Got any idea why this doesn't' work?

**petesmc** · Jan 4, 2001 16:32

<HTML>
<HEAD>
<TITLE>List</TITLE>
</HEAD>
<BODY> 
<basefont face="verdana">
<table width="60%" align="center">
<tr><td>
<?php

include("articles.inc");
mysql_select_db('petesmc_articles');

// SELECT

$select = mysql_query("SELECT AID, PID, articles.Description, Title, Name FROM articles, title, authors " .
" WHERE TID=title.ID AND AuthorID=authors.ID AND SubID=$id ");

if (!$select) {
echo("Error retrieving category from database! ".
"Error: " . mysql_error());
exit();
}

// WRITE ARRAY AND DISPLAY

while ($list= mysql_fetch_array($select)) {
$aid = $list["AID"];
$pid = $list["PID"];
$name = $list["Name"];
$title = $list["Title"];
$description = $list["Description"];
echo("<a href='article.php?aid=$aid&pid=$pid'>$title " .
"</a> By $name $description");
}
?>

</td></tr></table>
</body></html>

Example of the script, doesn't work obviously, it should display a record:

http://216.74.101.27/list.php?id=1

**TWTCommish** · Jan 4, 2001 16:35

Where's the username/password? Is it in the articles.inc file?

**Robo** · Jan 4, 2001 16:44

Yes, make sure the user/pass is present.

And also, in this query:

$select = mysql_query("SELECT AID, PID, articles.Description, Title, Name FROM articles, title, authors WHERE TID=title.ID AND AuthorID=authors.ID AND SubID=$id");

Try typing that query in PHPMyAdmin, and see if any results comes up.

You might have to use:

SELECT AID, PID, articles.Description, Title, Name FROM articles, title, authors WHERE articles.TID=title.ID AND articles.AuthorID=authors.ID AND articles.SubID=$id

Notice in the WHERE bit, I define the table's name in both the column names, you might have to do that when there's 2+ tables.

Also, try putting the WHERE queries in the same order as the FROM queries. eg:

FROM articles, title, authors WHERE articles.tid=title.id AND articles.authorID=authors.ID

Not:

FROM articles, title, authors WHERE articles.authorID=authors.ID AND articles.tid=title.id

**petesmc** · Jan 4, 2001 17:04

Yes, username and pass are in articles.inc

In PHPMyAdmin i get the error:

Error
SQL-query:

mysql_query("SELECT AID, PID, articles.Description, Title, Name FROM articles, title, authors WHERE TID=title.ID AND AuthorID=authors.ID AND SubID=$id");

MySQL said: You have an error in your SQL syntax near 'mysql_query("SELECT AID, PID, articles.Description, Title, Name FROM articles, t' at line 1

<Edited by petesmc on 01-04-2001 at 05:08 PM>

**petesmc** · Jan 4, 2001 17:22

This might help

http://216.74.101.27/structure.html

**Robo** · Jan 4, 2001 18:05

In:

SELECT AID, PID, articles.Description, Title, Name FROM

Define which table those fields are in, like:

SELECT articles.aid, articles.pid, articles.description... FROM...

**petesmc** · Jan 4, 2001 18:17

OKay, done that. In PHPMyAdmin is get an error saying that:

SQL-query:

SELECT articles.AID, articles.PID, articles.Description, title.Title, authors.Name FROM articles, title, authors WHERE articles.TID=title.ID AND articles.AuthorID=authors.ID AND articles.SubID=$id
MySQL said: Unknown column '$id' in 'where clause'

Here is my current code:

--------------

<?php

include("articles.inc");
mysql_select_db('petesmc_articles');

// SELECT

$text = "SELECT articles.AID, articles.PID, articles.Description, title.Title, authors.Name FROM articles, title, authors WHERE articles.TID=title.ID AND articles.AuthorID=authors.ID AND articles.SubID=$id";

$select = mysql_query($text);

if (!$select) {
echo("Error retrieving category from database! ".
"Error: " . mysql_error());
exit();
}

// WRITE ARRAY AND DISPLAY

while ($list= mysql_fetch_array($select)) {
$aid = $list["AID"];
$pid = $list["PID"];
$name = $list["Name"];
$title = $list["Title"];
$description = $list["Description"];
echo("<a href='article.php?aid=$aid&pid=$pid'>$title " .
"</a> By $name $description");
}
?>

**petesmc** · Jan 4, 2001 18:35

Thanx allot guys, i got it working

**Robo** · Jan 4, 2001 18:37

You can have $id in the PHPMyAdmin command, 'cos $id is a variable, and you can define it in PHPMyAdmin, replace it with a number instead.

But if the query was done in a php file, and the id had been defined, then it should work. like filename.php?id=2

**petesmc** · Jan 4, 2001 18:44

Thanx so much

**petesmc** · Jan 4, 2001 18:45

Just need help with another script that no one could figure our earlier........if you wanna help just ask me..

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