So, I wanna change, using values from a comboBox, another comboBox values (from database MYSQL ). Don't know where the problem is but thinking about that function.
Sorry for my english btw, isn't my native language.
Here are my dropdown lists in HTML:
<select name="Discipline" id="DisciplineList">
<?php include "combo1.php";
?>
</select>
</div>
<div class="alegere2">
<?php echo "Alegeti lectia:" ;
?>
<select name="lectie" id="lectieList">
</select>
Here is the code that I use to fill the second dropdown list using the values from the first one:
< script type="text/javascript">
$(document).on("change","#DisciplineList",function(){
var val = $(this).val();
$.ajax({
url: "combo2.php",
data: {Discipline:val},
type: "GET" ,
dataType: "html",
success: function(result){
$("#lectieList").html(result);
}
});
});
< /script>
Here is the first comboBox fill: ( it fills corectly)
<?php
require("db.php");
$results = mysqli_query($db,"SELECT nume,id FROM discipline");
$nr_discipline=mysqli_num_rows($results);
while($nr_discipline > 0){
$row = mysqli_fetch_row($results);
echo '<option value="'.$row[1].'">'.$row[0].'</option>';
$nr_discipline--;
}
?>
Here is the 2nd comboBox code : (isn't working )
<?php
// Connects to your Database
require("db.php");
$id_discipline = $_GET['Discipline'];
$Query= "SELECT nume,id FROM lectii WHERE id_disciplina =2";
$lectie = mysqli_query($db,$Query);
$nr_lectie = mysql_num_rows($lectie);
while ($nr_lectie > 0) {
$row = mysql_fetch_row($lectie);
echo '<option value="'.$row[1].'">'.$row[0].'</option>';
$nr_lectie--;
}
?>
I don't know where is the problem. It looks like the function is not working at all.
$("#lectie").html(result);
here lectie is class not id replace lectie by lectieList that is the id
Related
Is that possible to select variable in DB (such as : $username) and display their info in textbox ?
Example:
Select : $username
after the selection, in the text input , will show:
$username , $usernameEmail , $usernameContactNumber
and click on "Register", it will insert to database with the information
I have no idea how does this work. I have no problem in inserting value to database.
The only problem that I am unable to solve right now is the condition I needed.
When I choose $username (example: userA),
all the information of userA will be appearing in the text input below
$query = "SELECT * FROM userdatabase";
$result1 = mysqli_query($db,$query);
<select>
<?php while($row1 = mysqli_fetch_array($result1)):; ?>
<option value="<?php echo $row1[0];?>"><?php echo $row1[2];?></option>
<?php endwhile;?>
</select>
The code above is the selection (PHP
<script>
$(document).ready(function(){
$("form#get").submit(function(event) {
event.preventDefault();
var input = $("#username");
$.ajax({
type: "POST",
url: "userregisteraccount.php",
success: function(data) { input.val(data); }
});
});
});
</script>
code above is the ajax
so that when i choose "userName" , "userName" data will be show in the registerform
So I'm having this issue with geting a value from a dropdown list in HTML into a variable so that I can do the mysql query. This will be a little tricky to explain but I will do my best. (Do not be shy in correcting my english or my expressions so that the question becomes more concrete and easy to understand).
So I have this dropdown list that gets his values from a mysql query.
<td>Designação atual :</td> <td><select name="desig_act" id="desig_act">
<?php
while ($row2 = mysql_fetch_assoc($result4)) {
echo "<option size=30 value=".$row2['new_name_freg'].">".$row2["new_name_freg"]."</option>";
}
?>
</select>
This "connects" with this query:
$sql4 = ("SELECT DISTINCT new_name_freg FROM freguesias WHERE codg_cc = '$pesq'");
$result4 = mysql_query($sql4, $link);
This query will populate the dropdown list with values. What I'm seeking to do is to populate another dropdown list. For examples I select a list of countrys, and when I select on country it should appear all it's citys in the other dropdown list.
I have been searching guys. Belive me I have.
