Sample Code
<script>
config = {
locateFile: filename => `sql-wasm.wasm`
}
initSqlJs(config).then(function(SQL)
{
//Create the database
const db = new SQL.Database();
//create first table - Questions (Parent Table)
let question = "CREATE TABLE Questions(qid INT NOT NULL, qdesc CHAR, PRIMARY KEY(qid));\
INSERT INTO Questions VALUES(1, 'set1');\
INSERT INTO Questions VALUES(2, 'set2');\
INSERT INTO Questions VALUES(3, 'set3');\
INSERT INTO Questions VALUES(4, 'set4');\
INSERT INTO Questions VALUES(5, 'set5');";
db.run(question);
const result = db.exec("SELECT * FROM Questions");
console.log(result);
//create second Table - Qimages (Child Table)
let qimage = "CREATE TABLE Qimages(img_id INT AUTO_INCREMENT, imgurl VARCHAR,FOREIGN KEY (qid) REFERENCES Questions(qid));\
INSERT INTO Qimages VALUES('https://www.w3schools.com/howto/img_woods.jpg',1);\
INSERT INTO Qimages VALUES('https://www.w3schools.com/howto/img_5terre.jpg',1);\
INSERT INTO Qimages VALUES('https://www.w3schools.com/howto/img_mountains.jpg',2);";
db.run(qimage);
const res = db.exec("SELECT * FROM Qimages");
console.log(res);
</script>
Below is the Error I'm Getting
uncaught (in promise) Error: unknown column "qid" in foreign key definition
As the message says, there is no qid column in the Qimages table. I suspect that you want:-
CREATE TABLE Qimages(img_id INT AUTO_INCREMENT, imgurl VARCHAR,qid INTEGER REFERENCES Questions(qid));
Thus the value of the qid column (now defined), that uses the column level version of the Foreign Key constraint declaration, must have a corresponding value in the qid column of the Question table.
Alternately you could use:-
CREATE TABLE Qimages(img_id INT AUTO_INCREMENT, imgurl VARCHAR,qid INTEGER, FOREIGN KEY (qid) REFERENCES Questions(qid));