How to update if exist, otherwise insert in php? - php

Question

I want to check whether the data is existing or not. If data exists, update the table called "user_star_rate", otherwise insert the data into the table. Inserting is working properly, but the updating is not working.
Here is my code.
$jsqla7 = mysql_query("select * from user_star_rate where product_id='$product_id' and email='$visit_email'") or die(mysql_error());
$jfeta7 = mysql_fetch_assoc($jsqla7);
if($jfeta7 != null) {
$rate = "UPDATE user_star_rate SET rate_value='$rate_value' WHERE product_id='$product_id' and email='$visit_email'" ;
} else {
$rate = "INSERT INTO user_star_rate (email, product_id, rate_value) VALUES ('$visit_email','$product_id','$rate_value')" ;
}

Answer 1

If I understand what your saying, you should really be using a replace into and it would also decrease your code significantly.
Your code would become:
$query = "REPLACE INTO user_star_rate(product_id, email) VALUES('$product_id', '$visit_email')";
mysql_query($query) or die(mysql_error());
If it already exists then it will update it, else it will insert it. You should pay special attention to the docs in regards to foreign keys and auto incrementing ids.

Answer 2

Try this :
$tot = mysql_num_rows($jfeta7);
if($tot > 0){
$rate = "UPDATE user_star_rate SET rate_value='$rate_value' WHERE product_id='$product_id' and email='$visit_email'" ;
} else {
$rate = "INSERT INTO user_star_rate (email, product_id, rate_value) VALUES ('$visit_email','$product_id','$rate_value')" ;
}

Answer 3

Try:
$jsqla7 = mysql_query("select count(*) from user_star_rate where product_id='$product_id' and email='$visit_email'") or die(mysql_error());
$count = mysql_num_rows();
if($count) {
$rate = "UPDATE user_star_rate SET rate_value='$rate_value' WHERE product_id='$product_id' and email='$visit_email'" ;
} else {
$rate = "INSERT INTO user_star_rate (email, product_id, rate_value) VALUES ('$visit_email','$product_id','$rate_value')" ;
}