how can i set it that the section drop down list shows all sections available for a certain subject, depending on the previous drop down list that contains subjects. For example: if i chose the subject A , it should show me all sections of subject A available. Currently the drop down of the subject works, it fetches all subjects available but im unable to make the second drop down fetch the sections since i want to make such that:
select section from class WHERE (name of subject chosen above is equal to the name column in the database.
Please ignore my code style since im a beginner after all. Thank you in advance.
My code:
<div>
<label for="subjects" accesskey="o">Subject</label>
<?php
$conn = new mysqli('localhost', 'afhfhdfhf', 'fhfhhfhfhfhf', 'fhfhfhfh')
or die ('Cannot connect to db');
$result = $conn->query("select name from class");
echo "<html>";
echo "<body>";
echo "<select name='subject'>";
while ($row = $result->fetch_assoc()) {
unset($id, $name);
$name = $row['name'];
echo '<option value="subject">'.$name.'</option>';
}
echo "</select>";
echo "</body>";
echo "</html>";
?>
</div>
<br>
<div>
<label for="section" accesskey="o">Section</label>
<?php
$conn = new mysqli('localhost', 'fgfgfgfg', 'rfgfgfgfg!~fgfgf', 'fgfgfgf')
or die ('Cannot connect to db');
$result = $conn->query("select section from class WHERE name = '$name'");
echo "<html>";
echo "<body>";
echo "<select name='id'>";
while ($row = $result->fetch_assoc()) {
unset($id, $name);
$name = $row['name'];
echo '<option value="">'.$name.'</option>';
}
echo "</select>";
echo "</body>";
echo "</html>";
?>
</div>
<br>
(n.b. this really should have been a comment, but it was too long for that, so answer it is :-)
You've got a few problems there... first, you are echoing out <html> and <body> tags, you already have those on the page, so you don't need to do that.
The second problem is that $name doesn't really exist in the way you are looking in the second query. In fact, it'll be the value of the last name you echo'd out in the first loop.
So, the fundamental issue is that PHP runs on the server and then is passed to the client and you want to limit a select based on user input that happens in the browser. So what you need to do is either on selection of the subject, submit back to the server to get the section OR you could use something like jQuery to do an AJAX call back to the server to retrieve this info (either as JSON that you can parse and use, or as HTML that you can plop into the form). 2nd option is definitely the way to go.
Have a google about this kinda thing and you'll find LOADS of examples and come on back here if you hit a road block :-)
Related
I am writing code to register a new project via a html form. I want to be able to click a dropdown box which pulls the values from a table on the database.
At the moment a dropdown box displays but with no values.
PLEASE NOTE: I am a learning the basics so apologies if this is a simple question/answer scenario.
My code is below, any help is appreciated, I connect to the database via a php include script. The table is called 'customers' and the item I want to list is 'name';-
<?php
$result = mysql_query("SELECT customers FROM name");
echo "<select name='client'>";
while($row = mysql_fetch_assoc($result))
{
echo "<option value = '".$row[name]."'>".$row[name]."</option>";
}
echo "</select>";
?>
"The table is called 'customers' and the item I want to list is 'name';" -
Do SELECT name FROM customers instead of SELECT customers FROM name
Using mysql_error() to mysql_query()
would have shown you the error that the table name does not exist.
Plus,
[name] are missing quotes inside them => ['name'] which are being treated as constants.
in
echo "<option value = '".$row[name]."'>".$row[name]."</option>";
as caught and kudos to devdesign
echo "<option value = '".$row['name']."'>".$row['name']."</option>";
However, you are using a deprecated MySQL library. If you are still not getting results, then this could mean that you need to use (and should use) mysqli_ or PDO instead.
Here are a few links on the subject:
mysqli with prepared statements
PDO with prepared statements.
Your code should be. Also note the single quotes within $row['name']. You missed that.
<?php
$result = mysql_query("SELECT name FROM customers");
echo "<select name='client'>";
while($row = mysql_fetch_assoc($result))
{
echo "<option value = '".$row['name']."'>".$row['name']."</option>";
}
echo "</select>";
?>
I'm having an issue in passing a value from a dropdown list that is in a separate PHP file that is being used by jquery.
I ended up getting the values from the dropdown list that isn't posting correctly by using jquery based on the value selected in the first dropdown list. The dropdown list in question is populating correctly, but with the way I have it setup I cannot post the value to the submit PHP page.
I'm pretty sure it has to do with the way I have it setup; however, I'm very new to jquery and was looking for some guidance.
The main PHP Page (the small areas in question)
<select name="department_list" id="department_list" onchange="$('#singleUser').load('get_users.php?nid='+this.value);">
...
