What is wrong with my query?

[marukochan]

Can someone point out to me what is wrong with this code?

 

It suppose to print out the company id with "Awarded" status, but instead it prints out ALL the company id regardless it is "Awarded" or "Bidding".

 

$status = "Awarded";
$query = mysql_query ("SELECT * FROM project WHERE project_status = '$status'");
$result= mysql_num_rows($query);

while($row=mysql_fetch_array($query))
{
print $row["company_id"];
echo ("<br>");
}

 

Thanks.

[beboo002]

while($row=mysql_fetch_array($query))

 

change here something like

 

 

while($row=mysql_fetch_array($result))

[Barand]

With this data

[pre]

id, company_id, project_status

1,      1,      'Awarded'

2,      2,      'Bidding'

3,      3,      'Awarded'

4,      4,      'Bidding'

[/pre]

 

your code gave

 

1

3

 

So I'd check your data

 

 

@beeboo

if the query result is in $query, why  do you suggest using $result, which would contain the number of rows?

[beboo002]

while($row=mysql_fetch_array($query)) this query gives error like this

Warning: Supplied argument is not a valid MySQL result resource in c:\apache\htdocs\adminajax\home1.php on line 352

 

so he is not found any result

and sceond qus. is when arise when query gives any result

 

[Barand]

It may give an error with your code. It doesn't with his.