Develop Reference

Database generated dropdown giving more variables a value from the same database when selected

I hope the title wasn't too cryptic, and I am sure there will be other threads like this one but cannot seem to put my finger on the correct terminology.
I have a form that has 3 dropdowns. Let's call them 'sales consultant', sales email & sales team.
What I am looking to do is give a value to 'sales email' & 'sales team' when the sales consultant is chosen from the dropdown 'sales consultant'
The catch, I am using mysql to pull the data for each dropdown. See below:
<div id="sc">
Sales Consultant
<input class="" type="text" list="salesconsultant" name="salesconsultant" placeholder="Start typing consultants name" required />
<datalist id="salesconsultant">
<?php
$sql = "SELECT consultant, id FROM salesconsultants";
$result = $conn->query($sql);
if (!$result) die($conn->error);
if($result->num_rows > 0 ) {
while($row=$result->fetch_assoc()) {
echo "<option value=\"".$row["id"]."\">".$row["consultant </option>";
}
echo "</datalist>";
} else {
echo"record 0";
}
?>
</div>
The other rows I have in the database are 'salesemail', 'id', 'consultant' and 'teamid'.
So when sales consultant 1 is selected it will give the variable $salesemail the value of consultant 1's 'salesemail' and so on.
Is this possible?

Just change below line to :
echo '<option value="'.$row["id"].'".','."'.$row["salesemail"].'">'.$row["consultant"].'</option>';
In above code you are separating both value by , And you can get these values like this :
$result = $_POST['yourselectbox'];
$r1 = explode(',', $result);
echo "id: ". $r1[0]."<br />";
echo "email: ". $r1[1]."<br />";
Hope this will work!

display selected value from database

I have a form where user selects the category while adding the product.
When user want to edit the product, i am displaying all the previously populated values but could not able to figure out how to display the category he selected.
addproduct.php (displaying the categories from the database)- this code is working fine and can see all the categories in dropdown
<?php
require'dbconn.php';
$subject = mysql_query("select * from categories", $link);;
while($subjectData = mysql_fetch_array($subject)){
echo $subjectData['value'];?>
<option value="<?php echo $subjectData['name'];?>"><?php echo
$subjectData['name'];?>
</option>
In the edit product i want to display all the categories like above, but want to display the selected category in the form which i could not able to do.
editproduct.php (rough draft code) -- not working
<?php
require'dbconn.php';
$subject = mysql_query("select * from categories", $link);;
while($subjectData = mysql_fetch_array($subject)){
echo $subjectData['value'];?>
<option select="<?php echo $cat;?>"value="<?php echo $subjectData['name'];?>"><?php echo
$subjectData['name'];?>
</option>
$cat - category value(previously selected) pulled from database

require'dbconn.php';
$subject = mysql_query("select * from categories", $link);;
<option value="<?php echo $cat;?>"><?php echo $cat;?></option>
while($subjectData = mysql_fetch_array($subject)){
echo $subjectData['value'];?>
<?php if($cat!=$subjectData['name']){?> <option value="<?
php echo $subjectData['name'];?>"><?php echo
$subjectData['name'];?>
</option>
<?php } ?>
Try using this code and please use mysqli as mysql is deprecated. previously selected category should be before while loop. Hope it helps

Two issues with your code:
You are using mysql functions, which are depreciated and don't even exist in the current version of PHP. Use mysqli or PDO functions.
The html you are generating is invalid syntax.
I'll leave the first issue to you to correct.
For the 2nd issue, all of the non-selected options in your dropdown will not have the selected attribute.
Only the selected item will have that attribute. The code below assumes that the variable $cat has the previously selected value, and each row has a
column named 'cat'. When $cat matches the value in the column 'cat', it will add selected='selected' to the option.
<?php
require 'dbconn.php';
$subject = mysql_query("select * from categories", $link);;
while($subjectData = mysql_fetch_array($subject)){
echo $subjectData['value'];
$selected = "";
if($cat == $subjectData['cat']) {
$selected = "selected='selected' ";
}
echo "<option ".$selected."value=".$subjectData['name'].">";
echo $subjectData['name'];
echo "</option>\n";
}
?>

showing my database tables using selection in php

i'm trying to display all my tables in database with this code.
<label>
<select class="selectpicker" name="tabel">
<?php
include "koneksi.php";
$koneksi ->select_db('testskripsi');
$sql = mysqli_query($koneksi, "show tables") ;
while($table = $sql->fetch_array(MYSQLI_ASSOC)) {
echo '<option name="opsi" value = "' . $table['Tables'] . '">' . $table['Tables'] . '</option>';
}
?>
</select>
</label>
here's my result
i think i don't know how to fill this array to make my tables displayed
' . $table['Tables'] . '