<div id="singleUser" class="singleUser">
</div>
The PHP page (get_users) used to fill the values (only the area in question)
echo '<p style="text-align:center">';
echo "
<br />
Select a person to receive the form
<br />";
echo "<select id='userSelect' class='userSelect'>";
if ($id == '50') {
echo "<option value='Done'>Done</option>";
echo "</select>";
}else {
echo "<option value='none'>Select user</option>";
try {
$db = new PDO("mysql:host=$hostname;dbname=$dbname", $username, $password);
$stmt = $db->prepare("SELECT dm.empid AS empid, u.name AS name FROM mytable dm
JOIN mytable2 u ON dm.empid = u.id
WHERE dm.deptid = :id
ORDER BY u.name");
$stmt->bindParam(':id', $id);
$stmt->execute();
while ($r = $stmt->fetch()) {
$empid = $r['empid'];
$userName = $r['name'];
echo "<option value='".$empid."'>".$userName."</option>";
}
echo "</select>";
echo "</p>";
$db = null;
}
catch (PDOException $ex) {
echo "An Error occurred!";
}
}//end else
In the submit page:
if(isset($_POST['userSelect'])){
$givenID = $_POST['userSelect'];
//the rest of my code
I do have the div code above within the form tags and have method="post". All of my other inputs post correctly, so I'm thinking it has to do with the way I have only the div tags within the main page. Again, I'm pretty new to all of this so any ideas or changes that I should make so it posts correctly would be greatly appreciated.
You forgot the name of the select when you write it with php:
change this:
echo "<select id='userSelect' class='userSelect'>";
to
echo "<select id='userSelect' name='userSelect' class='userSelect'>";
i think the error is in the PHP file generating the user select it is missing the name attribute name="userSelect"
echo "<select id='userSelect' class='userSelect'>";
it should be
echo '<select id="userSelect" name="userSelect" class="userSelect">';
every form element with the name attribute gets posted with its value. if you do not enter the name attribute, the value can not be retrieved from the $_POST array. Also have in mind that disabled form inputs also do not get posted.
Edited the PHP quotes. Use Single qutes everyt time you do not need to insert PHP variable into the string. It is ~9 times faster than double quotes ;)
Hello i am new to php and i have tried to find a piece of code that i can use to complete the task i need, i currently have a page with a form set out to view the criteria of a course. also i have a dropdown menu which currently holds all the course codes for the modules i have stored in a database. my problem is when i select a course code i wish to populate the fields in my form to show all the information about the course selected. The code i am trying to get to work is as follows:
<?php
session_start();
?>
<? include ("dbcon.php") ?>
<?php
if(!isset($_GET['coursecode'])){
$Var ='%';
}
else
{
if($_GET['coursecode'] == "ALL"){
$Var = '%';
} else {
$Var = $_GET['coursecode'];
}
}
echo "<form action=\"newq4.php\" method=\"GET\">
<table border=0 cellpadding=5 align=left><tr><td><b>Coursecode</b><br>";
$res=mysql_query("SELECT * FROM module GROUP BY mId");
if(mysql_num_rows($res)==0){
echo "there is no data in table..";
} else
{
echo "<select name=\"coursecode\" id=\"coursecode\"><option value=\"ALL\"> ALL </option>";
for($i=0;$i<mysql_num_rows($res);$i++)
{
$row=mysql_fetch_assoc($res);
echo"<option value=$row[coursecode]";
if($Var==$row[coursecode])
echo " selected";
echo ">$row[coursecode]</option>";
}
echo "</select>";
}
echo "</td><td align=\"left\"><input type=\"submit\" value=\"SELECT\" />
</td></tr></table></form><br>";
$query = "SELECT * FROM module WHERE coursecode LIKE '$Var' ";
$result = mysql_query($query) or die("Error: " . mysql_error());
if(mysql_num_rows($result) == 0){
echo("No modules match your currently selected coursecode. Please try another coursecode!");
} ELSE {
Coursecode: echo $row['coursecode'];
Module: echo $row['mName'];
echo $row['mCredits'];
echo $row['TotalContactHours'];
echo $row['mdescription'];
echo $row['Syllabus'];
}
?>
however i can only seem to get the last entry from my database any help to fix this problem or a better way of coding this so it works would be grateful
Thanks
The main error is in your final query, you're not actually fetching anything from the query, so you're just displaying the LAST row you fetched in the first query.
Some tips:
1) Don't use a for() loop to fetch results from a query result. While loops are far more concise:
$result = mysql_query(...) or die(mysql_error());
while($row = mysql_fetch_assoc($result)) {
...
