adding news

JD_

im modifying some code i found from a tutorial site which is a simple news system but it didnt have the adding news function but it said it would be similiar to the add comments fucntion anyways ive tried doing it like this

echo "<HR width=\"300\">";
    echo "<FORM action=\"{$_SERVER['PHP_SELF']}" .
         "?action=addnews&id=$id\" method=POST>\n";
    echo "Name: <input type=\"text\" " .
         "width=\"30\" name=\"name\"><BR>\n";
	echo "<TEXTAREA cols=\"40\" rows=\"5\" " .
         "name=\"title\"></TEXTAREA><BR>\n";
    echo "<TEXTAREA cols=\"40\" rows=\"5\" " .
         "name=\"newstext\"></TEXTAREA><BR>\n";
    echo "<input type=\"submit\" name=\"submit\" " .
         "value=\"Add Comment\"\n";
    echo "</FORM>\n";

Thats the form for posting a news item. the only trouble is i dont know about this $id as its automatically incremated

the fucntion is like this

function addnews($id) {
    global $db;

/* insert the news*/
$query = "INSERT INTO news" .
         "VALUES('{$_POST['name']}'," . "'{$_POST['title']}'," .
         "'{$_POST['newstext']}')";
mysql_query($query);

echo "News entered. Thanks!<BR>\n";
echo "<a href=\"{$_SERVER['PHP_SELF']}" .
     "?action=show&id=$id\">Back</a>\n";
	 echo "$name"."$title"."$newstext";
}

the news mysql table is like this as you can see im missing the postdate from up above but how do i add this or will it add it automatically?

id int(10) UNSIGNED No auto_increment
postdate timestamp(14) Yes NULL
title varchar(50) No
newstext text No

any help much appreciated.

kevinsequeira

id int(10) UNSIGNED No auto_increment

this means the id will be automatically generated and incremented by one wrt to the earlier id in the sytem.
so u dont need to worry about it while adding the news
reg
kevin

JD_

ok i modified it so it looks like this

case 'addnews':
        {
    global $db;

/* insert the news */
$query = "INSERT INTO news (name, title, newstext)
VALUES ('$name','$title','$newstext')";
    mysql_query($query);
	mysql_close();

echo "News entered. Thanks!<BR>\n";
echo "<a href=\"{$_SERVER['PHP_SELF']}" .
     "?action=show&all>Back</a>\n";
}

but it isnt entering the data and i know its def in those variable cause if i do echo "$title" etc it shows what i inputted. what am i doing wrong?

and also if possible how would i make the link "Back" link to the news post e.g ?action=show&id=$id $id being the correct number

kevinsequeira

from the original code u posted
change function addnews($id)

function addnews($name,$title,$newstext) {
global $db;
/ insert the news/
$query = "INSERT INTO news (name,title,newstext) VALUES('$name','$title','$newstext')";
mysql_query($query);
echo "News entered. Thanks!<BR>\n";
$get_id_query = "select id from news where name='$name' AND title='$title' AND newstext='$newstext'";
$get_id_result = mysql_query($get_id_query);
$row = mysql_fetch_array($get_id_result);
echo "<a href=$_SERVER['PHP_SELF']?action=show&id=$row['id']>Back</a>";
}

and call the function like this
addnews($POST['name'],$POST['title'],$_POST['newstext']);

reg
kevin

JD_

edit: ok ive just noticed theres no name in the table news thats why it wasnt working. ive added it works great just the back link isnt adding the id just links like action=show&id=

kevinsequeira

before this line
$get_id_result = mysql_query($get_id_query);

do this
echo $query."<BR>";
echo $get_id_query."<BR>";

and after
mysql_query($query);
do this

if(mysql_error()){
echo mysql_error();
}

what is the output for each echo statement

reg
kevin

JD_

News entered. Thanks!
INSERT INTO news (name,title,newstext) VALUES('name in here','Title goes here','some text here')
select id from news where name='name in here' AND title='Title goes here' AND newstext='some text here'

kevinsequeira

debug ...

run the query
select id from news where name='name in here' AND title='Title goes here' AND newstext='some text here' directly in the table and see what the output is ie see if the news is entered into your database

then try changing this

$get_id_query = "select * from news where name='$name' AND title='$title' AND newstext='$newstext'";
$get_id_result = mysql_query($get_id_query);
if(mysql_num_rows($get_id_result)>0){
echo "found possible match";
}else{
echo "id not found";
}
$row = mysql_fetch_array($get_id_result);

reg
kevin

JD_

ok changed to code like u said and it did the "found possible match" but the link was still the same with the id being empty.

i ran the query in phpmyadmin and it found the aproppiate ID

cant see whats wrong.

kevinsequeira

post the edited code and what is the error you are getting now ???
if you are not getting any errors and only id doesnt show in the link try this

echo "<a href=$_SERVER['PHP_SELF']?action=show&id=".$row['id'].">Back</a>";
reg
kevin

JD_

ok to clairfy things the news is working perfectly entering in the db correctly and displaying correctly the only one things that isnt working is the link "Back" which should link to the correct ID but instead shows no id

function addnews($name,$title,$newstext) { 
    global $db; 
/* insert the news*/ 
$query = "INSERT INTO news (name,title,newstext) VALUES('$name','$title','$newstext')"; 
mysql_query($query);

echo "News entered. Thanks!<BR>\n"; 
$get_id_query = "select * from news where name='$name' AND title='$title' AND newstext='$newstext'"; 
$get_id_result = mysql_query($get_id_query); 
if(mysql_num_rows($get_id_result)>0){ 
echo "found possible match"; 
}else{ 
echo "id not found"; 
} 
$row = mysql_fetch_array($get_id_result); 
echo "<a href=$_SERVER['PHP_SELF']?action=show&id=$row['id']>Back</a>"; 
}

and the Case

case 'addnews':
        addnews($_POST['name'],$_POST['title'],$_POST['newstext']);

edit: and i dont get any error at all it just dosent ad the ID onto the link

kevinsequeira

try this

$row = mysql_fetch_array($get_id_result);
$the_id=$row['id'];
// to debug print $the_id
echo $the_id;
// if not displaying try this instead $the_id=$row[0];
// notice that ive changed the echo statement
echo "<a href=$_SERVER['PHP_SELF']?action=show&id=".$the_id.">Back</a>";

JD_

woohooo works !! linking correctly nice one i can see what you did i think taking the $row['id']; out of the link and putting it in another variable then adding that in the link. thanks for all your help and learning me some debugging methods.