get data from database with php and mysql (create report) - php

Question

I have a create a report form which has 2 dropdown lists and table in above. I want to filter the data and make report with dropdown list options. The first dropdown list has all table names from database and second one has "item" column, and every table has same items values. Like when we make report from table 3 and item option 1. We should make report.
Here is my code for HTML:
<div class="row">
<div class="col-lg-12">
<div class="ibox float-e-margins">
<div class="ibox-title">
<h5>Report by Items</h5>
</div>
<div class="ibox-content">
<form method="post" class="form-horizontal">
<div class="form-group"><label class="col-sm-4 control-label"Stocks</label>
<div class="col-sm-4"><select class="form-control m-b" name="Stock">
<option></option>
<option>Stock</option>
<option>Stock1</option>
<option>Stock2</option>
</select>
</div>
</div>
<div class="form-group"><label class="col-sm-4 control-label">Items</label>
<div class="col-sm-4">
<select class="form-control m-b" name="Items">
<option></option>
<option>Item1</option>
<option>Item2</option>
<option>Item3</option>
<option>Item4</option>
</select>
</div>
</div>
</div>
<div class="col-lg-25">
<div class="ibox float-e-margins">
</div>
</div>
<div class="form-group">
<div class="col-sm-4 col-sm-offset-5">
<button class="btn btn-white" type="submit">Cancel</button>
<button class="btn btn-primary" type="submit" name=submit>Run Report</button>
</div>
</div>
</form>
</div>
</div>
</div>
</div>
</div>
PHP and table code I try is:
<div class="ibox-content">
<table class="table table-striped table-bordered table-hover dataTables-example" >
<thead>
<tr>
<th>No</th>
<th>Quantity</th>
<th>Date</th>
<th>Sold</th>
<th>Total</th>
</tr>
</thead>
<tbody>
<?php
$mysqli = new mysqli( 'localhost', 'user', 'pass', 'mis_db' );
if (mysqli_connect_error()) {
echo mysqli_connect_error();
exit();
}
if($_POST['Stock']=='Stock1')
{
$stck1 = 'Stock1';
}
if($_POST['Stock']=='Stock2')
{
$stck1 = 'Stock2';
}
if($_POST['Stock']=='Stock1')
{
$stck1 = 'Stock3';
}
if($_POST['Items']=='Item1')
{
$itm = 'Item1';
}
if($_POST['Items']=='Item2')
{
$itm = 'Item2';
}
if($_POST['Items']=='Item3')
{
$itm = 'Item3';
}
if (isset($_POST['submit'])) {
$query = 'SELECT * FROM .$stck1 where Items=.$itm';
$data = mysqli_query($mysqli, $query) ;
if (!$data) {
echo("Error description: " . mysqli_error($mysqli));
} else {
while ($row = mysqli_fetch_array($data)) {
echo "<tr>
<td>" . $row['No'] . "</td>
<td>" . $row['Qty'] . "</td>
<td>" . $row['date'] . "</td>
<td>" . $row['Sold'] . "</td>
<td>" . $row['Total'] . "</td>
</tr>";
}
}
}
?>
</tbody>
<tfoot>
</tfoot>
</table>
</div>
I have doubt on MySQL query.

Answer 1

You have some syntax errors and your PHP can be simplified quite a bit.
As you can see I removed the if statements for each item in the dropdowns, not needed. You also had syntax errors in your $query, no . are required for concatenation so I've removed them. Finally, the $stck1 variable in the $query changed to simply $stck.
Try this:
<?php
$mysqli = new mysqli('localhost', 'user', 'pass', 'mis_db');
if (mysqli_connect_error()) {
echo mysqli_connect_error();
exit();
}
$stck = $_POST['Stock'];
$itm = $_POST['Items'];
if (isset($_POST['submit'])) {
$query = "SELECT * FROM $stck WHERE Items = $itm";
$data = mysqli_query($mysqli, $query) ;
if (!$data) {
echo("Error description: " . mysqli_error($mysqli));
} else {
while ($row = mysqli_fetch_array($data)) {
echo "<tr>
<td>" . $row['No'] . "</td>
<td>" . $row['Qty'] . "</td>
<td>" . $row['date'] . "</td>
<td>" . $row['Sold'] . "</td>
<td>" . $row['Total'] . "</td>
</tr>";
}
}
}
?>