}
2) Add another one of these while loops to your final query, since it's just being executed, but not fetched.
For me i would use some javascript(NOTE: i prefer jQuery)
An easy technique would be to do this(going on the assumption that when creating the drop downs, your record also contains the description):
Apart from creating your dropdown options like this <option value="...">data</option>, you could add some additional attributes like so:
echo '<option value="'.$row['coursecode'].'" data-desc="'.$row['description'].'">.....</option>
Now you have all your drop down options, next is the javascript part
Let's assume you have included jQuery onto your page; and let's also assume that the description of any selected course is to be displayed in a <div> called description like so:
<div id="course-description"> </div>
<!--style it how you wish -->
With your javascript you could then do this:
$(function(){
$("#id-of-course-drop-down").change(function(){
var desc = $(this).children("option").filter("selected").attr("data-des");
//now you have your description text
$("#course-description").html(desc);
//display the description of the course
}
});
Hope this helps you, even a little
Have fun!
NOTE: At least this is more optimal than having to use AJAX to fecch the description on selection of the option :)
needing some advice.
I am wanting to include 4 drop down lists on a website, which all contain data from different fields in a mysql table.
I then want to be able to press a submit button and display the required data on a webpage.
I am having trouble with knowing what programming language to use and also finding it difficult to find any tutorials. Any help would be greatly appreciated.
Thanks
You mean a HTML dropdown list or just a select dropdown in a form?
If you mean a select dropdown in a form you could do something like this:
<?PHP
$query = mysql_query("SELECT value FROM table ORDER BY value ASC") or die(mysql_error());
$result = mysql_num_rows($result);
// If no results have been found or when table is empty?
if ($result == 0) {
echo 'No results have been found.';
} else {
// Display form
echo '<form name="form" method="post" action="your_result.php">';
echo '<select name="list" id="lists">';
// Fetch results from database and list in the select box
while ($fetch = mysql_fetch_assoc($query)) {
echo '<option id="'.$fetch['value'].'">'.$fetch['value'].'</option>';
}
echo '</select>';
echo '</form>';
}
?>
And then in your_result.php you should fetch the data from your MySQL database based on value from the (when you fetch use mysql_real_escape_string):
<?PHP $_POST['list']; ?>
You could also do everything in just one file, but thats up to you. Try to use google, there are dozens of tutorials out there.
i am trying to use the mysql data items into the select combo box. it basically works well but the problem is when there are multiple combo boxes it is a lot of load to the server since adding each combo box takes a lot of time. i am trying to figure out a better way. may be pull date once into an array just for the session and place it in the combo boxes.
The logic is basically it is a quotation form where about 3500 items will be shown as drop down and user will select and then enter price and other details. the rows are dynamically added or deleted by the user.
i am currently using the following code:-
<?php
$con = mysql_connect('blah blah blah');
if (!$con) {
die ('Could not connect: ' . mysql_error());}
$db = mysql_select_db('blah',$con);
$extract1 = mysql_query("query") OR die (mysql_error());
$numrows1 = mysql_num_rows($extract1);
echo "<select name='itemname' title='select Item Name'>";
echo "
<option>Select Item Description</option>
";
while ($row1=mysql_fetch_assoc($extract1))
{
$ic=$row1['ItemName'];
echo "
<option>$ic</option>
";
}
echo "</select>";
mysql_close($con);
?>
Don't echo your option do this in your while statement:
$ic[]=$row1['ItemName'];
then outside of the while loop anywhere on the page:
foreach($ic as $i){
echo "<option>".$i."</option>";
}
First off, Don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.
<?php
$con = mysql_connect('blah blah blah') or die(mysql_error());
$db = mysql_select_db('blah',$con) or die(mysql_error());
$result = mysql_query("query MAYBE NARROW DOWN TO MORE RELEVANT RESULT SET") or die (mysql_error());
$option = '<select size="1" name="optionBox">';
if(mysql_num_rows($result)>=1){
while ($row=mysql_fetch_assoc($result)){
$option .="<option selected value=\"".$row['ItemName']."\">".$row['ItemName']."</option>\n";
}
}else{
$option .='<option selected value="0">No items to list</option>';
}
$option .='</select>';
echo $option;
mysql_close($con);
?>
Yes, if your data changes infrequently enough, it may be a good idea to put the data into an array on the session, and render it from there. Depending on the frequency of change of your data, you might be able to get away with rendering it to a non-session data element (for example, a file on your filesystem) and populating your comboboxes from there (or just rendering all your choices into the combobox elements in that file, and using that data directly); that depends on the frequency by which your data is updated, of course.