mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in ......HELP!

The problem is that mysqli_query() returns a boolean instead of a result. This will probably be boolean false, because a succesful SELECT query would return a result. All this means that your query cannot be run. It might have a wrong syntax or the database might have changed making it fail somehow. Try calling mysqli_error() after mysqli_query to see what went wrong.

Is there a reason that a query will not run on the .php page but will run successfully with the MySQL Workbench or MS Access? Does my SELECT Query look right to you?

Hmmm... The query looks alright to me, and if it works in the MySQL Workbench and/or MS Access it should be fine. The problem could be in $dbc not containing the proper MySQL connection, have you checked that?

Ok I recoded the page and put the database connection syntax on the same code, but just as before I get the same result with the query not displaying inner joins. It works fine with the just the Select statement, From and Where clause but INNER JOINS will still not work. Has anybody ever heard of this?

<?php

$db_host = 'localhost';
$db_user = 'cge';
$db_pwd = '04252';

$database = 'cookk';
$table = 'customers';

if (!mysql_connect($db_host, $db_user, $db_pwd))
    die("Can't connect to database");

if (!mysql_select_db($database))
    die("Can't select database");

	


// THIS QUERY WORKS PERFRECT IN MS ACCESS BUT NOT IN MySQL?!?!? 

$result = mysql_query("SELECT CustomerID, OldCustomerID, CONCAT_WS(', ', LastName, FirstName, MiddleName) AS 'Customer Name',CONCAT ((LEFT(FirstName, 1)), (LEFT(MiddleName, 1)), LastName) AS Username, zlu_Cars.Description AS Car, zlu_CarColor.Description AS 'Car Color', zlu_Computers.Description AS Computer, IsLaptop, zlu_Race.Description AS Race, zlu_Residence.Description AS Residence, zlu_BirthMonth.Description
FROM ((((((customers 
JOIN zlu_CarColor ON Customers.CarColorID = zlu_CarColor.CarColorID) JOIN zlu_Cars ON Customers.CarID = zlu_Cars.CarID) JOIN zlu_Computers ON Customers.ComputerID = zlu_Computers.ComputerID) JOIN zlu_Race ON Customers.RaceID = zlu_Race.RaceID) JOIN zlu_Residence ON Customers.ResidenceID = zlu_Residence.ResidenceID) JOIN zlu_BirthMonth ON Customers.BirthMonthID = zlu_BirthMonth.BirthMonthID)
WHERE Customers.CustomerID BETWEEN 500 and 600
ORDER BY LastName"); 




if (!$result) {
    die("Query to show fields from table failed");
}

$fields_num = mysql_num_fields($result);



echo "<table align='center' cellspacing='3' cellpadding='3' width='75%' border=1><tr>";

for($i=0; $i<$fields_num; $i++)
{
    $field = mysql_fetch_field($result);
    echo "<td>{$field->name}</td>";
}
echo "</tr>\n";

while($row = mysql_fetch_row($result))
{
    echo "<tr>";


    foreach($row as $cell)
        echo "<td>$cell</td>";

    echo "</tr>\n";
}
mysql_free_result($result)

?>

Select allOpen in new window

It's strange indeed. Have you tried working in the mysql_error() function? If so, what does it say after the SELECT with the JOINS fails?

No luck so far and mysqli_error() function isn't isolating the issue....I'll see what tomorrow brings.

@Ray

 Thanks for the code this is what displayed after it ran:

SELECT CustomerID, OldCustomerID, CONCAT_WS(', ', LastName, FirstName, MiddleName) AS 'Customer Name',CONCAT ((LEFT(FirstName, 1)), (LEFT(MiddleName, 1)), LastName) AS Username, zlu_Cars.Description AS Car, zlu_CarColor.Description AS 'Car Color', zlu_Computers.Description AS Computer, IsLaptop, zlu_Race.Description AS Race, zlu_Residence.Description AS Residence, zlu_BirthMonth.Description FROM ((((((customers JOIN zlu_CarColor ON Customers.CarColorID = zlu_CarColor.CarColorID) JOIN zlu_Cars ON Customers.CarID = zlu_Cars.CarID) JOIN zlu_Computers ON Customers.ComputerID = zlu_Computers.ComputerID) JOIN zlu_Race ON Customers.RaceID = zlu_Race.RaceID) JOIN zlu_Residence ON Customers.ResidenceID = zlu_Residence.ResidenceID) JOIN zlu_BirthMonth ON Customers.BirthMonthID = zlu_BirthMonth.BirthMonthID) WHERE Customers.CustomerID BETWEEN 500 and 600 ORDER BY LastName