Please try Tables_in_testskripsi instead of Tables.here is the correct code of yours.
<label>
<select class="selectpicker" name="tabel">
<?php
include "koneksi.php";
$koneksi ->select_db('testskripsi');
$sql = mysqli_query($koneksi, "show tables") ;
while($table = $sql->fetch_array(MYSQLI_ASSOC)) {
echo '<option name="opsi" value = "' . $table['Tables_in_testskripsi'] . '">' . $table['Tables_in_testskripsi'] . '</option>';
}
?>
</select>
</label>
do let me know if it was helpful for you

To get table names,
$tables=$this->db->query("
SELECT t.TABLE_NAME AS myTables
FROM INFORMATION_SCHEMA.TABLES AS t
WHERE t.TABLE_SCHEMA = 'database name'
AND t.TABLE_NAME LIKE '%a%'
")->result_array();
foreach($tables as $key => $val) {
echo $val['myTables']."<br>";// myTables is the alias used in query.
}

Onchange form filter with dropdown

I am trying to create a filter for a gallery that I've created. The gallery has 5 filters using dropdown menu's. When a item is selected from one of the 5 filters it has to filter the images. When a second filter is selected it has to filter the results of the first filter and so on.
I am using the onchange='this.form.submit()' script but I don't know how to assign a certain action to it when an item is selected. This is my code at the moment of writing:
<td>
Kleur:
<form method="POST">
<select name="kleur" onchange='this.form.submit()'>
<option> -- Geen optie -- </option>
<?php while ($line1 = mysqli_fetch_array($result1, MYSQLI_ASSOC)) { ?>
<option value="<?php echo $line1['kleur']; ?>"> <?php echo $line1['kleur']; ?>
</option>
<?php } ?>
</select>
</form>
<?php
if (isset($_POST['submit'])) {
$kleur = $_POST['kleur'];
$SQL = "SELECT * FROM `rozen` WHERE `kleur` LIKE '$kleur'";
$result = mysqli_query($connection, $sql);
echo $result;
}
?>
</br>
</td>
The following part doesn't seem to work:
<?php
if (isset($_POST['submit'])) {
$kleur = $_POST['kleur'];
$SQL = "SELECT * FROM `rozen` WHERE `kleur` LIKE '$kleur'";
$result = mysqli_query($connection, $sql);
echo $result;
}
?>
Does anyone know how to use this script? and perhaps explain how to save the selected item in the dropdown menu too?

You can add an attribute with all the information you need to filter to your images
<img filterInfo="Kleur|Geur|Bloemvorm|Gezondheid|Type|Zoeken">
then set a class for all your filters to catch the changes
$(".filters").on("change",function(){
var kleur = $('[name=Kleur]').val();
var Geur = $('[name=Geur]').val();
...
...
...
$.each($('#gallery img'),function(i,v){
var attrs = $(v).attr("filterInfo").slice("|");
if((kleur == "" || kleur == attrs[0]) && (Geur == "" || == attrs[1]) .... other filters)
$(this).show(); //or fadeIn();
else
$(this).hide(); //or fadeOut();
});
});

Only getting last record from mysql database

I am stuck on a code where I want to fetch data from MySQL into an array.. I have a form containing color and size select boxes, an onclick javascript function is triggered and it created two more select boxes like above, I have managed to get data into the javascript code where the code for creating new select boxes is written.
But I am only getting the last inserted record from both the tables. Although I have used a while loop.
Can someone help me out and I am also getting name of all select boxes likes name="color[]" , I want to insert records into a bridge table containing ids of color and size. below is my code please help ..
I will clear it up , each time I click add more button it should create 2 new dropdown lists, one for color and 2nd for size, both dropdowns should have the distinct data from database. so the ids for each record would be same in every dropdown list, I want to add more than 1 records in bridge table which contains product_id,color_id and size_id, so if I go for 3 dropwdown boxes , and i select blue color and small size in the first, then for second dropdown i again select blue color and size medium, as for the last dropdown which was also generated by the javascript function . i choose black color and large size. so from the dropdown it will get ids of size,color and it would be inserted accordingly.. so when i display the product and color blue is selected i would only see the sizes which were added to the color blue at the time of adding the product.. i hope this clears everything :)
$result=mysql_query("SELECT * FROM color,size");
while($row=mysql_fetch_array($result)) {
?>
<script>
var room = 1;
function add_fields() {
room++;
var objTo = document.getElementById('room_fileds')
var divtest = document.createElement("div");
divtest.innerHTML = '<div class="label">Room ' + room + ':</div><div class="content"><span>Color: <select name="color[]"><option value="<?php echo $row['color_id']; ?>"><?php echo $row['color']; ?></option></select></span><span>Size: <select><option value="<?php echo $row['size_id']; ?>"><?php echo $row['size']; ?></option></select></span></div>';
objTo.appendChild(divtest)
}
</script>
<?php
}
HTML code
<div id="room_fileds">
<div>
<div class='label'></div>
<div class="content">
<input type="button" class="btn btn-success" id="more_fields" onclick="add_fields();" value="Add More" /> <br /><br />
<select name="color[]" class="form-control">
<option value="0">Select Color</option>
<?php
$result=mysql_query("SELECT * FROM color");
while($row=mysql_fetch_array($result)){
?>
<option value="<?php echo $row['color_id'] ?>"><?php echo $row['color']; ?></option>
<?php } ?>
</select>
<select name="size[]" class="form-control">
<option value="0">Select Size</option>
<?php
$result=mysql_query("SELECT * FROM size");
while($row=mysql_fetch_array($result)){
?>
<option value="<?php echo $row['size_id'] ?>"><?php echo $row['size']; ?></option>
<?php } ?>
</select>
</div>
</div>
</div>

