I have some combobox with onchange event, and they're reset each other when selected the orther one of them, does any suggest how to retain the value on the page? this my script :
<form method="POST" name="form1" action="<?php $_SERVER['PHP_SELF'];?>">
<table border="0">
<tr>
<td colspan="6"></td>
</tr>
<tr>
<td>
<select name="select_petugas1" style="width:18px;" onchange="this.form.submit('select_petugas1');"> //first combobox
<option></option>
<?php include 'dbconn.php';
$sql_peg1="SELECT * FROM users"; $result_peg1=$conn->query($sql_peg1);
while( $row_peg1=$result_peg1->fetch_assoc() ){
echo "<option>".$row_peg1['nama']."</option>";
}
?>
</select>
</td>
<td>
<?php
if(isset($_POST['select_petugas1'])){
$select_petugas1=$_POST['select_petugas1'];
echo "<input type='text' name='select_petugas1' value='".$select_petugas1."'>"; // Throw 1st result into the text box
$sql_NIP1="SELECT NIP FROM users WHERE nama='$select_petugas1'";
$result_NIP1=$conn->query($sql_NIP1);
$row_NIP1=$result_NIP1->fetch_assoc();
$NIP1=$row_NIP1['NIP'];
?>
</td>
<td> NIP</td>
<td>:</td>
<td><input type="text" name='NIP1' value="<?php echo $NIP1; ?>"></td>
</tr> <!-- child of first result -->
<tr>
<td colspan="5" bgcolor="blue"></td>
</tr>
<tr>
<td>
<select name="peg_2" style="width:18px;" onchange="submit(this)"><!--2nd combobox-->
<option></option>
<?php
$sql_peg2="SELECT nama FROM users";
$result_peg2=$conn->query( $sql_peg2 );
while ($row_peg2=$result_peg2->fetch_assoc()){
echo "<option value='".$row_peg2['nama']."'>".$row_peg2['nama']."</option>";
}
?>
</select>
</td>
<td>
<?php
if( isset($_POST['peg_2']) ){
$peg_2=$_POST['peg_2'];
echo "<input type='text' name='peg2' value='".$peg_2."'>"; // 2nd result throw into 2nd texbox
$sql_NIP2="SELECT NIP FROM users WHERE nama='$peg_2'";
$result_NIP2=$conn->query($sql_NIP2);
$row_NIP2=$result_NIP2->fetch_assoc();
?>
</td>
<td> NIP</td>
<td>:</td>
<td><input type='text' name='NIP2' value="<?php echo $row_NIP2['NIP'];?>"> <!--2nd child of result-->
<?php
}
}
if(isset($_POST['NIP2'])){
$NIP2=$_POST['NIP2'];
echo "<br /> NIP2 :".$NIP2."<br />";
}
mysqli_close($conn);
?>
</td>
</tr>
</table>
</form>
<form method="POST" name="wilayah" id="wilayah" action="<?php $_SERVER['PHP_SELF'];?>">
<table border="1">
<tr>
<td>
<select name="select_provinsi" onchange="submit(this)" style="width:18;">
<option selected>PROVINSI</option>
<?php
include 'dbconn.php';
$sql_prov="SELECT * FROM wilayah GROUP BY provinsi";
$result_prov=$conn->query($sql_prov);
echo "";
while($row_prov=$result_prov->fetch_assoc()){
$provinsi=$row_prov['provinsi'];
echo "<option value='".$provinsi."'>".$provinsi."</option>";
}
?>
</select>
<?php
if(isset($_POST['select_provinsi'])){
$select_provinsi=$_POST['select_provinsi'];
echo "
<input type='text' name='select_provinsi' value='".$select_provinsi."' placeholder='PROVINSI'>
</td>
</tr>";
$sql_kabkota="SELECT * FROM wilayah WHERE provinsi='$select_provinsi' GROUP BY kab_kota";
$result_kabkota=$conn->query($sql_kabkota);
?>
<tr>
<td>
<select name="select_kabkota" style="width:18px;" onchange="submit(this)"><option>KAB/KOTA</option>
<?php
while($row_kabkota=$result_kabkota->fetch_assoc()){
echo "<option>".$row_kabkota['kab_kota']."</option>";
}
?>
</select>
<?php
}
if(isset($_POST['select_kabkota'])){
$select_kabkota=$_POST['select_kabkota'];
?>
<input type="text" name="kab_kota" value="<?php echo $select_kabkota;?>">
<?php
}
mysqli_close($conn);
?>
</td>
</tr>
</table>
</form>
hope any suggestion for resolved of my problem with them,,
onchange="submit(this)" means that you want to submit the form when the value of the combobox changes. So, when the form is sent, the page reloads and you get the default value of your form.
