" '. $selected.'>'. $row['name'] .'</option>';
//^Change this line
}
echo "</select>";
?>
In jQuery you can do it like
$('#dropdown').val('<?php echo "My val"; //The value goes here ?>');
Assuming that you mean you want to retain the selection after the form is submitted, you can do this inside the while loop:
$selected = (isset($_POST['app']) && $_POST['app'] == $row['id'] ? 'selected' : '');
echo "<option id=".$row['id']." value = ".$row['id']." ".$selected.">".$row['name']."</option>";
Related
I am trying to get the value of the selected option from my select. And I am trying to see it's output through a javascript echo. Here's what I've got so far. I am not getting the value
<form method="post" action="">
<select class="form-control" name="empSel" id="empSel">
<?php
$sql2 = "SELECT * FROM employee";
$result = mysql_query($sql2) or die("Couldn't execute sql2");
while ($row2 = mysql_fetch_assoc($result)) {
?>
<option value="<?php echo $row2['lastname'] ?>"><?=
/*$row2['user_surname']." ".*/
$row2['id']."-".$row2['lastname'] ?></option>
<?php
}
?>
</select>
<Label> Confirm</Label>
<div class="form-group col-md-6">
<input type="submit" class="btn btn-block btn-info"
name="submit"/>
</div>
</form>
<?php
if (isset($_POST['submit'])){
$userid = $_POST['empSel'];
echo '<script type="text/javascript"> alert('.$userid.')</script>';
$userid = preg_replace('/\D/', '', $userid);
$sql2 = "SELECT * FROM employee where id ='userid'";
$result = mysql_query($sql2) or die("Couldn't execute sql2");
while ($row = mysql_fetch_assoc($result)) {
echo '<script type="text/javascript"> alert("")</script>';
}
}
?>
An javascript alert doesn't pop out on this code. However, when I switch the value in the echo to a different variable an alert pops up. What does it mean? Do I properly get the value of my select and the page refreshed instantly that I didn't get to see it? Thanks
Edit:
An example of the option value would be 1-Lastname
And here's what I've tried.
<?php
if (isset($_POST['empSel'])){
$userid = $_POST['empSel'];
$userid = intval($userid);
echo '<script type="text/javascript"> alert('.$userid.')</script>';
}
?>
Now the javascript alert shows, but it echo 0. I think I am still not getting the value of my selected option
<option value="<?php echo $row2['id'] ?>">
This fixed it. In my previous code the option value was the surname...
I have created two select boxes each have different country names with flag images using select2 plugin and values are fetching from Database below is the code.
JS CODE
$(".currencyconverterselect").select2({
templateResult: addflag,
templateSelection: addflag
});
function addflag(opt) {
if (!opt.id) {
return opt.text;
}
var $opt = $(
'<span><img src="./img/flags/' + $(opt.element).attr('data-country-code') + '.png" class="userPic" /> ' + $(opt.element).text() + '</span>'
);
return $opt;
};
HTML CODE
<label style="color:white">Currency from</label>
<select name="fromcurrency_cc" id="fromcurrency_cc" class="form-control charts_currency" style="width:100%" required="required">
<?php
$sql = "SELECT fc.country_id,fc.code,fc.name,cc.country_code AS countrycode FROM fx_currency fc LEFT JOIN fx_country cc ON fc.country_id = cc.id ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
if ($row['code'] == 'AED')
{ ?>
<option data-country-code='<?php echo $row['countrycode']; ?>' value="<?php echo $row['code']; ?><?php echo $row['countrycode']; ?>" selected>
<?php echo $row['name']; ?>(
<?php echo $row['code']; ?> )</option>
<?php } else { ?>
<option data-country-code='<?php echo $row['countrycode']; ?>' value="<?php echo $row['code']; ?><?php echo $row['countrycode']; ?>">
<?php echo $row['name']; ?>(
<?php echo $row['code']; ?> )</option>
<?php }}
} else {
echo "0 results";
}
?>
</select>
I have also created a button below is the code
</i>
I want to swap the value of FROM currency to TO Currency and vice versa by clicking the button.
I have tried below code
$("#swapvalues_btn1").on('click', function() {
var pickup = $('#fromcurrency_cc').val();
$('#from').val($('#tocurrency').val());
$('#to').val(pickup);
});
Select2 will listen for the change event on the that it is attached to. If you make any external changes that need to be reflected in Select2 (such as changing the value).
Just use below code
$("#swapvalues_btn1").on('click', function() {
var fromcurrency = $('#fromcurrency').val();
var tocurrency = $('#tocurrency').val();
$('#fromcurrency').val(tocurrency).trigger('change');
$('#tocurrency').val(fromcurrency).trigger('change');
});
For More follow the link
https://select2.github.io/options.html#events
I am stuck in getting value from drop down. Drop down is dynamically filled from sql server database.
