I have a Simple CMS i made but i cant anderstand how to connect the content to the category
i need another mysql table for cat?
how can i add this in to the form?
which functions will show the cat in the Add content page?
matanc244 0 Newbie Poster
diafol
Show your exisitng code, your description is unclear.
matanc244 0 Newbie Poster
for example -
the form
<form action="add.php" method="post">
title: <input type="text" name="title">
discription: <input type="text" name="discription">
content: <input type="content" name="content">
<input type="submit">
</form>
i want to add to the form a global categories
the add.php
<?php
require 'config.php';
// Get values from form
$title=$_POST['title'];
$description=$_POST['description'];
$content=$_POST['content'];
// Insert data into mysql
$sql="INSERT INTO $content (titel, discription, content )VALUES('$title', '$discription', '$content')";
$result=mysql_query($sql);
// if successfully insert data into database, displays message "Successful".
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='addp.php'>Back to main page</a>";
}
else {
echo "ERROR";
}
mysql_close($con);
?>
diafol
Ok, I think I know what you want. I'd do this...
1) Create a new categories table in MySQL: cat_id [PK, int], category [varchar]
2) Run a SELECT query on a category table and extract both fields. In the while loop, you can build the select dropdown:
$select = '';
while($d = mysql_fetch_assoc($result)){
$select .= "<option value='{$d['cat_id']}'>{$d['category']}</option>";
}
3) Insert the variable in the required place, e.g.
<select name="cat">
<?php echo $select;?>
</select>
4) You pick up the category id:
$cat_id=$_POST['cat'];
5) By now you should have realised that you need a cat_id field in your content table.
6) INSERT SQL after you clean all the post variables with mysql_real_escape_string()
INSERT INTO content (title, description, cat_id )VALUES('$title', '$description', $cat_id)
** You've spelled some fields incorrectly
OK, that's my 2p. Note that you just have the ONE content table, with each record linked (related) to a category
matanc244 0 Newbie Poster
there is an error
Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in /home/content/60/10533160/html/cms/in.php on line 3
diafol
means you didn't run a query before the mysql_fetch_assoc
matanc244 0 Newbie Poster
still not working
<?php
require 'config.php';
$select = 'mysqli_query("SELECT * FROM categories")';
while($d = mysql_fetch_assoc($result)){
$select .= "<option value='{$d['cat_id']}'>{$d['category']}</option>";
}
?>
<html>
<body>
<form action="main.php" method="post">
Firstname: <input type="text" name="title">
Lastname: <input type="text" name="discription">
Age: <input type="content" name="content">
<select name="cat_id">
<?php echo $select;?>
</select>
<input type="submit">
</form>
</body>
</html>
NardCake 30 Posting Pro in Training
Line 7 $select = 'mysqli_query("SELECT * FROM categories")'; Why do you have the mysqli_query function itself in quotes?
matanc244 0 Newbie Poster
So what is the correct way to write the query
NardCake 30 Posting Pro in Training
You have quotes surrounding the entire function, so it's a string, simply remove the quotes so it reads: $select = mysqli_query("SELECT * FROM categories");
That should fix your problem.
matanc244 0 Newbie Poster
not working
<?php
require 'config.php';
$select = mysqli_query("SELECT * FROM categories");
while($d = mysql_fetch_assoc($result)){
$select .= "<option value='{$d['cat_id']}'>{$d['category']}</option>";
}
?>
ERROR
Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/content/60/10533160/html/cms/in.php on line 7
Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in /home/content/60/10533160/html/cms/in.php on line 8
Tinnin 8 Posting Whiz in Training
<?php
$query = "SELECT * FROM categories";
$result = mysqli_query($query, $dbc);
while ($row = mysqli_fetch_assoc($result)) {
// Do stuff
}
?>
diafol
As tintin shows, you need to include the connection ($dbc). I'm assuming this is set in your config.php file. Of course it may be called something other than $dbc. Just change the name of the variable in that case.
matanc244 0 Newbie Poster
still give the same error
Warning: mysqli_query() expects parameter 1 to be mysqli, string given in /home/content/60/10533160/html/cms/in.php on line 6
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null given in /home/content/60/10533160/html/cms/in.php on line 7
line 6
$result = mysqli_query($query, $con);
line 7
while ($row = mysqli_fetch_assoc($result)) {
diafol
I'm really sorry mat - my mistake, the parameter order in Tinnin's post is wrong - yours should be:
$result = mysqli_query($con, $query);
Anyway, you can check this out yourself at http://php.net/manual/en/mysqli.query.php and look at the 'Procedural Style' not the 'OOP Style'. However, I would suggest maybe moving towards OOP if your site needs it.
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