P.s: Please do not get mad if I change the question a couple of times when I see that you guys show me a way to explain it better. Sorry if my english isn't perfect. Thank you guys for the help.
You can do it with ajax and jquery. I try to write little example
<!-- index.php -->
<select name="desig_act" id="desig_act">
<?php while ($row2 = mysql_fetch_assoc($result4)): ?>
<option value="<?=$row2['new_name_freg']?>">
<?=$row2["new_name_freg"]?>
</option>
<?php endwhile; ?>
</select>
<!-- select tag for countries -->
<select name="country" id="country"></select>
write a little script to return countries as json
<?php //ajax-countries.php
$link = mysql_connect(); // connect as usual
$query = ("SELECT * FROM countries");
$result = mysql_query($query, $link);
$json = array();
while ($row = mysql_fetch_assoc($result)) $json[] = $row;
echo json_encode($json);
?>
Then you can have some sort of script like:
// script.js
$("#desig_act").change(function(){
$.getJSON( "ajax-countries.php", function( data ) {
$.each( data, function(key, val) {
$("#desig_act").append("<option val='" + key + "'>" + val + "</option>");
});
});
I hope it can be useful
1: Create a PHP script to return the data
Essentially just generate the value based off the $_GET input.
2: Create a json request in jquery
Calls the PHP file which will return the data and you will use that data to add more values to the select.
<?php
//Step 1 - The posted ajax data that will return our json request.
if(isset($_GET['fetchrow'])) //Is our ajax request called on page load? If yes, go to this code block
{
//Other stuff like DB connection
$pesq = mysql_escape_string($_GET['fetchrow']); //Put our variable as the variable sent through ajax
$sql4 = ("SELECT DISTINCT new_name_freg FROM freguesias WHERE codg_cc = '$pesq'"); //Run our query
$result4 = mysql_query($sql4, $link); //Please change from mysql_* to mysqli
$data = array(); //The array in which the data is in
while($row = mysql_fetch_assoc($result4)) //Look through all rows
{
array_push($data, $row); //Put the data into the array
}
echo json_encode($data); //Send all the data to our ajax request in json format.
die; //Don't show any more of the page if ajax request.
}
?>
<html>
<head>
<script type='application/javascript' src='https://cdnjs.cloudflare.com/ajax/libs/jquery/3.0.0-alpha1/jquery.min.js'></script> <!--Include jquery -->
<script>
//Step #2:
//The jquery script calls ajax request on change of the first select
$( "#desig_act" ).change(function() {
$.getJSON('thisfilename.php', {fetchrow:$("#desig_act").val()}, function(data){ //Get the json data from the script above
var html = '';
var len = data.length;
for (var i = 0; i< len; i++) { //Loop through all results
html += '<option value="' + data[i].new_name_freg + '">' + data[i].new_name_freg + '</option>'; // Add data to string for each row
}
$('#otherselect').html(html); //Add data to the select.
});
});
</script>
</head>
<body>
<!-- Your html code -->
<td>Designação atual :</td> <td><select name="desig_act" id="desig_act">
<?php
while ($row2 = mysql_fetch_assoc($result4)) {
echo "<option size=30 value=".$row2['new_name_freg'].">".$row2["new_name_freg"]."</option>";
}
?>
</select>
</td>
<!-- The new select -->
<select name='otherselect' id='otherselect'>
</select>
<!-- Rest of html code -->
</body>
</html>
In a file following code is written which populate the value of combobox :
<select id="company" required="required" class="form-control" value = "select" name="subject_code" placeholder="Select" >
<?php
//Iterating list of subjects available to be filled .
echo "<option> Select </option>";
foreach ($subjects_to_fill as $subject_id => $subject_name) {
# code...
echo "<option value=".$subject_id."> ".$subject_name." </option>";
}
?>
</select>
On selecting a particular item from the combobox, I want to display the faculty_name dynamically from faculty_table on the basis of $subject_id.
table structure:
faculty_table(faculty_id,faculty_name,subject_id)
subject_table(subject_id,subject_name,faculty_id)
You will need to use ajax for this task.