Dropdown 1 displays product name and it is dynamically filled.
Dropdown 2 displays environment name and it is filled by HTML.
I am getting value of environment but not product.
Please help me. Thanks
Here is the code:
<form action="" method="post">
//Dropdown 1
<p>Product Name:
<select name="productname">
<option value="">Select</option>
<?php
if( $conn )
{
$sql_dd = "SELECT ProductName from Product";
$stmt = sqlsrv_query( $conn, $sql_dd );
if( $stmt === false) {die( print_r( sqlsrv_errors(), true) );}
$rows = sqlsrv_has_rows( $stmt );
if ($rows === false)
echo "There is no data. <br />";
else
{ while( $row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC))
echo "<option value=''>".$row['ProductName']."</option>";
}
}
?>
//Dropdown 2
Client Type:
<select name="environment" style="width: 10%;">
<option value="">Select</option>
<option value="en1">en1</option>
<option value="en2">en2</option>
<option value="en3">en3</option>
</select>
<input type="submit" class="theme-btn" value="Search" name="submit"/>
<?php
if(isset($_POST['submit']) )
{
$productname = $_POST['productname'];
$environment= $_POST['environment'];
echo "productname: ".$productname." environment: ".$environment;
}?>
The value is not added
this:
echo "<option value=''>".$row['ProductName']."</option>";
should be
echo "<option value='" . $row['ProductName'] . "'>". $row['ProductName'] ."</option>";
I need the code for getting the addr and pnum of the patient when I choose the pname in the combo box.
How can I do that?
<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 ?>"/>
In your while loop add the following (if you have that column otherwise replace with something similar)
$paddress = $row['paddress'];
You can store the needed information by using the data attribute in your options
<option id="<?php echo $pnum; ?>" data-pname="<?php echo $pname; ?>" data-paddress="<?php echo $paddress; ?>" value="<?php echo $pname; ?>"><?php echo $pname; ?></option>
Then change your getVal() function
function getVal() {
var options = document.getElementById("model").options;
if (options.selectedIndex != -1) {
var addr = document.getElementById('paddress');
var name = document.getElementById('pname');
addr.value = options[options.selectedIndex].getAttribute('data-paddress');
name.value = options[options.selectedIndex].getAttribute('data-pname');
}
}
Now change the id's of the input fields for the address and the name to paddress and pname. It is important to know to never have duplicate id's
I hope that helped
<?php
include_once 'database_connect.php';
$conn = new dbconnection();
$dbcon = $conn->connect();
if (!$dbcon) {
die("Fail".mysqli_error($dbcon));
}
?>
<html>
<head>
<title></title>
<script type="text/javascript"></script>
</head>
<body>
<form name="frm" method="post" action='<?php echo $_SERVER['PHP_SELF']; ?>'>
<table width="50%" border="1" cellpadding="3" cellspacing="3" align="center">
<?php
$value1 = array();
$select_query = "SELECT Distinct branch FROM subjects";
$result = mysqli_query($dbcon, $select_query);
if (!$result) {
die("Fail".mysqli_error($dbcon));
}
while ($row = mysqli_fetch_array($result)) {
$value1[] = $row['branch'];
}
?>
<tr>
<td>Branch
<td><select name="branch" id="branch" onchange="document.frm.submit();">
<option>Select Branch</option>
<?php
foreach ($value1 as $gets)
echo "<option value={$gets}>{$gets}</option>";
?>
</select>
<?php
$value2 = array();
if (isset($_POST['branch'])) {
$branch = $_POST['branch'];
$getsub_query = "SELECT sub_code FROM subjects where branch='$branch'";
$result2 = mysqli_query($dbcon, $getsub_query);
if (!$result2) {
die("Fail\n".mysqli_error($dbcon));
}
while ($row1 = mysqli_fetch_array($result2)) {
$value2[] = $row1['sub_code'];
}
}
?>
<tr>
<td>Subject Code
<td><select name="subcode" id="subcode">
<option>Subject Code</option>
<?php
foreach ($value2 as $gets)
echo "<option value={$gets}>{$gets}</option>";
?>
</select>
This code gets the first drop down list branch from data base. When we select value from it, second drop down list get filled from database . but problem is when i select option in first drop down list, the selected option does not remain their in first drop down list . but second drop down list fills correctly. i want that option i have selected should remain selected. Like its state should be changed. I think first drop down list gets filled again on form load.
You have to mention in your html that the option is indeed selected.
Try replacing
echo "<option value={$gets}>{$gets}</option>";
By
$selected = '';
if (isset($_POST['branch'] && $gets==$_POST['branch']) {
$selected = ' selected="selected"';
}
echo "<option value={$gets}".$selected.">{$gets}</option>";