To restore the chosen value, I would do something like :
<select name="select_kabkota" style="width:18px;" onchange="submit(this)">
<option>KAB/KOTA</option>
<?php
if(isset($_POST['select_kabkota']))
$select_kabkota=$_POST['select_kabkota'];
while($row_kabkota=$result_kabkota->fetch_assoc())
{
$selected = $select_kabkota == $row_kabkota['kab_kota'] ? 'selected="selected"' : '';
echo "<option ".$selected." >".$row_kabkota['kab_kota']."</option>";
}
?>
</select>
Related
I am trying to populate Material Description from the Material Number. Bot hte values are stored in same SQL tasble. So what I want when I select Material Maiterial Description shpul auto populate.
Fileds in table are Material & MaterialDescripotion
Below is code in in main file where data is fetched
<?php
include_once "dbConnect.php";
$sql = "SELECT * FROM DRLINK";
$result2 = mysqli_query($conn, $sql);
if (!$result2) {
printf("Error: %s\n", mysqli_error($conn));
exit();
}
$options = "";
while($row2 = mysqli_fetch_array($result2))
{
$options = $options."<option>$row2[1]</option>";
}
?>
<html>
<!DOCTYPE html>
<head>
<title>Dropdown Ajax</title>
</head>
<body>
<form action ="DSSTRsubmit.php" method="POST">
<table border="1">
<tr>
<td>Select Retailer</td>
</tr>
<tr>
<td>
<?php
echo "<select>";
echo $options;
echo "</select>";
?>
</td>
</tr>
</table>
<br/>
<br/>
<br/>
<table border="1">
<tr>
<td>Material</td>
<td>Material Description</td>
<td>Quantity</td>
<td>Unit of Measure</td>
</tr>
<tr>
<td>
<div class="Material">
<select name="Material" onchange="getId(this.value);">
<option value="">Select Country</option>
<?php
$query ="SELECT * FROM MATERIALLIST";
$results = mysqli_query($conn, $query);
foreach($results as $MATERIALLIST) {
?>
<option value="<?php echo $MATERIALLIST["Material"];?>"><?php echo $MATERIALLIST["Material"];?></option>
<?php
}
?>
</select>
</div>
</td>
<td>
<div class="MaterialDescription">
<select name="MaterialDescription" id="DesList">
<option value=""></option>
</select>
</div>
<script src="//code.jquery.com/jquery-1.12.0.min.js"></script>
<script>
function getId(val){
$.ajax({
type:"POST",
url:"getdata1.php",
data:"Material="+val,
success: function(data){
$("#DesList").html(data);
}
});
}
</script>
</td>
<td><input type="text" name="dquantity_name" /> </td>
<td><input type="text" name="duom_name" /> </td>
</tr>
</table>
<legend> </legend>
<p> <button type="submit" class="pure-button pure-button-primary">Send Stock</button>
<br>
<br>
<?php
echo "Distributor Page";
?>
</body>
</html>
below is the getdata1.php
<?php
include_once "dbConnect.php";
if(!empty($_POST["Material"])){
$Material= $_POST["Material"];
$query = "SELECT * FROM MATERIALLIST WHERE Material = $Material";
$results = mysqli_query($conn,$query);
foreach($results as $MaterialDescription){
?>
<option value="<?php echo $Des["Material"];?>"><?php echo $materialDescription ["MaterialDescription"];?></option>
<?php
}
}
?>
Iam able to select the material but on selection of material no material description auto populates.
Thanks for the help
PHP is case sensitive. Be careful about naming variables.
foreach contains upper case variable, while echo - lowercase. Also in echo you have space after variable and before opening bracket.
