[Resolved] foreach? getting arrays to appear as form menus

[ejwf]

Hello
I want an array to appear in a form as a drop down menu.
I am struggling with the foreach code.

So far I have got

<TD WIDTH="50%" HEIGHT="32"><FONT FACE="Arial, Helvetica, sans-serif">Client:</FONT></TD>
<TD WIDTH="50%"><FONT FACE="Arial, Helvetica, sans-serif"> 
<?php 
echo '<select name="client"> 
<option selected>choose</option> '
foreach($CoName as $value){ 
printf("<option>".$value."</option>"); 
} 
echo  '</select>'?>
</FONT></TD><TR>

but this doesn't work - Parse error: parse error, expecting ','' or';'' in the line foreach($CoName as $value)

By the way, the variable $CoName comes from a query earlier in the code that I have echo'd out to test and it does work......

Firstly is the use of foreach correct in this type of code request?

Secondly what should the correct syntax be, I have looked in the manual and cannot see any difference to what I have in my code.

thanks

[bad76]

Hi,
you are missing the ; after the echo before the foreach...

[ejwf]

thanks bad76,

added the ; so it now reads

echo '<select name="client"> 
<option selected>choose</option> ; '
foreach($CoName as $value){ 
printf("<option>".$value."</option>"); 
} 
echo  '</select>'

but still get the same parse error: parse error, expecting ','' or';'' in the line foreach($CoName as $value)

any ideas?
thanks

[Weedpacket]

Nononono, outside the string, so that PHP will read it!

echo '<select name="client">
<option selected>choose</option>';
foreach($CoName as $value){ 
printf("<option>".$value."</option>"); 
} 
echo  '</select>'

[ejwf]

thanks weedpacket - am I dumb or what !!?!

never mind, this has resolved the parse error, but the code doesn't quite do what I was wanting.
the output in the drop down menu just says 'choose' (the option I set up in the code) it doesn't display the array information from the query.

The query to define the array is as follows;

$query = "SELECT `CompanyRef` FROM `tblUser` WHERE `CLogIn` = '$username'";
$result = mysql_query($query)
	or die(mysql_error());
$row = mysql_fetch_array($result);
$Company = $row["CompanyRef"];

$query = "SELECT `CoName` FROM `tblClient` WHERE `CompanyRef` = '$Company' ";
$result1 = mysql_query($query)
	or die(mysql_error());
$row1 = mysql_fetch_array($result1);
$CoName = $row1["CoName"]; 

($username comes as a variable from the login page)

I know that there are 2 values in the array created by this query - ie 2 CoName's - and I wanted these 2 values (or however many there are in the array) to be shown in the drop down menu, but this isn't happening.

What do I need to do with this code;

<TD WIDTH="50%" HEIGHT="32"><FONT FACE="Arial, Helvetica, sans-serif">Client:</FONT></TD> 
<TD WIDTH="50%"><FONT FACE="Arial, Helvetica, sans-serif"> 
<?php  

echo '<select name="client"> 
<option selected>choose</option> ' ;
foreach($CoName as $value){  

printf("<option>".$value."</option>");  

}  

echo  '</select>'?> 
</FONT></TD><TR> 

to make this happen

thanks

[bad76]

$CoName = $row1["CoName"];

...But this don't do an array. This make a sngle assignement.

May be you means this:

 
$result1 = mysql_query($query)
    or die(mysql_error());
while($row1 =  mysql_fetch_array($result1))
  $CoName[] = $row1["CoName"];

This give you an array named $CoName and containing all the db entry in the resultset.

bye bye

[ejwf]

thank you for making that clear.
However I still need some advise on how to complete the code.
I have sucessfully created the array. I have tested the array by echoing it out and I know there are 2 values that have been recognised in the array.

However I still cannot get these values to show in the menu in the form........
the code I have so far
1) to create the array

$query = "SELECT `CompanyRef` FROM `tblUser` WHERE `CLogIn` = '$username'";
$result = mysql_query($query)
	or die(mysql_error());
$row = mysql_fetch_array($result);
$Company = $row["CompanyRef"];

$query = "SELECT `CoName` FROM `tblClient` WHERE `CompanyRef` = '$Company'";
$result1 = mysql_query($query)
	or die(mysql_error());
while($row1 =  mysql_fetch_array($result1)) 
  $CoName[] = $row1["CoName"];

2) to create the drop down menu in the form

<TR><TD WIDTH="15%">&nbsp;</TD>
<TD WIDTH="25%" HEIGHT="32"><FONT FACE="Arial, Helvetica, sans-serif">Client:</FONT></TD>
<TD WIDTH="35%"><FONT FACE="Arial, Helvetica, sans-serif"> 
<?php 
echo '<select name="client"> 
<option selected>choose</option>  ';
foreach($CoName[] as $value){ 
printf("<option>".$value."</option>"); 
} 
echo  '</select>'?>
<FONT FACE="Arial, Helvetica, sans-serif"> </FONT></FONT></TD>
<TD WIDTH="25%">&nbsp;</TD></TR>

Like I say, this doesn't work, all I get in the drop down menu is 'choose'.

Can anyone see where I am going wrong with this - thanks

[Weedpacket]

The $CoName[] in the foreach loop should be just $CoName (the [] is for putting things into the array).

Incidentally, you don't need two queries to do what you're doing - you should be able to drop the first query (and subsequent code) and change the second query to

SELECT CoName
FROM tblClient, tblUser
WHERE tblUser.CLogin='$username'
      AND tblUser.CompanyRef=tblClient.CompanyRef

[ejwf]

BIG THANK YOU...........

I have got the array working now in the drop down menu

BTW, very nice shortcut on the query code - works a treat.

[Napster]

<?php  

echo "<select name=\"client\"> 
<option selected>choose";' 
foreach($CoName as $value){  

echo "<option>$value";  

}  

echo  "</select>";

?>