Sql not querying 2 variables

[swatisonee]

Hi,

Please have a look at these 2 sql statements. Have commented under the 2nd as I am puzzled why the 2nd one does not run when I choose 2 variables to query.

Can any one shed some light ? THanks !

$result = mysql_query("SELECT * FROM `Visit_Report` WHERE Visitid= $a ");
$myrow = mysql_fetch_array($result);
$customer = $myrow["Company"];
$last = $myrow["Lastname"];

list($customer0, $customer1) = explode(",", $customer);

echo $customer0; 
echo $customer1;

// This works perfectly without errors

$sql1 = "SELECT * FROM `Customers` WHERE `Company` = '$customer0' ";

$result1 = mysql_query($sql1) or die (mysql_error());
$myrow1 = mysql_fetch_array($result1);
$street = $myrow1["Street"];

//This too works without a problem. 

BUT, if I modify this to

$sql1 = "SELECT * FROM `Customers` WHERE `Company` = '$customer0'  && `Lastname` = '$last' ";
echo $sql1;

$result1 = mysql_query($sql1) or die (mysql_error());
$myrow1 = mysql_fetch_array($result1);

$street = $myrow1["Street"];

print_r($myrow1); // returns blank screen 

print_r($myrow1); // returns all values in 1st sql

The query does not run although it echoes out correctly.. Clearly no problem with field names sicne the 1st version goes thru...I use ver 3.23.53 of mysql.

Please advise !

[jbatson]

Instead of

$sql1 = "SELECT * FROM Customers WHERE Company = '$customer0' && Lastname = '$last' ";

Try

$sql1 = "SELECT * FROM Customers WHERE Company = '$customer0' AND Lastname = '$last' ";

[swatisonee]

Nope, still no luck I'm afraid.

Any other ideas please ??

[obrienkev]

How do you mean it doesn't run?

Are there matching values in the database? Did you test for errors?

After $result1 try this...

$num_results = mysql_num_rows($result1) or die(mysql_error());
echo 'RESULTS: '$num_results.;

Let me know how it goes..

[swatisonee]

Nothing. My screen is completely blank when i ran it. Anything else to try ?

For sure there are matching values because the echo $sql1 gives the proper result :

SELECT * FROM `Customers` WHERE `Company` = 'Acfol Inc' AND `Lastname` = 'Joseph ' 

[obrienkev]

does the echoed query display??

If the screen is completely blank then most likely you are missing a semi-colon or "" somewhere.
Post all your code and mite be able to spot it.

[obrienkev]

Should you not have to loop through the query?

while($myrow1 = mysql_fetch_array($result1))
{
$street = $myrow1["Street"];
echo $street.'<br />';
}

[swatisonee]

No luck still I'm afraid. And I need to have the Name getting passed as I have the same company with different addresses for each of the persons in the company....

I reckon it may have some connection to the fact that $customer0 was a part of $customer, before I "exploded " it.

list($customer0, $customer1) = explode(",", $customer); 

echo $customer0; 

Although , when i echo $customer0, it comes up ok as is evident from the echo $sql1.

Anything else i can try pl ?

Thanks.

[swatisonee]

Full Code

<?php

mysql_connect("localhost", $dbname, $dbpasswd )
    	or die ("Unable to connect to server.");

mysql_select_db($database)
    	or die ("Unable to select database.");

$result = mysql_query("SELECT * FROM `Visit_Report` WHERE Visitid= $a "); 
$myrow = mysql_fetch_array($result); 
$customer = $myrow["Company"]; 
$last = $myrow["Lastname"]; 

list($customer0, $customer1) = explode(",", $customer); 

echo $customer0; 
echo $customer1; 

$sql1 = "SELECT * FROM `Customers` WHERE `Company` = '$customer0'  AND `Lastname` = '$last' "; 
echo $sql1; 

$result1 = mysql_query($sql1) or die (mysql_error()); 
$myrow1 = mysql_fetch_array($result1); 

$street = $myrow1["Street"]; 
// THIS WORKS IF I DONT USE THE $last in the above query)

printf("<tr>
<br><font color=orange>Meeting with</font>        : %s  <br>
<font color=orange>Customer</font>                : %s<br>
<font color=orange>Address</font>                 : %s : <br>
<font color=orange>Visit Id</font>                : %s<br>
<font color=orange>Discussion</font>                 : %s : <br>


$myrow1["Lastname]"],
$customer0,               // This works but if I replace it with  $myrow1["Company"] it doesnt work.

$myrow1["Street"],
$myrow["Visitid"],  // this comes from the 1st sql stmt 
$myrow["Disc"] );  // this comes from the 1st sql stmt 

?>

[swatisonee]

Some progress if i may call it that.

I changed

$myrow1 = mysql_fetch_array($result1);   

to

$myrow1 = mysql_fetch_assoc($result1);  

and now the echo says "Query was Empty" .

Can anyone shed light please? I checked the table, and the entire row pertaining to this record is complete and correct so there should have been an output.

Thanks.

$sql1 = "SELECT * FROM `Customers` WHERE `Company` = '".mysql_real_escape_string($customer0)."' AND  `Lastname` = '".mysql_real_escape_string($last)."' ";  

echo $sql1;  

$result1 = mysql_query($sql1) or die (mysql_error());  

$myrow1 = mysql_fetch_array($result1);  // blank screen
$myrow1 = mysql_fetch_assoc($result1);  // query was empty
$myrow1 = mysql_fetch_row($result1);  // blank screen