Error, dont understand – Help

@RedheadNov 07.2004

Hi everyone

I’m having a small problem with php i keep getting this error:

~~[QUOTE]~~

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/victoria/public_html/admin/postage-list.php on line 30

[/QUOTE]

here is line 30

[code=php]while ($row = mysql_fetch_array ($result)) [/code]

the funny thing is, is that it worked great till i move it to a differnet Hosting company.

Can someone please help me

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PHP

8 Comments(s) _↴

@solavarNov 08.2004 — #The error is coming from the SQL above this line.

Let's have a look at your SQL statement,

eg:

$sql = "SELECT * FROM mytable";

$result = mysql_query($result);

@RedheadauthorNov 08.2004 — #This is what i have

[code=php]
 include ("../dbconnect.php");
 
 $sql = "SELECT * FROM postagestamps"; 
 
 $result = mysql_query($sql, $db); 
 
 while ($row = mysql_fetch_array ($result)) 
 
 [/code]

and i keep getting the error if i remove the $db from the result page i still get the same error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/victoria/public_html/admin/postage-list.php on line 30
[/QUOTE]

@solavarNov 08.2004 — #Let's see if MySQL will report the cause of error...

[code=php]
 
 include ("../dbconnect.php");
 
 $sql = "SELECT * FROM postagestamps";
 
 if(! $result = mysql_query($sql, $db))
 {
 echo "The error is: " .mysql_error();
 
 // What is $db
 echo "<br /> The value of '$db' is: $db";
 }
 else
 {
 
  while ($row = mysql_fetch_array ($result))
  {
   echo "...works ok";
   }
 }
 
 [/code]

The problem may be coming from the included file. $db is probably empty, thereby nullifying the result resource.

@RedheadauthorNov 08.2004 — #I also tryed this insert statement with PHP myAdmin version 2.6.0-pl2

[code=php]
 $sql = "INSERT INTO postagestamps (psname, psdiscription, categorie, subcategorie, imageid) VALUES ('psname', 'psdiscription', 'categorie', 'subcategorie', 'imgfile_name')" 
 [/code]

and got this error

#1064 - 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 '$sql = "INSERT INTO postagestamps (psname, psdiscription, categ
[/QUOTE]

so i guess thing have chaned in the new version ?

@solavarNov 08.2004 — #Hi redhead,

I think your variables are missing the $prefix...

[code=php]
 
 //You have...
 
 $sql = "INSERT INTO postagestamps (psname, psdiscription, categorie, subcategorie, imageid) VALUES ('psname', 'psdiscription', 'categorie', 'subcategorie', 'imgfile_name')";
 
 //...must be
 
 $sql = "INSERT INTO postagestamps (psname, psdiscription, categorie, subcategorie, imageid) VALUES ('$psname', '$psdiscription', '$categorie', '$subcategorie', '$imgfile_name')";
 
 [/code]

Also, make sure that you've got a SEMI-colon at the end of your SQL statement.

Good luck

solavar

@RedheadauthorNov 08.2004 — #Ok here is my problem with a select statement

[code=php]
 include ("../dbconnect.php");
 
 $sql = "SELECT * FROM postagestamps"; 
 
 $result = mysql_query($sql, $db); 
 
 echo "
 Stamps Of Victoria Postage Stamps Page
 
 <table width=50%> 
 
 <tr>
 
 <td>Postage Stamp Name</td>
 
 <td>Postage Stamp Discription</td>
 
 <td>&nbsp;</td>
 
 <td>&nbsp;</td>";
 
 while ($row = mysql_fetch_array ($result)) 
 
 { 
 
 echo "
 <tr>
 <td>$row[psname]</td> 
 <td>$row[psdiscription]</td> 
 <td><a href="edit-postagestamps.php?id=$row[id]">Edit</td>
 <td><a href="delete-postagestamps.php?id=$row[id]">Remove</td>
 </tr> 
 "; 
 } 
 
 echo "</table>"; 
 echo"<a href=adminlist.php>Go back to the main menu</a>";
 ?>
 [/code]

This is what i have and i'm getting this :

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/victoria/public_html/admin/postage-list.php on line 30

[/QUOTE]

The first echo statment does show up.

with my insert statement this is what i have:

[code=php]
 include ("../dbconnect.php");
 
 $sql = "INSERT INTO postagestamps ( id , psname , psdiscription , categorie , subcategorie , imageid ) 
 VALUES ('$psname', '$psdiscription', '$categorie', '$subcategorie', '$imgfile_name')"; 
 
 
 $result = mysql_query($sql, $db); 
 
 echo " 
 The following Other Stamp was added to the other stamp page 
 
 <table border=1> 
 <tr> 
 <td>Other Stamp Name</td> 
 <td>$psname</td> 
 </tr> 
 <tr> 
 <td>Pther Stamp Discription</td> 
 <td>$psdiscription</td> 
 </tr> 
 </table> 
 
 "; 
 
 echo "<img src=../psimages/$imgfile_name>";
 echo"<p><a href=adminlist.php>Go back to the main menu</a>.</p>";
 [/code]

when i run this in IE, everthing seams to be working but when i check, PHPMyAdmin nothing appears in the DB. I have checed all spelling for tables, db name, fields etc etc etc..

Like i sayid i had all this working on a different server with a different hosting company. and when i moved over to new server and new hosting comapny nothing works! ?

MAN OH MAN i'm so confused.

@solavarNov 08.2004 — #Uhmm, quite a puzzle.

Okay, get MySQL to display any error when SQL is executed.

Wherever you have:

$result = mysql_query($sql, $db);

do this instead...

[code=php]
 
 $result = mysql_query($sql, $db) or die(mysql_error());
 
 [/code]

If MySQL is failing to INSERT the data, it will tell you why.

@yunaNov 10.2004 — #I think you're missing the final semicolon:

<i>
 </i>$sql = "SELECT * FROM postagestamps";
 $result = mysql_query($sql, $db);

should be

<i>
 </i>$sql = "SELECT * FROM postagestamps;";                                                    <br/>
 ^
 $result = mysql_query($sql, $db);

It's often easy to forget that BOTH the SQL server (except Oracle!) and PHP need a final semicolon.

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