Here is my code there is plenty more but these work independently
echo $row['Queue'];
echo "Modify";
what I want to do is this
echo "$row['Queue']";
this doesn't work and gives a syntax error. any tips on how to get this to work?
thanks!
This would be:
echo "" . $row['Queue'] . "";
echo "".$row['Queue']."";
Related
I know similar questions have already been asked, but somehow I am unable to figure out the mistake in my code.
I'm making a .bat file with the following code
echo off
echo %cd%
set curr_directory = "%cd%"
echo $curr_directory
pause
OUTPUT is :
C:\Users\MyDesktop>echo off
C:\Users\MyDesktop>
$curr_directory
Press any key to continue . . .
So what I dont get is why the value of variable curr_directory is not being printed.
What i eventually want to do is use the variable to change the directory something like this: cd $curr_directory
Thanks
I don't know where to start. EVERYTHING about your code is wrong. This should work:
#echo off
echo %cd%
set curr_directory=%cd%
echo %curr_directory%
pause
In batch you access variables via %var% and not $var. Further, DO NOT PUT SPACES behind =. SET x=123 will store 123 in x but SET x= 123 will store _123 (_ means space) in x.
EDIT: As SomethingDark stated, the first line should be #echo off except you actually want the message echo off to be printed. And yes, SET x = 123 means %x % is 123
use %curr_directory% instead of $curr_directory
Avoid spaces inbetween like the below one
"set curr_directory = %cd%"
below should work
echo off
echo %cd%
set curr_directory=%cd%
echo curr_directory is %curr_directory%
pause
I have written a simple sh file to retrieve data from Oracle SQL but getting error. Following is my code:
. $HOME/.profile
function assignVariables
{
ID="finapp"
PASS="finapp"
MAIL_BODY_PATH="/rbluat/BACKEND/Finacle/FC10.2.9/app/CDCI_LOGS/"
}
echo $ID
echo $PASS
function getDatatrans
{
TRANID=`sqlplus -s $ID/$PASS#rbluat <<EOF
SELECT DISTINCT TRAN_ID,DTH_INIT_SOL_ID,TRAN_DATE,DEL_FLG FROM TBAADM.DTD WHERE PSTD_FLG='N' AND ENTRY_USER_ID='FIVUSR' and del_flg='N' and tran_date=(select db_stat_date from tbaadm.gct)AND REF_NUM IN (SELECT PYMT_REF_NUM FROM TBAADM.PORD WHERE STATUS IN ('A','H'));
exit;
EOF`
}
assignVariables
getDatatrans
echo $TRANID
I am getting output as :
[YOU HAVE NEW MAIL]
SELECT DISTINCT TRAN_ID,DTH_INIT_SOL_ID,TRAN_DATE,DEL_FLG FROM TBAADM.DTD WHERE PSTD_FLG='N' AND ENTRY_USER_ID='FIVUSR' and del_flg='N' and tran_date=(select db_stat_date from tbaadm.gct)AND REF_NUM IN (SELECT PYMT_REF_NUM FROM TBAADM.PORD WHERE STATUS IN ('A','H')) few.sh test.sh ERROR at line 1: ORA-00942: table or view does not exist
`
Here few.sh and test.sh are the file names present in the current working directory. few.sh is the file where I have written this code. I have no idea how these names are coming. I am working in KSH. I tried googling about it but found no clue.
The output of sqlplus in evaluated in the command
echo $TRANID
When TRANID has a * in it, ksh will show the files it can find.
You should use quotes to avoid evaluation:
echo "$TRANID"
When you are editing you code, you might as well add {} (not needed here, good habit):
echo "${TRANID}"
Likewise:
echo "${ID}"
echo "${PASS}"
...
-s "${ID}/${PASS}#rbluat"
I have a problem with my script.
I want to create a variable from 2 variables and then echo the result
$a = "server1"
$server_server1 = "Success"
echo ("`$server_$a")
The result of the echo should by Success
Thx in advance
You can use the Get-Variable Cmdlet:
(get-variable "server_$a").Value
Which from your example, will give you Success.