Add a <div> and below script in your current file.
<div id="faculty"></div>
<script type="text/javascript">
function get_faculty()
{
var id = document.getElementById("company").value;
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_faculty.php",
data: dataString,
success: function(html)
{
$("#faculty").html(html);
}
});
}
</script>
Now create another get_faculty.php file in same folder which will be called on onchange event of company dropdown.
Write below code in that file
if($_POST['id'])
{
$id=$_POST['id'];
$con=mysqli_connect("localhost","username","pass","dbname");
$sql=mysqli_query($con,"SELECT faculty_name from faculty_table where subject_id = ".$id);
while($row=mysqli_fetch_array($sql))
{
echo $row['faculty_name'];
}
}
Dont forget to write mysqli_connect credentials and query accordingly.
Fancybox script POST my form data to go.php page then open go.php in fancybox iframe
<script>
$(document).ready(function() {
$("#fancybox-manual-b").click(function() {
$.post('go.php', $(this).parent().serialize())
.done(function() {
$.fancybox.open({
href : 'go.php',
type : 'iframe',
padding : 5
});
});
});
});
</script>
<select name="country">
<option value="US">US</option>
<option value="EU">EU</option>
</select>
<input type="button" value="Calculate" id="fancybox-manual-b"/>
in go.php, I receive the POST data from the form correctly and when I try to insert this data into DB, it's been inserted correctly too!
But now I want to select * from my DB table to echo all the data in the table where column = POST data, but this query doesn't work!
<?php
$country = $_POST[country];
mysql_query("INSERT INTO calculator (country) VALUES ('$country')"); //Works Correctly
$result = mysql_query("SELECT * FROM calculator WHERE country='$country' ORDER BY ID ASC");
while($row = mysql_fetch_array($result)){
echo $row['ID']." ".$row['country'];
} //Nothing appear
?>
Please checkout this method,
$country = $_POST[country];
if(mysql_query("INSERT INTO calculator (country) VALUES ('$country')"));
{
$result = mysql_query("SELECT * FROM calculator WHERE country='$country' ORDER BY ID ASC");
while($row = mysql_fetch_array($result)){
echo $row['ID']." ".$row['country'];
}
}
I used the same code you shared. But added an additional if loop on mysql_query part. It will make sure your select works after data being inserted.
We have to use jquery ajax to send form data to the php page with POST method then receive back any printed data from the php page in the ajax instead of the console.
We will open the fancybox iframe into ajax success instead of the console.
<script type="text/javascript">
$(document).ready(function() {
$("#fancybox-manual-b").click(function() {
$.ajax({
type: 'POST',
url: 'go.php',
data: $(this).parent().serialize(),
success: function(data){
$.fancybox(data);
},
fail:function(data){console.info(data)}
});
});
});
</script>
i tried to automatically select the value of multiselect box using data retrieved from database, but its not working...
this is the code for html multiselect
<select name="category" id="category" multiple="multiple" class="select validate[required]" style="width:100%">
</select>
at page load i dynamically load options for multiselect from database (mysql)
$.ajax({
url:'search/category.php',
type:'POST',
data:{cat_id:1}, //1 means jobs category
async:true,
success: function(data){
$("#category").html(data);
}
});
but if i have to auto select value in multiselect its not working
<?php
$qry ="select tags from posts where id='$id'";
$res = mysqli_query($con,$qry);
$row = mysqli_fetch_assoc($res);
$tags = $row['tags'];
?>
<script>
$(function(){
$("#category").val(<?php echo $tags; ?>); //example: .val(3);
}
</script>
ajax is correct but you need to echo select options correctly
in category function try this :
foreach ($tags as $tag=>$val) {
echo "<option value=".$val." > ".$tag."</option>";
}
You are just missing the quotes in your $("#category").val(); line. Check this fiddle out.
$("#category").val( "'" + <?php echo $tags; ?> + "'");
should get it to work