MYSQL insert update - on duplicate key update - getting error

Question

I getting error "Query error!" with this code:

$result = $connection->query("INSERT INTO EMP_TBLE (NAME, AGE, CATEGORY, UPDATE_COUNT) VALUES('$NAME', '$AGE', $CAT, 1
ON DUPLICATE KEY UPDATE UPDATE_COUNT = UPDATE_COUNT + 1"));
echo "<br>".$result;die;

I also check different example like this example was running success but I found error in above code for my demo project.

Where do you get $NAME and $AGE etc from? If these are from user input, be careful about SQL injection. You should be using parameter binding here. — halfer
– halfer, Commented Sep 10, 2015 at 13:36
Print your query and post it into the question. See what you are running. Most probably, you'll find the error yourself. — user4035
– user4035, Commented Sep 10, 2015 at 13:41

mario.van.zadel · Accepted Answer · 2015-09-10 13:40:57Z

1

The closing brace for VALUES is missing. There is also a non-needed brace at the end of your code line.

Please change

$result = $connection->query("INSERT INTO EMP_TBLE (NAME, AGE, CATEGORY, UPDATE_COUNT) VALUES('$NAME', '$AGE', $CAT, 1 ON DUPLICATE KEY UPDATE UPDATE_COUNT = UPDATE_COUNT + 1"));

to

$result = $connection->query("INSERT INTO EMP_TBLE (NAME, AGE, CATEGORY, UPDATE_COUNT) VALUES('$NAME', '$AGE', $CAT, 1) ON DUPLICATE KEY UPDATE UPDATE_COUNT = UPDATE_COUNT + 1");

answered Sep 10, 2015 at 13:40

mario.van.zadel

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Comments

Julio Soares · Accepted Answer · 2015-09-10 13:44:28Z

0

Apparently you did not close the parentesis in the right place:

$result = $connection->query("INSERT INTO EMP_TBLE (NAME, AGE, CATEGORY, 
UPDATE_COUNT) VALUES('$NAME', '$AGE', $CAT, 1)
ON DUPLICATE KEY UPDATE UPDATE_COUNT = UPDATE_COUNT + 1");

Also... you should at least escape all the variables before concatenating in queries like this to avoid SQL injection

answered Sep 10, 2015 at 13:44

Julio Soares

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