get user selected option var from $_post - php - php

<script>
//this script show / hide lanes depend on user school choice
$(document).ready(function () {
toggleFieldsA();
toggleFieldsB();
toggleFieldsC();
toggleFieldsD();
toggleFieldsE();
//this will call our toggleFields function every time the selection value of School field changes
$("#school").change(function () {
toggleFieldsA();
toggleFieldsB();
toggleFieldsC();
toggleFieldsD();
toggleFieldsE();
});
});
function toggleFieldsA() {
if ($("#school").val() == 'School of Economics')
$("#a").show();
else
$("#a").hide();
}
function toggleFieldsB() {
if ($("#school").val() == 'School of Computer Science')
$("#b").show();
else
$("#b").hide();
}
function toggleFieldsC() {
if ($("#school").val() == 'School of Behavioral Sciences')
$("#c").show();
else
$("#c").hide();
}
function toggleFieldsD() {
if ($("#school").val() == 'School of Government and Politics')
$("#d").show();
else
$("#d").hide();
}
function toggleFieldsE() {
if ($("#school").val() == 'School of Nursing')
$("#e").show();
else
$("#e").hide();
}
</script>
<!---this part is course registration form using jquery mobile, javascript, html, and php---->
</head>
<body>
<?php
if (!empty($_POST)) {
//getting other details from the order form (later using files - now only demo capabilities)
$school_name = $_POST[school_name];
echo $school_name;
$lane_name = $_POST[lane_name];
echo $lane_name;
$year = $_POST[year];
echo $year;
} else {
?>
<div id='school_div'>
<span><label>School</label></span>
<?php
$sql = "SELECT school_name FROM school";
$result = mysql_query($sql);
echo "<select id='school' name='school_name'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['school_name'] ."'>" . $row['school_name'] ."</option>";
}
echo "</select>";
?>
</div>
<!---End of School part---->
<div id='a'>
<span><label>Lane</label></span>
<?php
$sql = "SELECT lane_name FROM lane WHERE `lane_school_id` = 1 ORDER BY `lane_name` ASC";
$result = mysql_query($sql);
echo "<select name='lane_name'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['lane_name'] ."'>" . $row['lane_name'] ."</option>";
}
echo "</select>";
?>
</div>
<div id='b'>
<span><label>Lane</label></span>
<?php
$sql = "SELECT lane_name FROM lane WHERE `lane_school_id` = 2 ORDER BY `lane_name` ASC";
$result = mysql_query($sql);
echo "<select name='lane_name'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['lane_name'] ."'>" . $row['lane_name'] ."</option>";
}
echo "</select>";
?>
</div>
<div id='c'>
<span><label>Lane</label></span>
<?php
$sql = "SELECT lane_name FROM lane WHERE `lane_school_id` = 3 ORDER BY `lane_name` ASC";
$result = mysql_query($sql);
echo "<select name='lane_name'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['lane_name'] ."'>" . $row['lane_name'] ."</option>";
}
echo "</select>";
?>
</div>
<div id='d'>
<span><label>Lane</label></span>
<?php
$sql = "SELECT lane_name FROM lane WHERE `lane_school_id` = 4 ORDER BY `lane_name` ASC";
$result = mysql_query($sql);
echo "<select name='lane_name'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['lane_name'] ."'>" . $row['lane_name'] ."</option>";
}
echo "</select>";
?>
</div>
<div id='e'>
<span><label>Lane</label></span>
<?php
$sql = "SELECT lane_name FROM lane WHERE `lane_school_id` = 5 ORDER BY `lane_name` ASC";
$result = mysql_query($sql);
echo "<select name='lane_name'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['lane_name'] ."'>" . $row['lane_name'] ."</option>";
}
echo "</select>";
?>
</div>
<!---End of Lane part---->
/*
I am trying to get user selected option from a form but when i try to get the info by using $_post i'am getting the lest option. i am using hide and show to filter the relevant "lanes" base on "school" choice
I'm getting always 'BA in Nursing'
*/

If I have understood your question correctly, you are trying to send the select values to the php script. For this you will need an html form, if you're not interested in A
<form method='POST'><!-- PUT THE PHP that creates the select options here!--></form>
Also, it's best to use quotes to access the $_POST super global array associative indexes:
$school_name = $_POST['school_name'];
$lane_name = $_POST['lane_name'];
$year = $_POST['year'];
Also, you could try print_r($_POST) to see what values you are receiving from the post!
In addition, the mysql extension is deprecated. It would be best if you were to switch to mysqli or PDO instead!

