T_CONSTANT_ENCAPSED_SRING error

gene02

<?php
// Select tables
$query1 = mysql_query "SELECT Load_ID , origin_state , origin_city , dest_state , dest_city , contact , phone , email
FROM loads
WHERE status = 1
LEFT JOIN customer ON loads . Cust_ID = customer . Cust_ID
ORDER BY Load_ID" ;
the error is showing line17 ORDER BY Load_ID
$result = mysql_query($query1);

while ($name_row = mysql_fetch_array($result))
{
print ("$name_row[0] $name_row[1] $name_row[2]<br>\n");
}

// if there aren't any loads, then lets generate a user-friendly notice that there aren't any
if ($result <0){
print("<p style=\"font-weight: bold; color: red;\">There Are Currently No Loads In The Database</p>");
}
?>

Thanks for any help you may give the database connection is in a include file and is connecting

halojoy

... i can not delete this post 🙁

gene02

Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in /home/gene02/public_html/admin/editloads.php on line 17

halojoy

gene02 wrote:
Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in /home/gene02/public_html/admin/editloads.php on line 17

If they are variables: $Cust_ID
It probably because you do not use '$Cust_ID', but only $Cust_ID

They can be CONSTANTS
You should always put 'Cust_ID' ... encapsule with single quotes
or variables single quotes: '$Cust_ID'

Wait!!!
You use function without ( .... )

$query1 = mysql_query (
"SELECT Load_ID , origin_state , origin_city , dest_state , dest_city , contact , phone , email
FROM loads
WHERE status = 1
LEFT JOIN customer ON loads . 'Cust_ID' = customer . 'Cust_ID'
ORDER BY '$Load_ID'"
);

gene02

OK GREAT we have moved error to new line and it goes like this
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gene02/public_html/admin/editloads.php on line 21
and that comes from here
while ($name_row = mysql_fetch_array($result))

gene02

MAN I don't think it was this hard learning coldfusion

NogDog

This indicates that the query could not be processed by MySQL for some reason. Try some debugging to find out why:

$result = mysql_query($query1);
if($result == false)
{
   user_error("Query failed: ".mysql_error()."<pre>$query1</pre>");
}

PS: for the production version you'll want to make sure that that error data does not get displayed to the user if it should fail for any reason, this is just a quick-and-dirty debug technique.

halojoy

gene02 wrote:
OK GREAT we have moved error to new line and it goes like this
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gene02/public_html/admin/editloads.php on line 21
and that comes from here
while ($name_row = mysql_fetch_array($result))

Man, 😃 try to use some information from [man]mysql_error[/man]

For almost any mysql function you can use this (many many programmers do)

$result = mysql_query( $query1 ) OR exit( 'DB Error: ' . mysql_error() );

This is from the php [man]mysql_query[/man] manual

$result = mysql_query( 'SELECT * WHERE 1=1' );
if ( !$result ) {
    exit( 'Invalid query: ' . mysql_error() );
}

gene02

Changed MYSQL_FETCH_ARRAY to

while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
printf("Load_ID: %s Cust_ID: %s", $row[0], $row[1]);

Thanks for your help I am sure I will need more help to get this thing you call PHP hmmmm!

I must say good response time here also thanks again