Replace your first code with the following:
<script>
var colors = [];
var sizes = [];
var room = 1;
<?php
$result = mysql_query("SELECT * FROM color");
while ($row = mysql_fetch_array($result)) { ?>
colors.push(['<?php echo $row['color_id'] ?>', '<?php echo $row['color'] ?>']);
<?php }
$result = mysql_query("SELECT * FROM size");
while ($row = mysql_fetch_array($result)) { ?>
sizes.push(['<?php echo $row['size_id'] ?>', '<?php echo $row['size'] ?>']);
<?php } ?>
function add_fields() {
room++;
var objTo = document.getElementById('room_fileds');
var divtest = document.createElement("div");
var html = '<div class="label">Room ' + room + ':</div><div class="content"><span>Color: <select name="color[]" class="form-control">';
for (i = 0; i < colors.length; i++) {
html += '<option value="' + colors[i][0] + '">' + colors[i][1] + '</option>';
}
html += '</select></span><span>Size: <select name="size[]" class="form-control">';
for (i = 0; i < sizes.length; i++) {
html += '<option value="' + sizes[i][0] + '">' + sizes[i][1] + '</option>';
}
html += '</select></span></div>';
divtest.innerHTML = html;
objTo.appendChild(divtest);
room++;
}
</script>

$divs='';
$result=mysql_query("SELECT * FROM color,size");
$room=1;
while($row=mysql_fetch_array($result)) {
$divs.='<div><div class="label">Room ' . $room . ':</div><div class="content"><span>Color: <select name="color[]"><option value="'.$row['color_id'].'">'.$row['color'].'</option></select></span><span>Size: <select><option value="'.$row['size_id'].'">'.$row['size'].'</option></select></span></div></div>';
$room++;
} ?>
<script>
function add_fields() {
var objTo = document.getElementById('room_fileds');
objTo.appendChild(<?php echo $divs; ?>);
}
</script>

Try this code hope it works for you
<script>
var room = 1;
function add_fields() {
room++;
var objTo = document.getElementById('room_fileds');
var divtest = document.createElement("div");
<?php
$div_start='<div class="label">Room '."'+ room +'".' :</div><div class="content">';
$color='<span>Color: <select name="color[]">';
$size='<span>Size: <select>'
$result=mysql_query("SELECT * FROM color,size");
while($row=mysql_fetch_array($result)) {
$color.='<option value="'.$row['color_id'].'">'.$row['color'].'</option>';
$size.='<option value="'.$row['size_id'].'">'.$row['size']'.</option>';
}
$color.='</select> </span>';
$size.='</select> </span>';
$div_close='</div>';
$innerHTML=$div_start.$color.$size.$div_close;
?>
divtest.innerHTML ='<?php echo $innerHTML; ?>';
objTo.appendChild(divtest);
}
</script>

This is the final code , it would help if anyone wants to code a similar thing.. just change values and it would work as per your need... and i would like to thank all people who replied to the thread and specially Mohammad Anini .. it wouldnt have been possible without his help !!!
$query = mysql_query("INSERT INTO products (product_name,product_description, product_pic1, product_pic2, product_pic3,product_price,category_id,subcategory_id,product_status,color,size,product_slug, meta_keywords,entrydate)
VALUES ('$product', '$description' , '$image', '$image2', '$image3', '$product_price', '$category', '$subcategory', '$product_status', '$capture_field_vals', '$size', '$product_slug1', '$meta_keywords', Now() )") or die(mysql_error());
if ($query === TRUE) {
$lastid = mysql_insert_id();
$color["color"]=array();
$size["size"]=array();
foreach ($_POST['color'] as $key => $colorvalue) {
}
foreach ($_POST['size'] as $key => $sizevalue) {
}
$cid=array();
foreach($color as $rec){
$cid[]=$rec;
}
$sz=array();
foreach($size as $rec){
$sz[]=$rec;
}
$length=sizeof($color);
for($i=0;$i<$length-1;$i++){
$query2 = mysql_query("INSERT INTO bridge (product_id,color_id, size_id)
VALUES ('$lastid', '$cid[$i]' , '$sz[$i]')") or die(mysql_error());
// echo "<script>alert('New Product successfully Addedd');windows.location.replace('addedit_product.php');</script>";
}