Related

Multiple dropdowns from mysql

I want to make multiple dropdowns from data out of my mysql database. I want 4 dropdowns to be exact. This is what I have at this moment:
<?php
mysql_connect('#', '#', '#');
mysql_select_db('test');
$sql = "SELECT wie, waar, metwie, voeruig FROM data";
$result = mysql_query($sql);
echo "<select name='test'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['wie'] . "'>" . $row['wie'] . "</option>";
}
echo "</select>";
?>
For instance, this:
mysql_connect('#', '#', '#');
mysql_select_db('test');
$sql = "SELECT wie FROM data";
$result = mysql_query($sql);
echo "<select name='test1'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['wie'] . "'>" . $row['wie'] . "</option>";
}
echo "</select>";
$sql = "SELECT waar FROM data";
$result = mysql_query($sql);
echo "<select name='test1'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['waar'] . "'>" . $row['waar'] . "</option>";
}
echo "</select>";
$sql = "SELECT metwie FROM data";
$result = mysql_query($sql);
echo "<select name='test2'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['metwie'] . "'>" . $row['metwie'] . "</option>";
}
echo "</select>";
$sql = "SELECT voeruig FROM data";
$result = mysql_query($sql);
echo "<select name='test3'>";
while ($row = mysql_fetch_array($result)){
echo "<option value='" . $row['voeruig'] . "'>" . $row['voeruig'] . "</option>";
}
echo "</select>";
?>
As stated by HawasKaPujaari, avoid using mysql. Use mysqli. You could use a conditional switch statement like this:
<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "test");
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
/* Select queries return a resultset */
if ($result = $mysqli->query("SELECT wie, waar, metwie, voeruig FROM data")) {
printf("Select returned %d rows.\n", $result->num_rows);
while ($row = mysql_fetch_array($result)) {
switch ($row) {
case "wie":
echo "<select name='wie'>";
echo "<option value='" . $row['wie'] . "'>" . $row['wie'] . "</option>";
echo "</select>";
break;
case "waar":
echo "<select name='waar'>";
echo "<option value='" . $row['waar'] . "'>" . $row['waar'] . "</option>";
echo "</select>";
break;
case "metwie":
echo "<select name='metwie'>";
echo "<option value='" . $row['metwie'] . "'>" . $row['metwie'] . "</option>";
echo "</select>";
break;
}
case "voeruig":
echo "<select name='voeruig'>";
echo "<option value='" . $row['voeruig'] . "'>" . $row['voeruig'] . "</option>";
echo "</select>";
break;
}
/* free result set */
$result->close();
}
?>
php and mysql are different softwares. what you are doing here is connecting php with mysql using mysql_*() functions
in your case what you get is an array in php. you can use this array for whatever you want. if you want to print array as is as for debugging use:
echo "<pre>";
print_r($row);
echo "</pre>";
From this you will get array structure
then you ca use different dropdowns for array elements
Note: mysql_*() is not safe. Use mysqli_* or PDO.
Try code below. hope it help you
<?php
$wie=array();
$waar =array();
$metwie =array();
$voeruig =array();
while ($row = mysql_fetch_array($result)){
$wie[]=$row["wie"];
$waar[]=$row["waar"];
$metwie[]=$row["metwie"];
$voeruig[]=$row["voeruig"];
}
?>
<select name="wie">
<?php
foreach($wie as $k=>$v)
{
?>
<option value="<?php echo $v?>"><?php echo $v;?>
<php
}
?>
</option>
<select name="waar">
<?php
foreach($waar as $k=>$v)
{
?>
<option value="<?php echo $v?>"><?php echo $v;?> </option>
<php
}
?>
</select>
<select name="metwie">
<?php
foreach($metwie as $k=>$v)
{
?>
<option value="<?php echo $v?>"><?php echo $v;?></option>
<php
}
?>
</select>
<select name="voeruig">
<?php
foreach($voeruig as $k=>$v)
{
?>
<option value="<?php echo $v?>"><?php echo $v;?> </option>
<php
}
?>
</select>

