I'm wanting to print a variable after a string but the variable doesn't show.
I'm following a book called learning python the hard way.
In it, it asks to declare a variable:
cars = 100
Print ("how many cars are there"), (cars)
And print it after a string.
However, it only prints the string. I've tried doing it in different ways with brackets, parentheses, etc. to no avail. Can someone help me out please?
cars = 100
print ("text", (cars) ,"text")
Thanks anyways
Related
I'm trying to work out how to insert a variable that contains one character in the middle of a variable
For example:
$fullnameresult = 'John doe'
$comma = ','
write-output $full$($comma)nameresult
I am hoping for the output to be:
John,Doe
Apart from the obvious inserting comma into the $fullnameresult variable, is there another way of doing it? I know this might sound silly but theres a long story behind it and this is the easiest way I could think of explaining it
I am confused regarding variables.
I have code, where I have the following line:
$search = $Name.SelectedItem.Split('-')[$($Name.SelectedItem.Split('-').Count-1)]+'*'
This line does nothing else, as split up a selected item (I am working with a Dropdownbox) and transfers it to $search.
The funny thing is, it does exactly that, what I want it to do.
When I type $search, the result can be for example:
Rue de Rivoli*
When I continue in the code and use $search through several arrays, for some reason, it does not function, because it does not find anything in a foreach loop. I have no error message and even the Rue de Rivoli* existing in one of the arrays, it does not find anything.
When I replace the above code and give directly the name to the variable $search, as seen below
$search = 'Rue de Rivoli*'
my search in the array works.
What am I missing here? I am doing something wrong, but I do not know what it is, can someone help me please to understand?
Thank you very much,
Mike
As requested, here more of the code. It is a lot to deal with, that is why I shorten it.
Clear-Host
$search = $CreateNewUserFormDropDownBoxLocation.SelectedItem.Split('-')[$($CreateNewUserFormDropDownBoxLocation.SelectedItem.Split('-').Count-1)]+'*'
#$search = 'Rue de Rivoli*'
$AllLocations = (get-variable -Include USPennsylvaniaAve, USSixthStreet, USRodeoDrive, USOneMicrosoftWay,`
USNorthTantauAvenue, USMarketStreet, USMainStreet, USEmilyDrive,`
USCalle8, USBroadway, US18thStreetNW, UKOxfordStreet, UKDowningStreet,`
UKBondStreet, FRRuedeRivoli, FRChampsElysees, CHBahnhofstrasse,`
CA17thAvenue) | ? {$_.value -is [array]}
Foreach ($Array in $AllLocations)
{
if ($array.value -like $search)
{break}
}
$result = "`$$($array.name)"
$result
This is about to become a function and does nothing else, as from the selecteditem, it takes it apart and add's the * behind it, so I can search
for a name with a wildcard.
I have several arrays and therefore I included only the necessary ones. Next step is to loop through the arrays and as soon as it found the item, it stops and gives the result to result.
This is my test code and it runs and does what I want, besides the line after Clear-Host. The code is correctly resolved and added to $search but does not work.
Below that code of line, I have my cheat line, where I add directly the correct result to the variable and it works fine.
As commented, this should solve the problem.
(I'm adding this as answer too, so the OP can accept it. Otherwise this question will remain seemingly unsolved)
When hardcoding the search string $search = 'Rue de Rivoli*' works, but using a Split() to get the search string does not, then usually the string you obtain using the split is surrounded by whitespace characters. If you leave these in, the string will appear to look just fine, but when using as comparison it won't work.
If for instance the complete $CreateNewUserFormDropDownBoxLocation.SelectedItem string is:
"François Exemple - Rue de Rivoli"
Then, using $CreateNewUserFormDropDownBoxLocation.SelectedItem.Split("-")[-1] will return:
" Rue de Rivoli"
Note the space in front.
By simply performing a Trim() you will get rid of that space.
The line therefore should be:
$search = ($CreateNewUserFormDropDownBoxLocation.SelectedItem.Split('-')[-1]).Trim() + '*'
I am still learning perl and have all most got a program written. My question, as simple as it may be, is if I want to hardcode a string to a field would the below do that? Thank you :).
