Jump to content

Recommended Posts

Hello!

 

I have a php page which shows a list of customers displayed in an html table format, The first column in the table contains a link for the user to select the row which then navigates to another php page, which is the customer order page. On this page, I want to get certain information based on the name of the customer to prepopulate the customer fields on this form, so I have some php code near the top of the page, as per attached document.

 

I have checked to ensure that the $_GET['business_name'] contains a value. The query runs fine in myssql and returns only one record when I provide a value for the business_name, however when I run the code in php, the $result variable shows 'Resource id ='7' type='mySQL result'.

 

I am a total newbie on PHP/MySQL and inherited this application. I have tried to find solutions on line through various forums, but no luck. Perhaps I am not asking/looking for the right question.

 

Anyway, any assistance would be greatly appreciated.

 

 

 

 

 

customer_order_php.php

Link to comment
https://forums.phpfreaks.com/topic/288695-query-issue-in-php/
Share on other sites


“<?php include("session.php"); ?>

<!--get customer information for customer name selected in view_customer_list-->
<?php

$business_name = $_GET['business_name'];

$this_query = "Select business_name, account_num, main_contact, business_phone, business_email, business_suite, business_address, business_city, business_region, business_province, business_postal FROM orders WHERE sale_datetime in (Select Max(sale_datetime) from orders Group by business_name) and business_name = '$business_name' ORDER BY business_name ASC ";
$result = mysql_query($this_query);
$row_customer = mysql_fetch_array($result);

?>

Link to comment
https://forums.phpfreaks.com/topic/288695-query-issue-in-php/#findComment-1480523
Share on other sites

Add this right after the query call and before the fetch call:

 

 

if (!$result)
{
   echo "Query did not run - ".MySQL_error();
   exit();
}

 

I really don't like your where clause.  Didn't know you could use a select as the object of an 'in' clause

Link to comment
https://forums.phpfreaks.com/topic/288695-query-issue-in-php/#findComment-1480524
Share on other sites

give the following a try:

while ($row = mysql_fetch_assoc($result)){
       DO WHAT YOU NEED WITH EACH ROW OF THE DATA HERE
}

By using mysql_fetch_assoc you can use the name of the column (eg. $row['account_num']) instead of referencing the key in the array.

 

You could even build your html and set it to a variable like this:

<?php
$business_name = $_GET['business_name']; 
$this_query = "Select business_name, account_num, main_contact, business_phone, business_email, business_suite, business_address, business_city, business_region, business_province, business_postal FROM orders WHERE sale_datetime in (Select Max(sale_datetime) from orders Group by business_name) and business_name = '$business_name' ORDER BY business_name ASC ";

$mytablehtml = '';
$result = mysql_query($this_query) or die(mysql_error()); 
if (!$result)
{
   echo "Query did not run - ".MySQL_error(); //This will spit out any errors with the sql statement
   exit();
}
while ($row = mysql_fetch_assoc($result)){
   $mytablehtml .= '<tr><td>'.$row['business_name'].'</td><td>'.$row['account_num'].'</td></tr>';
}
?>
<html>
<head></head>
<body>
   <table>
     <thead><tr><th>Business Name</th><th>Account Number</th></tr></thead>
     <tbody><?php echo $mytablehtml;?></tbody>
   </table>
</body>
</html>	

Keep in mind, when using the if (!result) function it will output the error with the sql statement on the page and that is it. This is a great way to debug your sql statements as it will pinpoint what the problem is. Personally I like to remove the mysql_error() part of echo statement in real applications and just echo a text stating what the problem might be as the mysql_error() statement contains information regarding file paths and server info.

 

Hope that helps!

Edited by joallen
Link to comment
https://forums.phpfreaks.com/topic/288695-query-issue-in-php/#findComment-1480529
Share on other sites

Thanks to you both for the time you took to reply and for your suggestions, neither of which worked unfortunately. I am going to go back to the drawing board and rethink my approach to this as I know that there has to be a more simpler way of doing this. gnerjm is right, the query being used (while it works in MySQL) needs to change as I think this is the problem. :happy-04:

Link to comment
https://forums.phpfreaks.com/topic/288695-query-issue-in-php/#findComment-1480608
Share on other sites

Deprecated!  Deprecated.  Say it 3 times fast.

 

In a way MySQL is "depreciated" or ever under-appreciated but the true term for what has happened to the interface is "deprecated'.  To quote that well-known mantra - "there is no I in deprecated".

Link to comment
https://forums.phpfreaks.com/topic/288695-query-issue-in-php/#findComment-1480617
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.