1054Unknown column 'zlu_Cars.CarID' in 'on clause'

Warning: mysql_num_fields(): supplied argument is not a valid MySQL result resource in .......\view_customers.php on line 105

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in .....\view_customers.php on line 118

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in.........\view_customers.php on line 128

Are you sure that the table zlu_Cars has a column called CarID (case sensivite, so carID or CarId and so on won't work) in the database cookk?

@WabbitSeason you were correct it was the spelling I didn't make the CarID and ComputerID plural. I'm still getting the errors Warning: mysql_num_fields(): supplied argument is not a valid MySQL result resource in .......\view_customers.php on line 105

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in .....\view_customers.php on line 118

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in.........\view_customers.php on line 128

 though but I think im getting closer . Here is the code as it stands now...

<?php
//.....
$database = 'cookek';
$table = 'customers';

if (!mysql_connect($db_host, $db_user, $db_pwd))
    die("Can't connect to database");

if (!mysql_select_db($database))
    die("Can't select database");


$result = "SELECT CustomerID, OldCustomerID, CONCAT_WS(', ', LastName, FirstName, MiddleName) AS 'Customer Name',CONCAT ((LEFT(FirstName, 1)), (LEFT(MiddleName, 1)), LastName) AS Username, zlu_Cars.Description AS Car, zlu_CarColor.Description AS 'Car Color', zlu_Computers.Description AS Computer, IsLaptop, zlu_Race.Description AS Race, zlu_Residence.Description AS Residence, zlu_BirthMonth.Description
FROM ((((((customers 
JOIN zlu_CarColor ON Customers.CarColorID = zlu_CarColor.CarColorID) JOIN zlu_Cars ON Customers.CarID = zlu_Cars.CarsID) JOIN zlu_Computers ON Customers.ComputerID = zlu_Computers.ComputersID) JOIN zlu_Race ON Customers.RaceID = zlu_Race.RaceID) JOIN zlu_Residence ON Customers.ResidenceID = zlu_Residence.ResidenceID) JOIN zlu_BirthMonth ON Customers.BirthMonthID = zlu_BirthMonth.BirthMonthID)
WHERE Customers.CustomerID BETWEEN 500 and 600
ORDER BY LastName"; 


/*echo PHP_EOL . $result;
$res = mysql_query($result);
if (!$res)
{
    echo mysql_errno();
    echo mysql_error();
}
*/
//

if (!$result) {
    die("Query to show fields from table failed");
}

$fields_num = mysql_num_fields($result);



echo "<table align='center' cellspacing='3' cellpadding='3' width='75%' border=1><tr>";

for($i=0; $i<$fields_num; $i++)
{
    $field = mysql_fetch_field($result);
    echo "<td>{$field->name}</td>";
}
echo "</tr>\n";

while($row = mysql_fetch_row($result))
{
    echo "<tr>";


    foreach($row as $cell)
        echo "<td>$cell</td>";

    echo "</tr>\n";
}
mysql_free_result($result)

?>

Select allOpen in new window

You're mixing up $res and $result. Try passing $res instead of $result to mysql_num_fields and so on. And please, do yourself and other coders a favour by renaming $result to $sql, as the name is very misleading in its current state :-).

Check line 30 - looks like $res and $result are confused.

A good practice for this kind of thing is to choose these names and not deviate from them:

$sql = SQL query string
$res = query results resource handle
$row = an array from the results set
$num = the number of rows in the results set

You can make variants on those, of course, but it's best to keep it simple so you don't confuse the variable names.

Great find.

@WabbitSeason

 Thanks for your help as well I wish I could award 2 Accepted Solutions you really hung in there with me, thanks again!

No problem, though AFAIK you can award an "Assisted Solution" to somebody who has been helpful, too.

You can accept two solutions and split points:
https://www.experts-exchange.com/help.jsp#hs=8&hi=100

I think you can also ask a moderator to re-open the question if you want to make a change.  I would have no objection at all, since I think you got help from both of us on this Q.

Best to all, ~Ray

Will do Ray thanks...

Waiting on Mods to hit be back about the reopen request...