$MaterialDescription
$materialDescription_["MaterialDescription"]
Im trying to display my database value into the textbox using drop down menu. which is i did and it is displaying. the problem here is that when i choose an item in the drop down list, it goes back to the first choice or last choice, the explanation i got was, my loop is selecting all of the items in the field causing the drop down menu to go back to the first choice when i click on other items. can you help me with the code on how to stop going back to the first choice when i select other options. Here is my whole code. i also use functions.
home.php
<?php
session_start();
include('dbconnect.php');
include('functions.php');
if(isset($_POST['brandname'])){
$id = $_POST['brandname'];
$result = mysql_query("SELECT * FROM tblstore WHERE brandname = '$id'");
while($row = mysql_fetch_array($result)){
$price = $row['price'];
$stocks = $row['stocks'];
}
}
?>
<html>
<body>
<form method="POST" name="">
<table align="center">
<tr>
<td>Choose here:</td>
<td>
<select name = "brandname" onchange = "this.form.submit()">
<?php dropdown() ?>
</select>
</td>
</tr>
<tr>
<td>Quantity:</td>
<td><input type="text" name="qty" id="qty" value="" /></td>
</tr>
<tr>
<td>Price:</td>
<td><input type="text" name="price" id="price" value="<?php echo $price ?>" disabled/></td>
</tr>
<tr>
<td>Stocks:</td>
<td><input type="text" name="stocks" id="stocks" value="<?php echo $stocks ?>" disabled/></td>
</tr>
<tr>
<td>Total:</td>
<td><input type="text" name="total" id="total" disabled/></td>
</tr>
<tr>
<td></td>
</tr>
</table>
</form>
<div align = "center">
hi' <?php echo $userRow['username']; ?> Sign Out
</div>
</body>
</html>
functions.php
<?php
function dropdown(){
$all = mysql_query("SELECT * FROM tblstore");
while($row = mysql_fetch_array($all)){
echo "<option value = '".$row['brandname']."' selected='selected'>" .$row['brandname'] . "</option>";
}
}
feel free to edit the whole code.. im a beginner in php and learning my way to it. thanks
Can add the multiple option if you need to select multiple
<select name="brandname" multiple>
<option value="Select">Select</option>
<?php
do {
?>
<option value="<?php echo $row['brandname']?>"> <?php echo $row['brandname'] ?></option>
<?php
} while ($row = mysql_fetch_assoc($all));
?>
</select>
I have some dropdown menus. If I select one of them I want to change the other's content with something that is reliable with what i selected. How can I do that ? With html and php.
For example I have one table
Year with id_year and year
Stuff with id_stuf and stuff
If I select an year in the first dropdown menu , in the other drodown menu will show only the stuff from that year .
This is my content with the dropdown menus
<div class="view">
<form name="tabel" method="post" action="insertexamen.php">
<table>
<tr>
<td>Data</td>
<td><input type="date" name="data" value="data" required="required"/><br></td>
</tr>
<tr>
<td>An</td>
<td>
<?php
$sql_year="SELECT * FROM an";
$rez_year = mysqli_query($link,$sql_year);
echo "<select name=\"year\" >";
while($year=mysqli_fetch_array($rez_year))
{
echo "<option value=\"".$year['id_an']."\">".$year['grupa']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Materie</td>
<td>
<?php
$sql_mat="SELECT * FROM materii";
$rez_mat = mysqli_query($link,$sql_mat);
echo "<select name=\"mat\" >";
while($mat=mysqli_fetch_array($rez_mat))
{
echo "<option value=\"".$mat['id_mat']."\">".$mat['numemat']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Profesor</td>
<td>
<?php
$sql_proff="SELECT * FROM profesor";
$rez_proff = mysqli_query($link,$sql_proff);
echo "<select name=\"proff\" >";
while($proff=mysqli_fetch_array($rez_proff))
{
echo "<option value=\"".$proff['id_prof']."\">".$proff['numep']." ".$proff['prenumep']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Asistent</td>
<td>
<?php
$sql_profff="SELECT * FROM profesor";
$rez_profff = mysqli_query($link,$sql_profff);
echo "<select name=\"profff\" >";
while($profff=mysqli_fetch_array($rez_profff))
{
echo "<option value=\"".$profff['id_prof']."\">".$profff['numep']." ".$profff['prenumep']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Sala</td>
<td>
<?php
$sql_room="SELECT * FROM sala";
$rez_room= mysqli_query($link,$sql_room);
echo "<select name=\"room\" >";
