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Old 08-02-2021, 10:10 AM  
Publisher Bucks
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Join Date: Oct 2018
Location: New Orleans, Louisiana. / Newcastle, England.
Posts: 1,129
Having PHP Issues with Displaying SQL Data on a Dynamic Web Page.

Any programmers able to see what I'm fucking up?

This is for a personal project I'm working on.

I should be able to open index.php to display a listing of 'recipes' in the database which will allow me to click through to viewmore.php where i have more information displayed on the page.

PHP Code:
<?php


// change the user_name and password
$db mysqli_connect("localhost""dbuser""dbpass");
 
// change the database_name
mysqli_select_db("dbname",$db);

$id $_GET['id'];

$result mysqli_query($con,"SELECT RecipeID FROM Recipes WHERE id = $id");

echo 
"<table width=100%>
<tr>
<th>ID</th>
<th>Title</th>
<th>Ingredients</th>
<th>Method</th>
<th>Keywords</th>
</tr>"
;

while(
$row mysqli_fetch_array($result))
{

$id $row['RecipeID'];

echo 
"<tr>";
echo 
"<td>" $row['RecipeID'] .  "</td>";
echo 
"<td> <a href='viewmore.php?id=$id'>" $row['Title'] . "</a> </td>";
echo 
"<td>" $row['Ingredients'] .  "</td>";
echo 
"<td>" $row['Method'] . "</td>";
echo 
"<td>" $row['Keywords'] . "</td>";
echo 
"</tr>";
}
echo 
"</table>";

?>
Which should, allow me to generate a link to viewmore.php?id=46 to show the data in the sql for that row, correct?

This is my viewmore.php page code...

PHP Code:
<?php

$id 
$_get['id'];

$result mysqli_query($con,"SELECT * FROM Recipes WHERE RecipeID='$id'");

echo 
"<table width=100%>
<tr>
<th>Title</th>
<th>Ingredients</th>
<th>Method</th>
<th>Keywords</th>
</tr>"
;

while(
$row mysqli_fetch_array($result))
{
echo 
"<tr>";
echo 
"<td> <a href='#'>" $row['Title'] . "</a> </td>";
echo 
"<td>" $row['Ingredients'] .  "</td>";
echo 
"<td>" $row['Method'] . "</td>";
echo 
"<td>" $row['Keywords'] . "</td>";
echo 
"</tr>";
}
echo 
"</table>";

?>
I'm sure its a simple issue that I'm missing but I've been up almost 48 hours and my mind just isn't seeing it :/

Any help would be greatly appreciated.
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