I have created 4 tables. The 4th one has just one field. I cannot insert data into any of them.
I have created the tables without any error feedback.
I am using the command likie:
INSERT INTO restaurant4 (lotno) VALUES ($lotno) ;
I get error messages each time and I don’t think any data has been inserted.
Why doesn’t this work?
Mike Smith
We dont have a crystal ball and cannot see your screen. How about posting your code and the actual error messages.
1 Like
The code is:
$sql = "INSERT INTO restaurant2 (lotno, fname)
VALUES ($lotno, $fname )" ;
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
The Error message is:
INSERT INTO restaurant2 (lotno, fname) VALUES (1111, John )
Unknown column ‘John’ in ‘field list’
Use prepared statements and your problem will be solved
I assume that when you say “prepared statements” you mean something like below:
Here is the code for line 110:
INSERT INTO restaurant2 (lotno, fname) VALUES ($lotno, $fname ) ;
Here is the error message:
Parse error:
syntax error, unexpected ‘INTO’ (T_STRING) in C:\PHP Samples\Restaurant\RestaurantStore1t.php on line 110
Mike Smith
No, a prepared statement is where you use replaceable parameters in the query, call the prepare()
method, and then call the execute()
method with values for those parameters.
Have a look at the code being discussed in this thread and you’ll see the idea: Use a checkbox to update a mysql table - #16 by droopsnoot
Prepared statements get around a lot of issues such as properly quoting values inside queries.
1 Like
Yes, it worked perfectly.
Thank you so much.
Mike Smith
system
Closed
October 15, 2018, 1:49am
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