while($room=mysqli_fetch_array($rez_room))
{
echo "<option value=\"".$room['id_s']."\">".$room['salaa']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td>Tip</td>
<td>
<?php
$sql_type="SELECT * FROM examen";
$rez_type= mysqli_query($link,$sql_type);
echo "<select name=\"type\" >";
while($type=mysqli_fetch_array($rez_type))
{
echo "<option value=\"".$type['id_tip']."\">".$type['tip']."</option>";
}
echo "</select>";
?><br>
</td>
</tr>
<tr>
<td><input name="submit" type="submit" value="Trimite"/></td>
<td><input name="reset" type="reset" value="Reset"/></td>
</tr>
</table>
Are you looking for something like that:
<html>
<title>dropdownlist</title>
<head>
<script language="Javascript" type="text/javascript" >
function choix(formulaire)
{
var j;
var i = form1.boite1.selectedIndex;
if (i == 0)
for(j = 1; j <3; j++)
form1.boite2.options[j].text="";
else{
switch (i){
case 1 : var text = new Array( "London","Manchester","Birmingham");
break;
case 2 : var text = new Array("Paris","Marseille","Lyon");
break;
case 3 : var text = new Array("Berlin","Munich","Francfort");
break;
}
for(j = 0; j<3; j++)
form1.boite2.options[j+1].text=text[j];
}
form1.boite2.selectedIndex=0;
}
</script>
</head>
<body>
<form name="form1">
<select name="boite1" onChange="choix(this.form)">
<option selected>country</option>
<option>England</option>
<option>France</option>
<option>Germany</option>
</select>
<select name="boite2">
<option selected>cities</option>
<option></option>
<option></option>
<option></option>
</form>
</select>
</body>
</html>
if you want absolutely to do it using only html and php you will need to use some ajax because php is serverside and html is slient side. so for what you asy I would recommend the code above
I am using this code to serach database and I am using 4 fields in this code but this code does not serach database value.
where problem in this code tell me plz edit my code for full working serach with 4 fields
my code :
<?php
{
include ('connection.php');
if(isset($_REQUEST['submit'])){
$optid = $_POST['OPRID'];
$optdec = $_POST['OPRDEFNDESC'];
$empid = $_POST['EMPLID'];
$empmail = $_POST['EMAILID'];
$query ="SELECT * FROM OPERATOR WHERE OPRID LIKE '%".$optid."%'
or OPRDEFNDESC LIKE '%".$optdec."%' or EMPLID LIKE '%".$empid."%'
or EMAILID LIKE '%".$empmail."%' ";
}
else{
$query="SELECT * FROM OPERATOR";
$objParse = oci_parse ($ora_conn, $query);
}
?>
<form action="multi.php" method="get" action="<?=$_SERVER['SCRIPT_NAME'];?>">
<table width="500" border="1" align="center">
<tr>
<th>Operator ID
<input name="OPRID" type="text" id="OPRID" value="";>
<tr>
<th>Operator Name
<input name="OPRDEFNDESC" type="text" id="OPRDEFNDESC" value="";>
<tr>
<th>Person ID
<input name="EMPLID" type="text" id="EMPLID" value="";>
<tr>
<th>Email ID
<input name="EMAILID" type="text" id="EMAILID" value="";>
<input type="submit" value="Search"></th>
</tr>
</table>
</form>
<table>
<tr>
<td>Operator ID</td>
<td>Operator Name</td>
<td>Person ID</td>
<td>Email ID</td>
</tr>
<?
$success = oci_execute($objParse);
//while($objResult = oci_fetch_array($objParse,OCI_BOTH))
while($objResult = oci_fetch_array($objParse, OCI_RETURN_NULLS+OCI_ASSOC))
{
?>
<tr>
<td><div align="center"><?=$objResult["OPRID"];?></div></td>
<td><?=$objResult["OPRDEFNDESC"];?></td>
<td><?=$objResult["EMPLID"];?></td>
<td><div align="center"><?=$objResult["EMAILID"];?></div></td>
<td align="center"><a href="Optr_Edit.php?OprID=
<?=$objResult["OPRID"];?>">Edit</a></td>
</tr>
<?
}
?>
</table>
<?
oci_free_statement($objParse);
oci_close($ora_conn);
}
?>
Try like this
<tr>
<td><div align="right"><strong>Password Encrypted:</strong></div></td>
<td>
<select name="txtENCRYPTED">
<option value="">Select</option>
<option <?php if ($objResult["ENCRYPTED"] == "Y") {echo 'selected';} ?>value="Y">Y</option>
<option <?php if ($objResult["ENCRYPTED"] == "N") {echo 'selected';} ?> value="N">N</option>
</select>
</td>
</tr>
You are doing it wrong
Select element do not have value attribute
You have value attribute only in options element.
For eg:
<select name="txtENCRYPTED" id="txtENCRYPTED">
<option value="">Select</option>
<option value="Y">Y</option>
<option value="N">N</option>
</select>
In your code you have provided the db-retrived-data in tags setting .
