Stumped on how to display the data from 2nd table

badgoat

Hiya,

In the code below, I am displaying the variable '$diwtitle', '$dropdownacctmgr', and '$joined'. which are being entered on this page. I cannot figure out how to display the variables '$acctmgr_phone', '$acctmgr_cellphone', and '$acctmgr_email', which are saved in a table named 'acctmgr'.

I am new, kindly asking for guidance. 🙂

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<html>
<head>
<title>Untitled</title>
</head>
<body>
<?php

$connect= mysql_connect("localhost","root","***") 
   or die("Could not connect to database in localhost !");
$result=mysql_select_db("testdiw") 
   or die("Could not select that database !");


 echo "SELECT * from testdiw";

$joined = date("Y-m-d h:i:s");
$sqlquery = "INSERT INTO diw VALUES('','". $diwtitle ."','". $dropdownacctmgr ."','". $joined ."')";

$sqlquery = "SELECT * from diw";
$queryresult = mysql_query($sqlquery) or die(" Could not execute mysql query !");

echo '<table border=1 align=center  width=500>
<TR>
		<TD colspan=2><b>General Deposition
Information></b></TD>
	</TR>
	<TR>
		<TD><b>DIW Title:</b></TD>
<TD>' .$diwtitle. '</TD>
	</TR>
		<TR>';
//		$sqlquery1 = "SELECT * from acctmgr, diw WHERE acctmgr_name = $dropdownacctmgr";
//		$queryresult1 = mysql_db_query($sqlquery1);

//$result = mysql_query( "SELECT * FROM acctmgr" );
echo '		<TD>Account Manager Information:</TD> <TD>' .$dropdownacctmgr. '<BR>' .$acctmgr_phone. '<BR>' .$acctmgr_cellphone. '<BR>' .$acctmgr_email. '</TD>
	</TR>
</TABLE>';

 ?>


</body>
</html>

lazzerous

Nowhere in your script are $acctmgr_phone, $acctmgr_cellphone or $acctmgr_email defined

Are they coming from the select statement?

That stetement is pulling all records from 'diw'

To loop through all the results, and output the selected data for each entry in the database, you would need to do this:

while ($row=mysql_fetch_array($queryresult){
    echo $row['acctmgr_phone'] . "<br>\n";
    echo $row['acctmgr_cellphone'] . "<br>\n";
    echo $row['acctmgr_email'] . "<br>\n";
    }

badgoat

Hi Lazzerous,

Thank you for the reply! The variables $acctmgr_phone, $acctmgr_cellphone or $acctmgr_email are stored in the acctmgr table. These variables were saved to the acctmgr table in a previous screen. I don't want them inserted into the diw table, I just want them echoed for display. How may I accomplish this?

badgoat

From my naive understanding of what needs to happen, I 'think' I need to put in a query like:

SELECT * FROM acctmgr WHERE acctmgr_name = $dropdownacctmgr

But I am not sure, and if that is the answer, I am unsure where to place it. 🙁

lazzerous

Then you would need somethign like this:

$get_acctmgr_info = mysql_query("SELECT acctmgr_phone, acctmgr_cellphone, acctmgr_email FROM acctmgr WHERE acctmgr_name = '$dropdownacctmgr'");
$row=mysql_fetch_array($get_acctmgr_info);

echo $acctmgr_phone . "<BR>\n";
echo $acctmgr_cellphone . "<BR>\n";
echo $acctmgr_email . "<BR>\n";

badgoat

Excellent! Does placement matter? Like, must I place this code at the top along with the rest of the mysql queries, or does it reside right before I want to echo these variables?

badgoat

I am getting an error message, saying that the variables from the acctmgr table are undefined. Am I missing something at the beginning?

lazzerous

ok something you HAVE to know right now

1: Read the F.A.Q. for these forums.. please

2nd - when you receive an error, ALWAYS include all related code AND the actual error message - we can't do much for you otherwise.

Would you go to the body shop for a quote without bringing the car?

badgoat

Here is the error message:

SELECT * from testdiw
Notice: Undefined variable: acctmgr_phone in c:\program files\easyphp1-7\www\diw4.php on line 40

Notice: Undefined variable: acctmgr_cellphone in c:\program files\easyphp1-7\www\diw4.php on line 40

Notice: Undefined variable: acctmgr_email in c:\program files\easyphp1-7\www\diw4.php on line 40

Here is the code as it stand with your help:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<html>
<head>
<title>Untitled</title>
</head>
<body>
<?php

$connect= mysql_connect("localhost","root","****") 
   or die("Could not connect to database in localhost !");
$result=mysql_select_db("testdiw") 
   or die("Could not select that database !");


 echo "SELECT * from testdiw";

$joined = date("Y-m-d h:i:s");
$sqlquery = "INSERT INTO diw VALUES('','". $diwtitle ."','". $dropdownacctmgr ."','". $joined ."')";

$sqlquery = "SELECT * from diw";
$queryresult = mysql_query($sqlquery) or die(" Could not execute mysql query !");


$get_acctmgr_info = mysql_query("SELECT acctmgr_phone, acctmgr_cellphone, acctmgr_email FROM acctmgr WHERE acctmgr_name = '$dropdownacctmgr'");
$row=mysql_fetch_array($get_acctmgr_info); 


echo '<table border=1 align=center  width=500>
<TR>
		<TD colspan=2><b>General Deposition
Information></b></TD>
	</TR>
	<TR>
		<TD><b>DIW Title:</b></TD>
<TD>' .$diwtitle. '</TD>
	</TR>
		<TR>';

echo '		<TD>Account Manager Information:</TD> <TD>' .$dropdownacctmgr. '<BR>' .$acctmgr_phone. '<BR>' .$acctmgr_cellphone. '<BR>' .$acctmgr_email. '</TD>
	</TR>
</TABLE>';

 ?>


</body>
</html>