retriving data from mysql

[juansoul]

Im having problem retrieving data from mysql database ,I succesfully created a table and enterd information via an html form but when i try to retrieve info from the database I get this error

Warning: Supplied argument is not a valid MySQL result resource in /home/www/bullzeyecreative/Display.php on line 28the error seem to be arround were the while function begins
can some please tell me whats wrong in here

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<?php
$Host = "localhost";
$User = "00000";
$Password = "00000";
$DBName = "00000";
$TableName = "Feedback";

$Link = mysql_connect ($Host, $User, $Password);

$Query = "SELECT * from $TableName";
$Result = mysql_db_query ($DBName, $Query, $Link);

print ("<table border=1 width=\"75%\" cellspacing=2 cellpadding=2>");
  print ("<tr>");
    print ("<td>Name</td>");
    print ("<td>Email</td>");
    print ("<td>Comments</td>");
  print ("</tr>");
 
while ($Row = mysql_fetch_array($Result)) {
  print ("<tr align=center valign=top>\n");
  print ("<td align=center valign=top>$Row[FirstName]</td>\n");
  print ("<td align=center valign=top>$Row[LastName]</td>\n");
  print ("<td align=center valign=top>$Row[Email]</td>\n");
  print ("<td align=center valign=top>$Row[Comments]</td>\n");
  print ("</tr>\n"); }
 
mysql_close ($Link);
  print ("</table>\n");
?>


[DoppyNL]

The query you executed failed for some reason.
perhaps you made a typo somewhere.

print out the query in your php-script to see what is actually executed by PHP.

Greetz Daan

[juansoul]

well cant you tell me by looking at the script i posted here where the typo is because i cant find it , and i dont know how to print the query

[DoppyNL]

you can print your query by simply using "print($Query);"

What actually is going wrong in your script is that the query ($Query) gets an error when executed so no values are returned.
after that you try to retrieve the data at the beginning of your loop php generates an error because $Result doesn't contain a correct query-result (the query failt).

You should check if the table "Feedback" exists in your database (case sensitive!) since this is probably the problem.


some other minor mistakes in your script:

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$Row[FirstName]

must be

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$Row['FirstName']

there must be quotes around strings, otherwise php thinks its a constant and generate a warning.

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print ("<td align=center valign=top>$Row['FirstName']</td>\n");

must be:

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print ("<td align=center valign=top>" . $Row['FirstName'] . "</td>\n");

you can't add a value from an array like you tried.

Code: Select all

mysql_close ($Link);

is not wrong, but also not necesary, php will do that for you when the script ends.

Greetz Daan

[Jay]

well cant you tell me by looking at the script i posted here where the typo is because i cant find it , and i dont know how to print the query

You've got
$Link = mysql_connect ($Host, $User, $Password);

$Query = "SELECT * from $TableName";
$Result = mysql_db_query ($DBName, $Query, $Link);

Try
$Link = mysql_connect ($Host, $User, $Password);
mysql_select_db($DBName); # This line is missing
$Query = "SELECT * from $TableName";
$Result = mysql_query ($Query);

That should work.

[juansoul]

Man Im still getting an error message on the line where the "while" function starts why????[/b]

[Jay]

It would help us to help you if you provided a bit more info. Did you try amending the code as I suggested?