The tag defines the menu. It can have the following settings
The name setting adds an internal name to the field so the program that handles the form can identify the fields.
The size option defines how many items should be visible at a time. Default is one item.
The multiple setting will allow for multiple selections if present.
The tag defines the single items in the menu.
The value setting defines what will be submitted if the item is selected.
one solution :-
<form method="post" action="" >
<select name="encrypt" value="encrypted" id='select'>
<option value="">Select</option>
<option value="<?php if($objResult["ENCRYPTED"]=='Y'){ echo 'Y'; } ?>">Y</option>
<option value="<?php if($objResult["ENCRYPTED"]=='N'){ echo 'N'; } ?>">N</option>
</select>
<input type="submit" value="submit" id='form'/>
</form>
</td>
</tr>
//script type jquery.js
//script type javascript
$(document).ready(function(){
$('form').submit(function(){
alert($('#select').val());
});
});
</script>
here I have another question, how can we update the database through html/php form with some of the data was populated using dropdown menu?
It is something like this:
User choose to update the asset, and the asset will display the form with the stored data, and they can just directly update the asset using that form again (this is including the dropdown menu).
Hope some of you can get back to me soon. Really appreciate it.
Thank you.
Edited:
<form method="POST" name="update" action="update.php">
<h2>EDIT</h2>
<table class="reference" cellspacing="0" cellpadding="0" border="1" width="60%" id="tablecss">
<tr>
<td> Category </td>
<td> <?php echo '<select name="categoryid">';
foreach ($category as $data)
{
echo '<option'.($row['name']==$data? ' selected' : '').'>'.$data.'</option>';
}
echo '</select>'; ?> </td>
</tr>
<tr>
<td> Brand </td>
<td> <?php echo '<select name="brandid">';
foreach ($brand as $data1)
{
echo '<option'.($row['name']==$data1? ' selected' : '').'>'.$data1.'</option>';
}
echo '</select>'; ?> </td>
</tr>
<tr>
<td> Location </td>
<td> <?php echo '<select name="locationid">';
foreach ($location as $data2)
{
echo '<option'.($row['name']==$data2? ' selected' : '').'>'.$data2.'</option>';
}
echo '</select>'; ?> </td>
</tr>
<tr>
<td> Staff </td>
<td> <?php echo '<select name="staffno">';
foreach ($staff as $data3)
{
echo '<option'.($row['name']==$data3? ' selected' : '').'>'.$data3.'</option>';
}
echo '</select>'; ?> </td>
</tr>
<tr>
<td> Supplier </td>
<td> <?php echo '<select name="supplierid">';
foreach ($supplier as $data4)
{
echo '<option'.($row['name']==$data4? ' selected' : '').'>'.$data4.'</option>';
}
echo '</select>'; ?> </td>
</tr>
<tr>
<td> Project </td>
<td> <?php echo '<select name="projectid">';
foreach ($project as $data5)
{
echo '<option'.($row['name']==$data5? ' selected' : '').'>'.$data5.'</option>';
}
echo '</select>'; ?> </td>
</tr>
<tr>
<td> <input type="hidden" name="assetid" value="<?php echo "$assetid"; ?>" > </td>
</tr>
</table>
<input type="submit" name="submit" value="Update" onclick="return confirm('Update this?');">
<input type="button" value="Back" onclick="history.back();">
</form>
So, basically this can't function well. They didn't capture the stored data, but it can update into the database. Any help?
This is pretty straight forward form processing and you could have found out how to do it with 1 minute of googling...
//formpage.php:
<?php
$data_to_populate_dropdown = array();
//run some sql query or whatever to populate that array
//..or just populate the dropdown statically, then there's no need for this
//if the form has been submitted
$errors = array();
if(isset($_GET['action']) && $_GET['action']=='process') {
if(isset($_POST['my_dropdown'])) {
$my_dropdown_value = mysql_real_escape_string($_POST['my_dropdown']);
//write this to db...
header('Location: formpage.php'); //redirect to avoid double posting;
} else {
$errors[] = 'Some data is missing';
}
}
if(!empty($errors)) {
foreach($errors as $error) {
echo $error;
}
}
?>
<form action="formpage.php?action=process" method="post">
<select name="my_dropdown">
<?php foreach($data_to_populate_dropdown as $key => $value) {
echo '<option value="'.$key.'">'.$value.'</option>';
} ?>
</select>
<input type="submit" value="OK" />
</form>
This code is not tested and quite abstract, but I think you get the idea.