category php code ?

[Davie33]

Hi guys hoping you can help with this code of mine.

I seem to be getting an error ,i know its the query but tried a few to fix it without joy.

<?php
$catcheck = yasDB_select("SELECT `name` FROM `categories` WHERE `id` = " . $row['category']);
$catname = $catcheck->fetch_array(MYSQLI_ASSOC);
if ($setting['seo']=='yes') { ?>
    <a href="<?php echo $setting['siteurl'];?>category/<?php echo $row['category'];?>/1.html" title="<?php echo ucfirst($catname['name']);?>"><?php echo ucfirst($catname['name']);?></a>
    <?php
} else { ?>
    <a href="<?php echo $setting['siteurl'];?>index.php?act=cat&id=<?php echo $row['category'];?>" title="<?php echo ucfirst($catname['name']);?>"><?php echo ucfirst($catname['name']);?></a>
    <?php
}
?>

this is the line thats causing the error.

$catcheck = yasDB_select("SELECT `name` FROM `categories` WHERE `id` = " . $row['category']);

error

    Error No: 0 - MySQL error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 
    Query: "SELECT `name` FROM `categories` WHERE `id` = "
    #0 /public_html/arcade/templates/minix_25/game.php(236): yasDB_select('SELECT `name` F...')
    #2 {main}

[Ch0cu3r]

$row['category'] is used in your query to return records from your table where the id matches. Problem is $row['category'] is not populating the query with a value as shown in the error message (query highlighted in red)

 

 

Error No: 0 - MySQL error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1
    Query: "SELECT `name` FROM `categories` WHERE `id` = " <-- no value provide for id resulting in the error above
    #0 /public_html/arcade/templates/minix_25/game.php(236): yasDB_select('SELECT `name` F...')
    #2 {main}

 

Wherw is $row['category'] defined? Are you sure that is the correct variable to use in the query?

[Davie33]

It not my code i had it made for me that's why i posted on here for help am kinda lost with it.

Sry am not an expert on php .

How can i fix it ?.

[maxxd]

What's the surrounding code? Given the variable name $row, it's possible the code is trying to run the select query inside a loop, which isn't a good idea. If that is what's actually happening, you're probably going to want to rework the initial query to use a join in order to bring back the entire record set with a single call to the database.

[Davie33]

Hi what am looking for is when i go to the game page that code above is to give the name of the category the game is from.

For some reason the code does work but still give out the error as thats whats on my site atm is that code.

Check my link below then go to the main game page on any game you will see the name of the cat the game is from.

[maxxd]

Right, and that's why we're asking you to supply the surrounding code. $row['category'] either isn't being set, or it's being set incorrectly. However, without the surrounding code, there's no way to tell where, how, or why.

[Davie33]

Am not to sure but i think this has solved it.

 

 

<?php
$catcheck = yasDB_select("SELECT `name` FROM `categories` WHERE `id` = " . $row['category']);
$catname = $catcheck->fetch_array(MYSQLI_ASSOC);
if ($id = $row['category']) {
if ($setting['seo']=='yes') { ?>
    <a href="<?php echo $setting['siteurl'];?>category/<?php echo $row['category'];?>/1.html" onMouseover="showhint('<b><?php echo ucfirst($catname['name']);?></b>',this, event, 'auto')"><?php echo ucfirst($catname['name']);?></a>
    <?php
} else { ?>
    <a href="<?php echo $setting['siteurl'];?>index.php?act=cat&id=<?php echo $row['category'];?>" onMouseover="showhint('<b><?php echo ucfirst($catname['name']);?></b>',this, event, 'auto')"><?php echo ucfirst($catname['name']);?></a>
    <?php
}
}
?>

[Davie33]

Fook me i already fixed it on another file thanks guys for the help anyway.

I really thought i had lost the file that i fixed .