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Query Not Executing Correctly
Posted: Wed Sep 22, 2004 4:54 am
by Harlequin
I have the following query which returns a "Query Empty" error:
Code: Select all
$Query = mysql_query("SELECT * FROM events
WHERE EventCode == 'PMJ';
ORDER BY EventTime DESC");
$Data = mysql_query($Query) or die("Error: " . mysql_error());
while($row = mysql_fetch_array($Query))
Can anyone see what I can't...? I've checked and double checked and it seems OK and there's definately data in the tale.
echo
Posted: Wed Sep 22, 2004 5:17 am
by phpScott
echo out the query and run it against the db directly
but I think it is the == try just using a single =
mysql isn't the same a php for checking if an item is equal.
Posted: Wed Sep 22, 2004 5:36 am
by Harlequin
Well, this is just driving me mad...
This works when I execute it against the database:
Code: Select all
SELECT * FROM events
WHERE EventCode = 'PMJ';
But this, in the PHP code doesn't:
Code: Select all
$Query = mysql_query "SELECT * FROM events
WHERE EventCode = 'PMJ'";
I get "Unexpected T_STRING
Grrr...
Posted: Wed Sep 22, 2004 5:38 am
by timvw
[php_man]types.string[/php_man]
notice that your syntax is wrong.
Posted: Wed Sep 22, 2004 5:43 am
by Harlequin
I know mate but I can't seem to get it right...!
I'll keep plugging away though...
two different codes
Posted: Wed Sep 22, 2004 6:06 am
by phpScott
OK you now have to different sets of codes
the second set
$Query = mysql_query "SELECT * FROM events
WHERE EventCode = 'PMJ'";
will not work because you need the () around the select statement.
try
Code: Select all
$Query = mysql_query ("SELECT * FROM events
WHERE EventCode = 'PMJ'");
Posted: Wed Sep 22, 2004 6:10 am
by Harlequin
I tried that Scott. Now I get a syntax error returned so I must have something screwed up later in the query where I am trying to present the results.
Thanks very much for the tip mate.
Posted: Wed Sep 22, 2004 6:22 am
by Harlequin
Scott.
I know this is basic stuff but I just can't sem to get it working properly:
I Execute the query:
Code: Select all
// Select Events For PMJ:
$Query = mysql_query ("SELECT * FROM events
WHERE EventCode = 'PMJ'");
Then I get the error (SQL Syntax":
Code: Select all
$Data = mysql_query($Query) or die("Error: " . mysql_error());
while($row = mysql_fetch_array($Query))
{
// Pull Strings From Table:
$EventDate=$rowї"EventDate"];
$EventDateBroken = explode("-", $EventDate);
What's causing the error...? Can you see it...?
Posted: Wed Sep 22, 2004 6:27 am
by John Cartwright
Code: Select all
<?php
$Data = mysql_query($Query) or die("Error: " . mysql_error());
while($row = mysql_fetch_array($Query))
{
// Pull Strings From Table:
$EventDate=$row["EventDate"];
$EventDateBroken = explode("-", $EventDate);
?>
should b
Code: Select all
<?php
$Data = mysql_query($Query) or die("Error: " . mysql_error());
while($row = mysql_fetch_array($Data))
{
// Pull Strings From Table:
$EventDate=$row["EventDate"];
$EventDateBroken = explode("-", $EventDate);
?>
just say it
Posted: Wed Sep 22, 2004 6:28 am
by phpScott
you are doing
Code: Select all
$Query = mysql_query("SELECT * FROM events
WHERE EventCode == 'PMJ';
ORDER BY EventTime DESC");
$Data = mysql_query($Query) or die("Error: " . mysql_error());
while($row = mysql_fetch_array($Query))
what you are doing is creating the $Query with mysql_query() in it.
then with $DATA you are agian using mysql_query().
try
Code: Select all
$Query = "SELECT * FROM events
WHERE EventCode == 'PMJ';
ORDER BY EventTime DESC";
$Data = mysql_query($Query) or die("Error: " . mysql_error());
while($row = mysql_fetch_array($Data))
$query now only holds the text for the query, it then gets submitted to mysql_query();
$Data will now hold your result set.
Posted: Wed Sep 22, 2004 6:29 am
by John Cartwright
Code: Select all
// Select Events For PMJ:
$Query = "SELECT * FROM events
WHERE EventCode = 'PMJ'";
and then
Code: Select all
<?php
$Data = mysql_query($Query) or die("Error: " . mysql_error());
while($row = mysql_fetch_array($Query))
{
// Pull Strings From Table:
$EventDate=$row["EventDate"];
$EventDateBroken = explode("-", $EventDate);
?>
should b
Code: Select all
<?php
$Data = mysql_query($Query) or die("Error: " . mysql_error());
while($row = mysql_fetch_array($Data))
{
// Pull Strings From Table:
$EventDate=$row["EventDate"];
$EventDateBroken = explode("-", $EventDate);
?>
Posted: Wed Sep 22, 2004 7:34 am
by Harlequin
Worked a treat
Now all i need to do is get the tables layed out correctly. For some reason when I insert a table below the code for the last table it's header text appears above the last table.

good luck
Posted: Wed Sep 22, 2004 7:54 am
by phpScott
good luck
view source of the web page is your friend.
Posted: Wed Sep 22, 2004 8:06 am
by Harlequin
AHA...!
I forgot to close the <table> TAG
Thanks once again...!