Following is the table i created for displaying the Restaurant Name, Location and Menu for table owners. Now each of the row for the column Menu have Button as values. My table is ready with perfect values.
Upon clicking the button corresponding to the each Restaurant, a new File(openmenu.php) will open and will echo the Restaurant Name, Mobile Number of that Restaurant and the menu. But so far, on clicking every Button ,I can only display above entries of the Last row of the table. Help Me Out. I am new to php.
<?php
include 'nav.php';
$sql = 'SELECT * FROM owners';
$query = mysqli_query($con, $sql);
if (!$query) {
die ('SQL Error: ' . mysqli_error($con));
}
?>
<html>
<head>
<link rel = "stylesheet" type = "text/css" href = "css/style.css">
<style>
.data-table{
width: 1024px;
margin-left: 150px;
text-align:center;
border: 1px solid firebrick;
background-color: white;
}
td,th{
border: 1px solid firebrick; padding: 3px 2px 1px 1px;
}
</style>
</head>
<body>
<div class="container">
<article>
<table class="data-table">
<thead>
<tr>
<th>Restuarant Name</th>
<th>Location</th>
<th>Menu</th>
</tr>
<tr>
</tr>
</thead>
<tbody>
<?php
while ($row = mysqli_fetch_array($query)){
$_SESSION['resphone'] = $row['resphone'];
$_SESSION['restaur'] = $row['restaur'];
echo '<tr>
<td>'.$row['restaur'].'</td>
<td>'.$row['loc'].'</td>
<td style="background-color:firebrick;"><form method="post" action="openmenu.php?id=$row[restaur]"><input value="<?php echo $restaur;?>" type="hidden">
<input type="submit" value="View"></form></td>
</tr>';
}
?>
</tbody>
</table>
</form>
</article>
</div>
</body>
</html>
<?php
include('nav.php');
?>
<html>
<head>
<link rel="stylesheet" href="css/style.css">
<style>
table, td {
border: none;
text-align: center;
text-align-last: center;
}
</style>
</head>
<body>
<div class="container">
<article>
<form method="get" align="center" action="" class="formwrap" enctype='multipart/form-data'>
<h1><?php $restaur = $_SESSION['restaur'];
echo $restaur ;?></h1>
<h1>Call to Order:</h1>
<?php $resphone = $_SESSION['resphone'];
echo $resphone;
?>
<br>
<br>
<?php
$sql = "select img from owners where restaur ='$restaur'";
$result = mysqli_query($con,$sql);
$row = mysqli_fetch_array($result);
$image_src2 = "upload/".$row['img'];
?>
<img src='<?php echo $image_src2; ?>' >
</form>
</article>
</div>
</body>
</html>
There are several ways you could solve this problem.
You could make a <form> with your view button ( if this view button not already has a form ) and give this a method like: POST or GET.
This would look like: <form method="POST"> or <form method="GET">
I think your view button could look something like this
<form method="post" action="OtherPHPpage.php">
<input type="hidden" name="view" value="<?php echo $row['resphone']; ?>">
<input type="submit" value="View">
</form>
This will redirect you to your other page and you will be able to post the value of the hidden input type
You can also use php sessions even though this is not the best solution in your case
session_start();
Put this on the top of your php page
// set the session
$_SESSION['variablename'] = $row['resphone'];
// retrieve the session variable
$whatever = $_SESSION['variablename'];