syntax-error

Common Lisp lambda expression error

由 社会主义新天地 提交于 2019-12-02 18:13:18
问题 I am trying to make a program, which will rewrite given row without duplicates in an output row. I am using this website as a compiler: https://www.tutorialspoint.com/execute_lisp_online.php And here is my code (SETQ X (LIST 2 -3 (LIST 4 3 0 2) (LIST 4 -4) (LIST 2 (LIST 2 0 2))-3))' (DEFUN SEARCHDEEP (WHAT WHERE) ;Function will find out if atom `WHAT`is in a row `WHERE` => works fine (COND ((NULL WHERE) NIL) (T (OR (COND ((ATOM (CAR WHERE)) (EQUAL WHAT (CAR WHERE))) (T (SEARCHDEEP WHAT (CAR

I keep getting an error next to the return VARCHAR(4)

由 戏子无情 提交于 2019-12-02 16:50:06
问题 ERROR IS BETWEEN THE VARCHAR2 AND THE (4) seems to be a syntax error. create or replace FUNCTION Employee_exists (p_employee_id IN NUMBER) RETURN VARCHAR2(4); AS BEGIN SELECT employee_id FROM employees WHERE employee_id = p_employee_id; RETURN 'true'; END Employee_exists; 回答1: There are actually two errors: You can't specify the length of the return type in a function or procedure specification ; is a statement terminator, as such it has no place between the return type and AS Your first line

SASS error when using % in property value together with for loop variable [duplicate]

由 ↘锁芯ラ 提交于 2019-12-02 15:08:46
问题 This question already has answers here : Adding a unit to a number in Sass (2 answers) Closed 3 years ago . I'm trying to write some SASS (.scss) to generate CSS for a donation barometer (think progress bar). The CSS i need looks something like this: .p-0:after { left: 0%; } .p-1:after { left: 1%; } [... up to 100] The SASS I have is this: @for $i from 0 through 100 { .p-#{$i}:after {left: #{$i}%;} } Which gives me this error: Syntax error: Invalid CSS after "...r {left: #{$i}%": expected

What is reason for syntax error of print statement of given python code [closed]

由 给你一囗甜甜゛ 提交于 2019-12-02 13:43:16
xString = input("Enter a number: ") x = int(xString) yString = input("Enter a second number: ") y = int(yString) print('The sum of ', x, ' and ', y, ' is ', x+y, '.', sep='') on executing above code, interpretor throwing syntax error saying syntax error as below. print(?The sum of ?, x, ? and ?, y, ? is ?, sum, ?.?, sep=??) SyntaxError: invalid syntax The single quote used in the print statement is ' with ascii value 39. >>> ord("'") 39 The one ’ used in the print statement in the question is not a quote ' but RIGHT SINGLE QUOTATION MARK' (U+2019) >>> u"’" u'\u2019' Since, you are using python

what's wrong with this include statement that i did in php? [closed]

由 女生的网名这么多〃 提交于 2019-12-02 13:34:42
<?php include("articles/article_6.html"); ?> i wanted the article to be posted on my website but it gives the error: Parse error: syntax error, unexpected 'version' (T_STRING) in /opt/lampp/htdocs/white_coats/articles/article_6.html on line 1 article_6.html has no php tags. its just a word file that was saved in .html so that it could be displayed on the website. it's just an article with some pictures. Then you don't include it. You read it's contents and print em. Include statements are nearly always for code includes and not plain html or text content. echo file_get_contents("articles

Python: SyntaxError

由 谁说胖子不能爱 提交于 2019-12-02 13:24:53
I'm writing a script to parse an email, but there is some SyntaxError on the for loop in the following part: def main(): writer = csv.DictWriter(open('features.csv', 'w'), list(EXTRACTS.keys())) for mail in os.listdir(MAILDIR): writer.writerow({ key: value(email.message_from_file(open(os.path.join(MAILDIR, mail), 'r'))) for key, value in EXTRACTS.items() }) Please help me out of this! EDIT: File "/IS/extraction.py", line 52 for key, value in EXTRACTS.items() ^ SyntaxError: invalid syntax You are running this on an older Python version that doesn't yet support dict comprehensions. The {key:

Bash if then else syntax [duplicate]

由 橙三吉。 提交于 2019-12-02 12:40:15
This question already has an answer here: Why should there be a space after '[' and before ']' in Bash? 4 answers I'm trying to do test automation with a bash script using if then else statements but I'm running into a few errors. For one, when I try to execute it I'm doing something wrong with the variable assignment with j and k, because it tells me that the command j and the command k aren't found when I try to execute. How do you correctly create variables? The most confusing thing though is when I try to execute the script I get an error telling me I have an unexpected token near fi, and

C# The name ' … ' doesn't exist in the current context

由 末鹿安然 提交于 2019-12-02 12:24:38
问题 I'm new in C# and I try to learn it from the very basics, but I stuck with classes. I made my first example to practice which is working properly but when I add a little complexity I get an error: "The name 'iArcher' doesn't exist in the current context." Please help to explain what's wrong and suggest a proper (and easy) solution. Thanks! using System; namespace Units { class Archer { public int id; public int hp; public float speed; public float attack; public float defence; public float

SYNTAX ERROR - Can't assign to operator in setting variables

由 久未见 提交于 2019-12-02 12:16:39
问题 I have this Syntax Error in IDLE: SyntaxError: can't assign to operator This then highlights the end of a line, line 2 of the following code: date = "Unknown" day-of-week = "Unknown" time = "Unknown" week = "Unknown" I would appreciate any help I can get with this :) 回答1: Python is interpreting day-of-week as "day" minus "of" minus "week". Try using day_of_week instead. Sample code to show this. >>> day = 3 >>> of = 2 >>> week = 4 >>> day-of-week -3 回答2: "Day-of-week" is an invalid variable

OUTPUT Inserted.row in mysqli

由 混江龙づ霸主 提交于 2019-12-02 11:46:11
问题 I have the following sql table: id|email|fbid When I perform the query INSERT INTO users(email,fbid) VALUES('randomvalue','otherrandomvalue') I want to get the id of the inserted row. To do so, I've tried to edit the query like this: INSERT INTO users(email,fbid) VALUES('randomvalue','otherrandomvalue') OUTPUT Inserted.id But I'm getting: 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'OUTPUT
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