Originally Posted by mgraphic

You are missing a ; on the line before it

Now I am getting this error:

syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING i

Referring line:

PHP Code:

while ( $row = mysql_fetch_array($subcats_query) ) echo "<li>$row['subcategory_name']"; 


When you enclose array vars in double quotes, you need to wrap them in {}

echo "<li>{$row['subcategory_name']}";

Originally Posted by mgraphic

When you enclose array vars in double quotes, you need to wrap them in {}

echo "<li>{$row['subcategory_name']}";

Now I get the same (Whitespace) error on this line:

PHP Code:

while ( $row = mysql_fetch_array($cat_query) ) echo "<li><a href=categories.php?id=$row['category_id']>$row['category_name']</a>"; 


Note I tried adding the open curly bracket just before the $row['category_id'] and the closing curly bracket right after the ] on category_name, and another error.

PHP Code:

while ( $row = mysql_fetch_array($cat_query) ) echo "<li><a href=categories.php?id={$row['category_id']}>{$row['category_name']}</a>"; 


http://www.php.net/manual/en/languag...string.parsing

Ok, I have figured that out, Finally. Now, your still with me on this I hope. Now what I can't seem to figure out is with the multi column code to add to it. I think the query name needs some changing, but I am having problems with it.

Here's the two pieces of code that I am trying to join:

Script #1:

PHP Code:

<?php


    $link = mysql_connect("localhost", "user", "pass");
    mysql_select_db("duckster_categories", $link);

    $query = "SELECT * from categories";
    $result=mysql_query($query);
 
    $cols=4;        // Here we define the number of columns
    echo "<table align=\"center\">";    // The container table with $cols columns
    do{
    echo "<tr>";
    for($i=1;$i<=$cols;$i++) {    
    
    // All the rows will have $cols columns even if the records are less than $cols

    $row=mysql_fetch_array($result);
    if($row){
    // $img = $row['image_path'];
?>

    <td>
    <table>
    <tr valign="top">
    <!-- <td><img src="http://tycoontalk.freelancer.com/images/<?=$img ?>" /></td> <!-- columns can have both text and images -->
    <td>
    
    <b><?="<a href=\"$row[url]\">".$row['category_name']?></a></a></b>
  

    <?=$row['email'] ?><br />
    <?=$row['password'] ?><br />
    </td>
    <td width="50">&nbsp;</td>    <!-- Create gap between columns -->
    </tr>
    </table>
    </td>

<?php
    }
    else{
    echo "<td>&nbsp;</td>";    //If there are no more records at the end, add a blank column
    }
    }
    } 
    while($row);
    echo "</table>";
?>

Script #2:

PHP Code:

 <?php 

// create conection to DB

$db_link = mysql_connect("localhost", "user", "pass");
$db_name = mysql_select_db("duckster_categories");

if ( isset($_GET['id']) && is_numeric($_GET['id'] )) {

$cat_id = $_GET['id'];
$subcats_query_sql = "SELECT * FROM categories, subcategories WHERE categories.category_id = $cat_id AND categories.category_id = subcategories.subcategory_cat_id";
$subcats_query = mysql_query($subcats_query_sql);

echo "<ul>";
while ($row = mysql_fetch_array($subcats_query) ) echo "<li>{$row['subcategory_name']}";
echo "</ul>";

// Outputs you a list of subcategories in this category

} else {

$cat_query_sql = "SELECT * FROM categories";
$cat_query = mysql_query($cat_query_sql); 

echo "<ul>";
while ( $row = mysql_fetch_array($cat_query) ) echo "<li><a href=\"categories.php?id={$row['category_id']}\">{$row['category_name']}</a></li>";
echo "</ul>";

// This is the categories list

} 

?>

Basically, I just want to have one page, where the categories with their correct links (which is already done in script #2) are show in multiple columns (as in script #1), instead of a list as it currently is.

Thank you again!
Brian

If you want the table layout from script 1, just copy the block of code that builds the table data layout from script 1 between the else { . . . } brackets in script 2

Originally Posted by mgraphic

If you want the table layout from script 1, just copy the block of code that builds the table data layout from script 1 between the else { . . . } brackets in script 2

Did that, all I get is a blank page.

