Vista - Why the Value not able to Pass to a Function

Hi,

Here I have a code that do not pass the value to the function, which is
below

$tempString='http://xxx.com'

$sqlString="select id from urlScreenLog where screennedURL='$tempString'"
$ScreenlogID = QueryScreenLog $sqlString

Function QueryScreenLog([string]$queyString)
{
$mycommand.CommandText = $queyString
$myreader = $mycommand.ExecuteReader()
}

in the function, $queyString is always "select id from urlScreenLog where
screennedURL=''", not like what I want

"select id from urlScreenLog where screennedURL='http://xxx.com'"


Any help? I appreciated it!

Peter

Try an 'escape character' (small tick below the top left 'esc' key on the
keyboard) before the single quotes eg

$sqlString = "select id from urlScreenLog where
screennedURL=`'$tempString`'"

--
Jon


"Peter Sun" <koarla@xxxxxx> wrote in message
news:%23tf9aE1OIHA.4440@xxxxxx

Looking at it again, I think it's more of an order issue....

Define your function before you call it - that's one of the idiosyncracies
of PowerShell

ie


Function QueryScreenLog([string]$queyString)
{
$mycommand.CommandText = $queyString
$myreader = $mycommand.ExecuteReader()
}

$sqlString="select id from urlScreenLog where screennedURL='$tempString'"
$ScreenlogID = QueryScreenLog $sqlString



--
Jon


"Jon" <Email_Address@xxxxxx> wrote in message
news:e$Hc$K1OIHA.5264@xxxxxx

Thank you fro you reply, Jon,

I looked back to the code a again, what I found the value is in the
$tempString,
but what refer to the $sqlString, then I found

$sqlString="select id from urlScreenLog where screennedURL=''".

I already put the function in front making call.

let me do more test and try different way.


Peter


"Jon" <Email_Address@xxxxxx> wrote in message
news:e64eiS1OIHA.5980@xxxxxx

Yeah give it another go. Post back if you can't get it resolved. There are
many helpful people in this group, who I'm sure will be able to get it
running smoothly, if not.

--
Jon


"Peter Sun" <koarla@xxxxxx> wrote in message
news:OOSFPi1OIHA.4688@xxxxxx

Hi, Jon,

Thank you so much, and I know here always have a lot of kinds of people and
helpful.

I figured the problem out, it come from the statement of calling function, I
was using

$ScreenlogID = QueryScreenLog $a,$sqlString, $b

So, powershell just pass the first one varible $a.

Thanks again.

Peter




"Jon" <Email_Address@xxxxxx> wrote in message
news:OkqFrs1OIHA.5360@xxxxxx

I should use
$ScreenlogID = QueryScreenLog $a $sqlString $b
No Comma


"Peter Sun" <koarla@xxxxxx> wrote in message
news:O1c%23qp2OIHA.5524@xxxxxx

Hi Peter

You can safely expand '$tempString' as long as it's embeded in a double quoted
string.
So this is acceptable:

PS> $tempString='http://xxx.com'
PS> $sqlString="select id from urlScreenLog where screennedURL='$tempString'"

PS> $sqlString
PS> select id from urlScreenLog where screennedURL='http://xxx.com'

As you can see $sqlString contains the expanded value of $tempString.
But you can't do it the opposite way, single quoted strings disables variable
expansion:

PS> $tempString='http://xxx.com'
PS> $sqlString='select id from urlScreenLog where screennedURL="$tempString"'

PS> $sqlString
select id from urlScreenLog where screennedURL="$tempString"


As Jon said, you should declare your functions prior to calling them.
Another issue, generally (depend on what your goal is), when passing arguments
to a function you should delimit each argument
with a space (positional parameter syntax), as in:

$ScreenlogID = QueryScreenLog $a $sqlString $b

To avoid arguments mismatch you can write the full syntax to your function:

$ScreenlogID = QueryScreenLog -queyString $sqlString

Which should be -queryString ;-) (typo)


Lastly, check out PowerShell help:

help about_parameter



Hope This Helps


-----
Shay Levi
$cript Fanatic
http://scriptolog.blogspot.com
Hebrew weblog: http://blogs.microsoft.co.il/blogs/scriptfanatic

--
Jon


"Peter Sun" <koarla@xxxxxx> wrote in message
news:%23OOy602OIHA.3556@xxxxxx

Sorry pressed the wrong button there.


Yes, that's correct separate them by spaces, or use the naming technique, as
Shay mentions below eg


#------------------------------------
function WhatisA {

param([string]$a)


"A is $a"

}

WhatisA -a "apple"

#------------------------------------

--
Jon