Inserting/updating data into MySql database using php - php

Question

I am trying to insert/update the MySql database depending on whether a post already exists on the database (I am checking this with a unique user_id). The following works:
$select_query = "SELECT * ";
$select_query .= "FROM test ";
$select_query .= "WHERE user_id = '$user_id'";
$check_user_id = mysqli_query($connection, $select_query);
$query = "INSERT INTO test (";
$query .= " user_id, name, message";
$query .= ") VALUES (";
$query .= " '{$user_id}', '{$name}', '{$message}'";
$query .= ")";
$result = mysqli_query($connection, $query);
if ($result) {
echo "Success!";
} else {
die("Database query failed. " . mysqli_error($connection));
}
However, when I use the following code with an if/else statement, it does not work anymore, although the console reports "Success!" (meaning $result has a value). Any help would be greatly appreciated. Thanks.
$select_query = "SELECT * ";
$select_query .= "FROM test ";
$select_query .= "WHERE user_id = '$user_id'";
$check_user_id = mysqli_query($connection, $select_query);
if (!$check_user_id) {
$query = "INSERT INTO test (";
$query .= " user_id, name, message";
$query .= ") VALUES (";
$query .= " '{$user_id}', '{$name}', '{$message}'";
$query .= ")";
} else {
$query = "UPDATE test SET ";
$query .= "name = '{$name}', ";
$query .= "message = '{$message}' ";
$query .= "WHERE user_id = '{$user_id}'";
}
$result = mysqli_query($connection, $query);
if ($result) {
echo "Success!";
} else {
die("Database query failed. " . mysqli_error($connection));
}

Answer 1

As i understand your code. you are trying to check if the user_id is existing in your database..
i made a simple code and i think its works for me..
$select_query = mysql_query("SELECT * FROM test WHERE user_id = '$user_id'") or die (mysql_error());
$result = mysql_num_rows($select_query);
if(!$result){
$query = mysql_query("INSERT INTO test (user_id, name, message) VALUES ('$user_id', '$name', '$message')");
if($query){
echo "Success!";
}
else
{
die (mysql_error());
}
}
else{
$query2 = mysql_query("UPDATE test SET name='$name', message='$message' WHERE user_id = '$user_id'")
}

Answer 2

mysql_query returns the operation identifier, not the actual result. This is why $check_user_id is always true, so you are always trying to update (even not existing!) rows.
you have to "read" the result ofmysql_queryby for example using
$check_user_id = mysql_num_rows( mysql_query($connection, $select_query) );
now it returns 0 (false) iff there were no results for q $select_query

Answer 3

This statement is giving you a resource to the result
$check_user_id = mysqli_query($connection, $select_query);
next you are checking for if(!$check_user_id) : this condition evaluates to false because of the negation !. Thus your condition goes to the else part and and never enters the if.
The $result always has value because you are calling it towards the end of the script.

Answer 4

Since you previously know the user_id, and assuming that is a primary key in the table, you could use "ON DUPLICATE KEY UPDATE" clause:
$query = mysql_query("INSERT INTO test (user_id, name, message)
VALUES ('$user_id', '$name', '$message')
ON DUPLICATE KEY
UPDATE name='$name', message='$message';
");
Same result with only one query.
Ref: http://dev.mysql.com/doc/refman/5.0/en/insert-on-duplicate.html

Answer 5

Use Code for data inserting in mysql.
$query = mysql_query("INSERT INTO test set user_id = '$user_id', name = '$name', message = '$message'");
if($query){
echo "Success!";
}