Here's what I did:

PHP Code:

 // Outputs you a list of subcategories in this category

} else {


echo "<table align=\"center\">";    // The container table with $cols columns
    do{
    echo "<tr>";
    for($i=1;$i<=$cols;$i++) {    
    
    // All the rows will have $cols columns even if the records are less than $cols

    $row=mysql_fetch_array($result);
    if($row){
    // $img = $row['image_path'];
?>

    <td>
    <table>
    <tr valign="top">
    <!-- <td><img src="http://tycoontalk.freelancer.com/images/<?=$img ?>" /></td> <!-- columns can have both text and images -->
    <td>
    
    <b><?="<a href=\"$row[url]\">".$row['category_name']?></a></a></b>
  

    <?=$row['email'] ?><br />
    <?=$row['password'] ?><br />
    </td>
    <td width="50">&nbsp;</td>    <!-- Create gap between columns -->
    </tr>
    </table>
    </td>

<?php
    }
    else{
    echo "<td>&nbsp;</td>";    //If there are no more records at the end, add a blank column
    }
    }
    } 
    while($row);
    echo "</table>";

// This is the categories list

}

You forgot the query and define the number of columns

Originally Posted by mgraphic

You forgot the query and define the number of columns

Ok, I've added the query, and the # of columns, now I am getting this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in categories.php

PHP Code:

<?php 

// create conection to DB

$db_link = mysql_connect("localhost", "user", "pass");
$db_name = mysql_select_db("db");

if ( isset($_GET['id']) && is_numeric($_GET['id'] )) {

$cat_id = $_GET['id'];
$subcats_query_sql = "SELECT * FROM categories, subcategories WHERE categories.category_id = $cat_id AND categories.category_id = subcategories.subcategory_cat_id";
$subcats_query = mysql_query($subcats_query_sql);

echo "<ul>";
while ($row = mysql_fetch_array($subcats_query) ) echo "<li>{$row['subcategory_name']}";
echo "</ul>";

// Outputs you a list of subcategories in this category

} else {

$cat_query_sql = "SELECT * FROM categories";
$cat_query = mysql_query($cat_query_sql); 

$cols=4;        // Here we define the number of columns
    echo "<table align=\"center\">";    // The container table with $cols columns
    do{
    echo "<tr>";
    for($i=1;$i<=$cols;$i++) {    
    
    // All the rows will have $cols columns even if the records are less than $cols

    $row=mysql_fetch_array($result);
    if($row){
    // $img = $row['image_path'];
?>

    <td>
    <table>
    <tr valign="top">
    <!-- <td><img src="http://tycoontalk.freelancer.com/images/<?=$img ?>" /></td> <!-- columns can have both text and images -->
    <td>

<?=$row['email'] ?><br />
    <?=$row['password'] ?><br />
    </td>
    <td width="50">&nbsp;</td>    <!-- Create gap between columns -->
    </tr>
    </table>
    </td>

<?php
    }
    else{
    echo "<td>&nbsp;</td>";    //If there are no more records at the end, add a blank column
    }
    }
    } 
    while($row);
    echo "</table>"; 

echo "<ul>";
while ( $row = mysql_fetch_array($cat_query) ) echo "<li><a href=\"categories.php?id={$row['category_id']}\">{$row['category_name']}</a></li>";
echo "</ul>";

// This is the categories list

} 

?>

You are calling mysql_fetch_array($result) when $result does not exist. Your query is assigned to the var $cat_query

Originally Posted by mgraphic

You are calling mysql_fetch_array($result) when $result does not exist. Your query is assigned to the var $cat_query

I really don't know the correct syntax, I need to relearn alot of my php, but this category index needs to be working...I hate to be a pain, I will learn it, but I need to learn from working examples.

Edit mysql_fetch_array($result) to mysql_fetch_array($cat_query)

Originally Posted by mgraphic

Edit mysql_fetch_array($result) to mysql_fetch_array($cat_query)

Thank you mgraphic for your continuing help on this issue. I have got it working.