See TechNet for more information on Get-Variable
short way to accomplish same result:
iex "`$server_$a"
So I just started learning UNIX yesterday, and I'm trying to create a basic script that asks for your contact details (name, address, phone number), and then stores that into a file called details.out.
This is driving me NUTS! Its such an easy/basic thing, yet I cant do it, and I've been stuck on it for a solid hour now...
after much googling and searching, I still can't find the answer. So this is what I've done so far, and was wondering where I am going wrong!
echo Please type your first and last name
read $firstname $lastname
echo Please type in your address
read $address
echo Please type in your phone number
read $phone
echo Thank you very much!
echo The details have been stored in '"details.out"'
cat >> details.out <<EOF
Name: echo $firstname echo $lastname
Address: echo $address
Phone Number: echo $phone
EOF
When I read "details.out" it it displays as follows:
Name: echo
Address: echo
Phone Number: echo
ANY help would be appreciated! (and if you get try and point me in the right directions as opposed to straight up giving me the answer, I would appreciate that!)
P.S I'm using Putty if that helps!
when you use read (or declaring variables), don't put $ sigil on the variable names
when you display a variable, always put double quotes around : ex. echo "$var"
when you use here-doc, no need to put echo command
when you use echo, use quotes :
"Double quote" every expansion, and anything that could contain a special character, eg. "$var", "$#", "${array[#]}", "$(command)". Use 'single quotes' to make something literal, eg. 'Costs $5 USD'. See http://mywiki.wooledge.org/Quotes http://mywiki.wooledge.org/Arguments and http://wiki.bash-hackers.org/syntax/words
Whenever you put a $ before a variable name, you're retrieving the current value of that variable. You don't want to do that in your read command. The variables are empty when the script starts, the empty values are put in place of the $firstname and $lastname and read is called with no arguments, causing it to read a line and discard it.
Setting a variable with assignment:
var=value
Setinng a variable with read:
read var
Neither of them use $var because they don't want to look at the current value, they want to replace it.
There's no need for those echos in the heredoc either. They aren't in command position, so they'll just get copied as part of the input to cat.
i am trying to render the view from module to project base view but it gives error.
I tried below combinations without any luck. It gives the error "DefaultController cannot find the requested view "appsMenu"."
echo $this->renderPartial("appsMenu",array("moduleName"=>""),true, true);
echo $this->renderPartial("//appsMenu",array("moduleName"=>""));
echo $this->renderPartial("views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("protected/views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("/protected/views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("views/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("/views/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("site/views/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("site/views/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("protected/views/site/appsMenu",array("moduleName"=>""));
echo $this->renderPartial("//protected/views/site/appsMenu",array("moduleName"=>""));
and tried with extensions too
echo $this->renderPartial("appsMenu.php",array("moduleName"=>""),true, true);
echo $this->renderPartial("//appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("protected/views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("/protected/views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("views/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("/views/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("site/views/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("site/views/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("protected/views/site/appsMenu.php",array("moduleName"=>""));
echo $this->renderPartial("//protected/views/site/appsMenu.php",array("moduleName"=>""));
I am in "Forms" module and trying to render a file "protected/views/site/appsMenu.php". Plz help me..
Use //:
$this->renderPartial("//site/appsMenu");
This can be seen in the documentation
absolute view within the application: the view name starts with double slashes '//'. In this case, the view will be searched for under the application's view path. This syntax has been available since version 1.1.3.
This nasty little bit did the trick for me
$this->renderPartial('//../modules/MyMod/views/MyCon/MyView');
Using // to alias to $root/protected/views and then putting that ../ bit in there to get me to $root/protected/views/../modules/$m/views/$c/$v which really means $root/protected/modules/$m/views/$c/$v
Of course, put reasonable values in for the $X and/or MyXXX values above.
require_once('./protected/modules/MyMod/views/MyCon/MyView.php');
Use above line if you are not able to render file using yii