Set select option using php when reloading the same page

I have a php select populated and given a matching value by using php to loop through the results of a sql query.
$result = mysqli_query($con, "select * from course")
echo "<form action='' method='post'>";
echo "<select name='CourseSelect'>";
echo "<option value='0'> - Select Course - </option>";
while($row = mysqli_fetch_array($result))
{
echo "<option value='" . $row['Title'] ." '>" . $row['Title'] . "</option>";
}
echo "</select>";
echo "<input name='SubmitCourse' type='submit'>";
echo "</form>";
That gives me a drop down list populated with all course titles,
upon submission I can access the selected value using $_POST['CourseSelect'];
However the drop down (select) resets itself to the default value when the page reloads.
How can I keep that option selected using php?
I know that I can use the selected keyword inside of an a select option to make that option the default selected option.
for example the second option would be selected when loading the page:
<select>
<option>One</option>
<option selected>Two</option>
<option>Three</option>
</select>
You can make it simple as
while($row = mysqli_fetch_array($result))
{
$select = '';
if( isset($_POST['CourseSelect']) && $_POST['CourseSelect'] == $row['Title'] ) $select = 'SELECTED';
echo "<option value='".$row['Title']."' ".$select.">" . $row['Title'] . "</option>";
}
you can use like below..
$result = mysqli_query($con, "select * from course");
$selected = "";
echo "<form action='' method='post'>";
echo "<select name='CourseSelect'>";
echo "<option value='0'> - Select Course - </option>";
while($row = mysqli_fetch_array($result))
{
$selected = $row['Title'] == $_REQUEST['CourseSelect'] ? "Selected" : "";
echo "<option value='" . $row['Title'] ." ' $selected>" . $row['Title'] . "</option>";
}
echo "</select>";
echo "<input name='SubmitCourse' type='submit'>";
echo "</form>";
try this:
echo "<option value='0'> - Select Course - </option>";
while($row = mysqli_fetch_array($result))
{
$selected = $_POST['CourseSelect'] == $row['Title'] ? 'selected' : '';
echo "<option value='{$row['Title']}' {$selected}>{$row['Title']}</option>";
}
you can do like this
echo "<form action='' method='post'>";
echo "<select name='CourseSelect'>";
if( isset($_POST['CourseSelect']) && $_POST['CourseSelect'] != "0")
{
echo "<option >".$_POST['CourseSelect']."</option>";
}
else
{
echo "<option value='0'> - Select Course - </option>";
}
while($row = mysqli_fetch_array($result))
{
if( isset($_POST['CourseSelect']) && $_POST['CourseSelect'] != $row['Title'])
echo "<option value='" . $row['Title'] ." '>" . $row['Title'] . "</option>";
}
echo "</select>";
echo "<input name='SubmitCourse' type='submit'>";
echo "</form>";
PHP Function
# select box
/*
Example:
Parameter 1:
$options[1] = 'Course 1';
$options[2] = 'Course 2';
$options[3] = 'Course 3';
Parameter 2:
$selectedOption = 2; The dropdown need to be selected
*/
function buildOptions($options, $selectedOption)
{
foreach ($options as $value => $text)
{
if ($value == $selectedOption)
{
$Return .='<option value="'.$value.'" selected="selected">'.stripslashes($text).'</option>';
}
else
{
$Return .='<option value="'.$value.'">'.stripslashes($text).'</option>';
}
}
return $Return;
}
Function Call
$result = mysqli_query($con, "select * from course");
while ($row = mysqli_fetch_array($result))
{
$UniqueId = $row['Title'];
$Value = $row['Title'];
$optionsArray[$UniqueId] = $Value; // Store the values into an array
}
$CourseSelect = isset($_POST['CourseSelect']) ? $_POST['CourseSelect'] : '0';
echo "<form action='' method='post'>";
echo "<select name='CourseSelect'>";
echo "<option value='0'> - Select Course - </option>";
echo buildOptions($optionsArray, $CourseSelect);
echo "</select>";
echo "<input name='SubmitCourse' type='submit'>";
echo "</form>";
Notes
You can use buildOptions() for all of your projects to display select box. I have been using this for years.