$out[45]="VUS";
In the other lines I use the below to define the values that are passed into the `$[out], but the one in question is hardcoded and the others come from a split.
my #vals = split/\t/; # this splits the line at tabs
my #mutations=split/,/,$vals[9]; # splits on comma to create an array of mutations
my ($gene,$transcript,$exon,$coding,$aa);
for (#mutations)
{
($gene,$transcript,$exon,$coding,$aa) = split/\:/; # this takes col AB and splits it at colons
grep {$transcript eq $_} keys %nms or next;
}
my #out=($.,#colsleft,$_,#colsright);
$out[2]=$gene;
$out[3]=$nms{$transcript};
$out[4]=$transcript;
$out[15]=$coding;
$out[17]=$aa;
Your line of code: $out[45]="VUS"; is correct in that it is defining that 46th element of the array #out to the string, "VUS". I am trying to understand from your code, however why you would want to do that? Usually, it is better practice to not hardcode if at all possible. You want to make it your goal to make your program as dynamic as possible.
I am a newbie to programming and I hope someone can explain this to me:
So I have two text files i.e. Scan1.txt and Scan2.txt that are stored in my computer. Scan1.txt contains:
Tom
white
black
mark
john
ben
Scan2.txt contains:
bob
ben
white
gary
tom
black
patrick
I have to extract the matched values of these two files and the unmatched values and print them separately. I somehow found the solution for this which works fine. But can someone please explain how exactly the match happens here. Looks like somehow just this line:
$hash{$matchline}++ in the code does the matching and increments the value of hash when the match is found. I understand the logic but I do not understand how this match actually happens. Can someone help me understand this?
Thank you in advance!
Here is the code:
open (F1, "Scan1.txt");
open (F2, "Scan2.txt");
%hash=();
while ($matchline= <F1> ){
$hash{$matchline}=1;
}
close F1;
while( $matchline= <F2> ){
$hash{$matchline}++;
}
close F2;
foreach $matchline (keys %hash){
if ($hash{$matchline} == 1){
chomp($matchline);
push(#unmatched, $matchline);
}
else{
chomp($matchline);
push (#matched, $matchline);
}
}
print "Matched Entries are >>\n";
print "```````````````````````\n";
print join ("\n", #matched) . "\n";
print "```````````````````````\n";
print "Unmatched Entries are >>\n";
print "```````````````````````\n";
print join ("\n", #unmatched) . "\n";
print "```````````````````````\n";
The code you mention above will give you a false result if a given word exists more than one time in the second file and not exists in the first.
this line:
$hash{$matchline}++
increments a different counter for each different word.
in the first loop it sets to 1 for the words in the first file.
so if a word exists in each file the counter will be at least 2.
the $hash itself is a set of counters.
A more generalized version of your problem is that of computing the set union or intersection between two sets. This link gives a very good treatment of the problem in general.
In your case, the set is nothing but the list of values from each file. The logic is, if a certain value was present in both files then $hash{matchline} == 2, because the value will be incremented in both the while loops. However, if the line was present in only one of the files, the value of $hash{matchline} == 1, since only one while loop will increment the value and not the other.
Also, Lajos Veres raises a very important point: if a certain word, say "Tom" is present twice in the same file, then the algorithm will fail. It is a subtle detail, which can be resolved in many ways- removing duplicates beforehand, using two hashes, etc.
Hope this helps.
I want to define a variable /Test123 in perl .I am trying something like this but i am getting error
my $uriReference="Test123";
##Here is where i am trying to create /Test123
my $routingUrl = "\"/".$uriReference."\"";
I think i am doing some mistake here Need help?
You can use qq() instead to emulate double quote interpolation:
my $url = qq("/$uriReference");
Although I do not see anything technically wrong with your assignment, it should assign "/Test123" to your variable. So perhaps the error is about something else. We would know if you had included your error message in your question. There is no reason you should ever not include error messages in questions.
You should also know that you do not need to break up your quotes, you can do this:
my $url = "\"/$uriReference\"";