Dynamically display for table with headings accordingly - php

Question

Firstly, the headings are stored in h1 tags. This is taken from a separate table named "menu_type". Which is linked through a "menu" table.
I am trying to display data on the base like this:
HEADING
Table Data
2nd Heading
Second Data
--- In a loop until it is all completed ---
Here is a like page of what it is doing
I believe I have the methods correct and can see what it is doing, it is printing the first heading, then a blank table, then the second heading and then the data from the first table.
See my code:
<?php
$query = "SELECT * FROM menu_type";
$result = mysqli_query($connect, $query);
$result_array = array();
$numRows = mysqli_num_rows($result); // returns the num of rows from the query above, allowing for dynamic returns
while($row = mysqli_fetch_assoc($result)){
$menuType = $row['type'];
$result_array[] = $menuType; // This array holds each of the menu_types from DB
}
$countArray = count($result_array);
for($i = 0; $i < $numRows; $i++){
$query = "SELECT * FROM menu WHERE type_id='$result_array[$i]'";
$result = mysqli_query($connect, $query) or die(mysqli_error($connect));
echo "
<div id='hide'>
<h1 id='wines' class='head-font text-center head'>$result_array[$i]</h1>
<table class='table table-hover table-responsive'>
<thead>
<tr>
<th>
Item
</th>
<th class='text-right'>
Price
</th>
</tr>
</thead>
<tbody>
<tr>
";
$menuQuery = "SELECT * FROM menu, menu_type WHERE type_id='$i' AND menu.type_id = menu_type.id";
$menuResult = mysqli_query($connect, $menuQuery) or die(mysqli_error($connect));
while ($row = mysqli_fetch_assoc($menuResult)) {
$name = $row['name'];
$description = $row['description'];
$price = $row['price'];
echo "
<td>$name - <small>$description</small></td>
<td class='text-right'>£$price </td>
";
}
echo
"
</tr>
</tbody>
</table>
";
}
// print_r($result_array[2]);
?>

Answer 1

You don't need these anymore, it's like you're repeating the query. It looks incorrect also.
$menuQuery = "SELECT * FROM menu, menu_type WHERE type_id='$i' AND menu.type_id = menu_type.id";
$menuResult = mysqli_query($connect, $menuQuery) or die(mysqli_error($connect));
The menu items are already in this query, you just have to loop through it;
$query = "SELECT * FROM menu WHERE type_id='$result_array[$i]'";
$result = mysqli_query($connect, $query) or die(mysqli_error($connect));
UPDATE 1: this code is incorrect, it doesn't insert the value to the array. I updated the code (after "try this").
$result_array[] = $menuType;
UPDATE 2: the $result is repeatedly used in mysqli functions, the index is being moved. What I did is copied the initial $result to $resultCopy. Try code again, haha
Use array_push($array,$value_you_insert) function, for inserting elements to an array.
Try this;
<?php
$query = "SELECT * FROM menu_type";
$result = mysqli_query($connect, $query);
$resultCopy = $result;
$result_array = array();
$numRows = mysqli_num_rows($result); // returns the num of rows from the query above, allowing for dynamic returns
while($row = mysqli_fetch_assoc($resultCopy)){
$menuType = $row['type'];
array_push($result_array,$menuType);
}
for($i = 0; $i < $numRows; $i++){
echo "
<div id='hide'>
<h1 id='wines' class='head-font text-center head'>".$result_array[$i]."</h1>
<table class='table table-hover table-responsive'>
<thead>
<tr>
<th>Item</th>
<th class='text-right'>Price</th>
</tr>
</thead>
<tbody>
";
$query = "SELECT * FROM menu WHERE type_id='$result_array[$i]'";
$result = mysqli_query($connect, $query) or die(mysqli_error($connect));
while ($row = mysqli_fetch_assoc($result)) {
$name = $row['name'];
$description = $row['description'];
$price = $row['price'];
echo"
<tr>
<td>".$name." - <small>".$description."</small></td>
<td class='text-right'>£".$price."</td>
</tr>
";
}
echo
"
</tbody>
</table>
";
}
?>