Form values from dropdown list not passing

I created a drop down list that has been populated by the database and now I'm having trouble retrieving the data. Normally, I would know how to retrieve the value of the drop down list if I had to manually name the data, but in this case, I'm not quite sure how I would name it.
Here is my current code:
<h1>Generate Reports</h1>
<form enctype="multipart/form-data" action="http://localhost/yiiFolder/index.php/create" method="post">
<table>
<tr>
<td><strong>Materials</strong></td>
<?php
mysql_connect('host', 'root', 'password');
mysql_select_db ("db");
$sql = "SELECT material_name FROM materials";
$result = mysql_query($sql);
echo "<td><select name='materials'>";
while ($row = mysql_fetch_array($result))
{
echo "<option value='" . $row['material_name'] . "'>" .
$row['material_name'] . "</option>";
}
echo "</select></td></tr> ";
$sql2 = "SELECT location_name From locations";
$result2 = mysql_query($sql2);
?>
<td><strong>Locations</strong></td>
<?php
echo "<td><select name='locations'>";
while ($row2 = mysql_fetch_array($result2))
{
echo "<option value='" . $row2['location_name'] . "'>" .
$row2['location_name'] . "</option>";
}
echo "</select></td></tr>";
?>
<tr>
<td><button name="submit" type=submit>Generate</button></td>
</tr>
</table>
</form>
<?php
$material = $row['material_name'];
$locations = $row2['location_name'];
$generate = $_POST['submit'];
if(isset($generate))
{
echo $material;
echo $locations;
}
?>
You're trying to capture value before the submit button is hit. Also, as Hanky pointed out you're using the wrong names while referring to select data. You should do this instead
if(isset($_POST['submit'])) // this code will run after the button is clicked
{
$material = $_POST['materials']; // and not material_name
$locations = $_POST['locations']; // and not location_name
echo $material;
echo $locations;
}
PS: You're following a very unsecure way of developing a web application. At the very least you need to switch to PDO and always escape the data.

How to use a select option value as a PHP variable in the next query?

I have the following code which provides a drop down list of all the rows in that specific table, this works fine. The code is below:
<?php
$con=mysqli_connect("localhost","user","pass","db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT ID, NAME FROM b_sonet_group ORDER BY ID DESC");
echo "<select>";
echo "<option value=''>Select Your Project</option>";
while($row = mysqli_fetch_array($result))
{
echo "<option value='" . $row['ID'] . "'>" . $row['NAME'] . "</option>";
}
echo "</select>";
mysqli_close($con);
?>
I now want a second drop down list that is determined by what ever is selected in the one above based on the ID. So, I want something like:
$result2 = mysqli_query($con,"SELECT ID, ALBUM_NAME FROM a_different_table WHERE ID=ID_FROM_QUERY_ABOVE");
echo "<select>";
echo "<option value=''>Select Your Album</option>";
while($row = mysqli_fetch_array($result))
{
echo "<option value='" . $row['ID'] . "'>" . $row['ALBUM_NAME'] . "</option>";
}
echo "</select>";
mysqli_close($con);
?>
I basically want to get the ID from the first drop down to provide the results in the second dropdown. Can this be done?
You can't "only" do this with Ajax, but you should do it with Ajax.
PHP way (not suggested, and untested). Basically use isset and if it is, more will be added to the form. The POST from the select, is the select name. So change the plain select tag which I did in the example below. This also requires them to submit it.
$result = mysqli_query($con,"SELECT ID, NAME FROM b_sonet_group ORDER BY ID DESC");
echo '<form id="project_form" method="post">';
echo "<select id='select_your_project' name = 'select_your_project'>";
echo "<option value=''>Select Your Project</option>";
while($row = mysqli_fetch_array($result))
{
echo "<option value='" . $row['ID'] . "'>" . $row['NAME'] . "</option>";
}
echo "</select>";
if(isset($_POST['select_your_project'])){
$result2 = mysqli_query($con,"SELECT ID, ALBUM_NAME FROM a_different_table WHERE ID='".$_POST['select_your_project']."'");
echo "<select id='select_your_album' name = 'select_your_album'>";
echo "<option value=''>Select Your Album</option>";
while($row = mysqli_fetch_array($result2))
{
echo "<option value='" . $row['ID'] . "'>" . $row['ALBUM_NAME'] . "</option>";
}
echo "</select>";
}
echo '<input type="submit" value="Submit">';
echo '</form>';
if(isset($_POST['select_your_album'])){
//do form submitted stuff here
}
Ajax way (two separate files, untested but gives you the idea)
//Main page (view) START
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
//this will trigger automatically when they change the first select box
$('#select_your_project').on('change', function(event){
if($(this).val() == 'select_your_project'){
$("#ajax_reply_div").empty()
}else{
var values = $(this).serialize();
$.ajax({
url: "php_data_file.php",
type: "post",
data: values,
success: function(data){
$("#ajax_reply_div").empty().append(data);
},
error:function(){
$("#ajax_reply_div").empty().append('something went wrong');
}
});
}
});
</script>
<form id="id_of_form">
<?php
echo "<select id='select_your_project' name='select_your_project'>";
echo "<option value='select_your_project'>Select Your Project</option>";
while($row = mysqli_fetch_array($result))
{
echo "<option value='" . $row['ID'] . "'>" . $row['NAME'] . "</option>";
}
echo "</select>";
?>
</select>
<div id="ajax_reply_div">
</div>
<input type="submit" value="Submit">
</form>
//Main page (view) END
//php_data_file.php START
if(isset($_POST['select_your_project'])){
$result2 = mysqli_query($con,"SELECT ID, ALBUM_NAME FROM a_different_table WHERE ID='".$_POST['select_your_project']."'");
//as a note it is better to only send an array back then build the HTML with jQuery, but this way is easier if you are new to jQuery/Ajax
echo "<select id='select_your_album' name = 'select_your_album'>";
echo "<option value=''>Select Your Album</option>";
while($row = mysqli_fetch_array($result2)){
echo "<option value='" . $row['ID'] . "'>" . $row['ALBUM_NAME'] . "</option>";
}
echo "</select>";
}
//php_data_file.php END

