Your SQL CTE syntax is incorrect: you have a comma after the last CTE’s close-paren. The only time you need a comma there is if you have another CTE following it. Try changing to with unit as ( select t.RefDate, … ), — yes comma here, between CTEs agg as ( select u.RefDate, … ) … Read more
You query has syntax error, for each “or”, you need to put city like instead of or. In addition, remember that in most of the dbms, it is case sensitive. If you want to ignore both upper case and lower case, do it like this. I am using mysql syntax, different dbms has different functions … Read more
The difference is that functionOne is a function expression and so only defined when that line is reached, whereas functionTwo is a function declaration and is defined as soon as its surrounding function or script is executed (due to hoisting). For example, a function expression: // TypeError: functionOne is not a function functionOne(); var functionOne … Read more
With the commented-out code uncommented, you’re trying to declare a constructor directly within a method. You can’t do that in Java. // You can’t do this buttonPlay.addListener(new ClickListener(){ // 1 public void clicked(InputEvent event, float x, float y) { // 2 public GameScreen(Create create) { // 3 this.create = create; // 3 } // 3 … Read more
Try adding white spaces between your query words and make sure you escape the input: $id = mysql_real_escape_string($id); $str = mysql_real_escape_string($str); $name = mysql_real_escape_string($name); $r=mysql_query(“INSERT INTO varta (id,data,name) VALUES (‘$id’,’$str’,’$name’);”); Or better yet – take a look at MySQLi or PDO and use prepared statements. solved Is my SQL syntax really wrong? [closed]
You have issued an if statement. So Naturally, PHP looks to complete the statement. Hence why your getting the error stating an expected closing bracket. Remove your if statement and your code should look like this: if(OS == “XP”){ header(‘Location: XP.html’); exit(); } PHP error reporting isn’t the best tool in the world. It alerts … Read more
Two things, just at a glance. As Rakesh’s answer states, platform00.upper() = input(“etc”) is not valid syntax, and should replaced with platform00 = input(“etc”).upper(). The way you have it, you’re attempting to set a value to a non-variable, specifically the return value of a function. For a second thing, you have a curly brace in … Read more
Your quotation marks around the command string are wrong. 3 solved Getting a Parse error: syntax error, unexpected ‘;’ [closed]
Oh my god, thanks @Orel Eraki, of course beside the unbalanced single-quotes you will have to follow proper SQL Insert Syntax and hava a form of ‘INSERT INTO …’, try it like this (see my change dirctly after “values(” as the one of Orel Eraki (no ‘table’ “keyword” after into): Sql_insert=”insert into itemmanag(itemID,purchesPrice,sellPrice,quantity,vendor,unitM )values(‘”+txt_itemID.getText()+”‘,'”+Item_Pprice.getText()+”‘,'”+txt_itemSprice.getText()+”‘,'”+txt_qunti.getText()+”‘,'” +jcombo1+”‘,'”+jcombo2+ … Read more
A C struct can be packed into flat binary data, which is what Python 2 calls a string. The struct module lets you take a string that represents one of these C structs, and “unpack” it into a Python data structure. To do so you call struct.unpack. You need to specify a format string (as … Read more
First of all, paste your code correctly… You are missing </table> and </form> in the end… Second of all, you were missing a simple apostrophe and a semi-colon in the line $result = mysquli_query($con, ‘select * from account where username=”‘.$username.'” and password=”‘.$password.'”) Also, you should use mysqli_query instead of mysqlui_query… Typo! Your code should be … Read more
loop through and create for item in a: item = [] or you can use list comprehension a = [[] for item in a] if your intended output is [[i], [j], [k]] then use a = [[item] for item in a] solved Is there a way to declare all the values in the list as … Read more
you should close the brackets on the second input n = int(input()) s = 1000000 a = [] for i in range(n): a.append(int(input()) <<<<<<<<<<<<<<<< here for i in range(0,(n-16)): for k in range(15,(n-16)): if (a[i]*a[i+k]) <= s: s = a[i]*a[i+k] print(s) solved python won’t run this program; says there is a syntax error
Wooble is correct, you are missing a closing bracket on one of your if statements. The following code should do the trick <?php //Funcao que vai abrir o arquivo php que faz conexao com o servidor require_once (“sistemadelogin.php”); $descricao = $_POST[descimagem]; $nomeFOTOG = $_FILES[‘fotoimagem’][‘name’]; $nomeFOTOP = $_FILES[‘miniImagem’][‘name’]; $tmpFOTOG = $_FILES[‘fotoimagem’][‘tmp_name’]; $tmpFOTOP = $_FILES[‘miniImagem’][‘tmp_name’]; $destinoG = … Read more
tight_contour is a std::vector object containing elements of type std::vector<cv::Point>. The (1) is constructing tight_contour to hold 1 initial default-constructed element. 0 solved What does (1) mean when declaring a vector?