weird error with mysql...mysql_fetch_array(): supplied argument is not...

[kelphyr]

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

 

$keyw=($_POST['keywords']);

$sql = "SELECT * FROM items_db WHERE name_japanese LIKE '%$keyw%'";

$result = mysql_query($sql);

 

while($row = mysql_fetch_array($result)){

...

 

i dont understand

[hitman6003]

Change:

 

$result = mysql_query($sql);

 

to

 

$result = mysql_query($sql) or die(mysql_error());

 

And see what the error is.

[beaux1]

Replace with:

while($row = @mysql_fetch_array($result)){

 

Tell me if it works, I did for me when I got those problems before.

[hitman6003]

Replace with:

while($row = @mysql_fetch_array($result)){

 

Tell me if it works, I did for me when I got those problems before.

 

That doesn't solve anything...it merely forces php to not display any errors.

[beaux1]

Oh shit, I did that on my site. Runs fine but eh, maybe I should look into it.

[kelphyr]

oh it says my table doesnt exist, ill check this and tell you whats up with that brb

[kelphyr]

okok, i got it my table name had a letter more.. now it works yay

[trq]

There is nothing weird about this error considering you've made the mistake of not checking your query succeeded before trying to use it. The general syntax for a select query should always be at least...

 

<?php

  $keyw=($_POST['keywords']);
  $sql = "SELECT * FROM items_db WHERE name_japanese LIKE '%$keyw%'";
  if ($result = mysql_query($sql)) {
    if (mysql_num_rows($result)) {
      while ($row = mysql_fetch_array($result)) {
        // display data.
      }
    } // else no results found.
  } // else query failed.

?>