How to submit a form which contains mysql generated dropdown lists

Ive got a property website im working on for a school project http://www.holidayaviemore.com
I have three dropdown lists which have all been generated from a mysql database.
I have added a submit button so that when the options have been selected the database info will display..this is where I am really stuck!
Im not sure what code of functions are needed to diplay the content on the page or on another page..any ideas? This is the code for the dropdownn boxes and form I have so far..
EDIT
Basically I want the options selected to be displayed on a webpage http://www.holidayaviemore.com shows the dropdown lists
<p>Select Options From Below to find property</p>
<form action="#" name="form" id="form" method="post">
<?php
$sql = "SELECT DISTINCT availability FROM properties";
$result = mysql_query($sql);
echo "<select name='availability'><option value=''>- Availability?--</option>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='". $row['availability']. "'>" . $row['availability'] ."</option>";
}
echo "</select>";
?>
<?php
$sql = "SELECT bedrooms FROM properties LIMIT 10";
$result = mysql_query($sql);
echo "<select name='bedrooms'><option value=''>--Please Choose Bedrooms Bedrooms--</option>";
$count = 1;
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['bedrooms'] . "'> " . $count . "</option>";
$count++;
}
echo "</select>";
?>
<?php
$sql = "SELECT sleeps_min FROM properties LIMIT 12";
$result = mysql_query($sql);
echo "<select name='sleeps_min'><option value=''>--Please Choose Guests--</option>";
$count = 1;
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['sleeps_min'] . "'> " . $count . "</option>";
$count++;
}
echo "</select>";
?>
<input type="submit" name="" value="GO" />
</form>
I think you need this. You need to see if form already submitted and add selected in option tag where it needed.
<p>Select Options From Below to find property</p>
<form action="#" name="form" id="form" method="post">
<?php
$sql = "SELECT DISTINCT availability FROM properties";
$result = mysql_query($sql);
echo "<select name='availability'><option value=''>- Availability?--</option>";
while ($row = mysql_fetch_array($result)) {
if(isset($_POST['availability']) and $row['availability'] == $_POST['availability']){
echo "<option value='". $row['availability']. "' selected>" . $row['availability'] ."</option>";
}
else{
echo "<option value='". $row['availability']. "'>" . $row['availability'] ."</option>";
}
}
echo "</select>";
?>
<?php
$sql = "SELECT bedrooms FROM properties LIMIT 10";
$result = mysql_query($sql);
echo "<select name='bedrooms'><option value=''>--Please Choose Bedrooms Bedrooms--</option>";
$count = 1;
while ($row = mysql_fetch_array($result)) {
if(isset($_POST['bedrooms']) and $row['bedrooms'] == $_POST['bedrooms']){
echo "<option value='" . $row['bedrooms'] . "' selected> " . $count . "</option>";
}
else{
echo "<option value='" . $row['bedrooms'] . "'> " . $count . "</option>";
}
$count++;
}
echo "</select>";
?>
<?php
$sql = "SELECT sleeps_min FROM properties LIMIT 12";
$result = mysql_query($sql);
echo "<select name='sleeps_min'><option value=''>--Please Choose Guests--</option>";
$count = 1;
while ($row = mysql_fetch_array($result)) {
if(isset($_POST['sleeps_min']) and $row['sleeps_min'] == $_POST['sleeps_min']){
echo "<option value='" . $row['sleeps_min'] . "' selected> " . $count . "</option>";
}
else{
echo "<option value='" . $row['sleeps_min'] . "'> " . $count . "</option>";
}
$count++;
}
echo "</select>";
?>
<input type="submit" name="" value="GO" />
</form>

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