SQL ERROR mysql_fetch_array(): not valid?

Question

I just can't figure out why i get the error message, I have tried removing the'' and the()

I have run the script in phpmyadmin and it says the problem with my syntax is at $result = ("SELECT * FROM 'test_prefixCatagory' ORDER by 'Cat'");

$result = ("SELECT * FROM 'test_prefixCatagory' ORDER by 'Cat'");


while($row = mysql_fetch_array($result))

$sCat = ($row['Cat']);
$sCatID = ($row['CatID']);
{
echo "<table>";
 echo "<tr valign='top'><td><b><a href='#".$sCat."'>".$sCat."</a></b><br>";
 // column 1 categories
 $result2 = ("SELECT * FROM `test_prefixSubCat` WHERE `CatID`=$sCatID");
 // sub-cats
 while($row2 = mysql_fetch_array($result2)) 
    {
  $sSub = ($row2['CatID']);
  $sSubID = ($row2['SubID']);
  echo "<dd><a href='#'>".$sSub."</a><br>";

 }

 echo "<br></td></tr>";
echo "</table>";
 }

Do anyone have an idea?

You should no longer listen to the one that told you to put brackets around everything like ($row2['CatID']). — MrCode
– MrCode, Commented May 17, 2012 at 14:02

d4rkpr1nc3 · Accepted Answer · 2012-05-17 14:14:08Z

1

Try this :

<?php
$result = mysql_query("SELECT * FROM `test_prefixCatagory ORDER by `Cat`");

while ($row = mysql_fetch_array($result)) {
$sCat = $row['Cat'];
$sCatID = $row['CatID'];
echo "<table>";
echo "<tr valign='top'><td><b><a href='#" . $sCat . "'>" . $sCat . "</a></b><br>";
// column 1 categories
$result2 = mysql_query("SELECT * FROM `test_prefixSubCat` WHERE `CatID`='".$sCatID. "'");
// sub-cats
while ($row2 = mysql_fetch_array($result2)) {
    $sSub = $row2['CatID'];
    $sSubID = $row2['SubID'];
    echo "<dd><a href='#'>" . $sSub . "</a><br>";

    }

        echo "<br></td></tr>";
        echo "</table>";
}
?>

edited May 17, 2012 at 14:14

answered May 17, 2012 at 13:48

d4rkpr1nc3

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3 Comments

andrew Over a year ago

actually both results need fixing according to d4rkpr1nc3's suggestion

d4rkpr1nc3 Over a year ago

yes, sorry, i've stop looking down after seeing the first line

Marinus Over a year ago

$result = mysql_query("SELECT * FROM test_prefixCatagory ORDER by Cat"); while($row = mysql_fetch_array($result)) { echo "<table>"; echo "<tr valign='top'><td><b><a href='#".$row['Cat']."'>".$row['Cat']."</a></b><br>"; $sCatID = ($row['CatID']); $result2 = mysql_query("SELECT * FROM test_prefixSubCat WHERE CatID=$sCatID"); while($row2 = mysql_fetch_array($result2)) { $sSub = ($row2['CatID']); $sSubID = ($row2['SubID']); echo "<dd><a href='#'>".$row2['Sub']."</a><br>"; Had to rearrange a few things in it. Now its working, your answer helped thanks.

triclosan · Accepted Answer · 2012-05-17 13:56:35Z

0

$result = ("SELECT * FROM `test_prefixCatagory` ORDER by `Cat`");

edited May 17, 2012 at 13:56

answered May 17, 2012 at 13:49

triclosan

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2 Comments

d4rkpr1nc3 Over a year ago

the "`" character is only necesary when tables or column names are sql keywords (values, relations, keys, rows, columns, order etc.).

triclosan Over a year ago

yes its strongly necessary in case of keywords, but harmless at all

MrCode · Accepted Answer · 2012-05-17 14:12:40Z

0

Not only do you need to add mysql_query but you also need to remove the single quotes from the table name and field name. You can use backticks if you wish but not single quotes around table names.

$result = mysql_query("SELECT * FROM `test_prefixCatagory` ORDER by `Cat`");

// other query:
 $result2 = mysql_query("SELECT * FROM `test_prefixSubCat` WHERE `CatID`=$sCatID");

When debugging MySQL problems, use mysql_error() to see a description of the problem.

edited May 17, 2012 at 14:12

answered May 17, 2012 at 14:07

MrCode

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