如果Record存在UPDATE MySql Table或使用PHP INSERT

Question

i have to check if record exists in table forex_rates_new, i need to check for 2 criterias that is it has to compare agency_id and currency_name. If agency_id and currency_name are same then the system has to UPDATE or else INSERT.

forex_rates_new
id  agency_id   currency_name    rate
1   1111          aaa            1.898
2   2190          aaa            0.787

That is in the database if agency_id with value 111 occurs and currency_name with USD occurs. Next time if the user enters the same value 111 and currency_name as USD then the record should go in updated mode or else insert. This is my query:

INSERT INTO forex_rates_new (`agency_id`,`currency_name`,`rate`,`logo`,`created_at`) 
VALUES('".$agency_id."','".$currency_name."','".$rate."','".$logo."', now())

any help is appreciated

Answer 1

i am giving you simple demonstration here how to achieve that. first you need to do select data.something like this.

"SELECT * FROM request WHERE agency_id='$agency_id' ";

Now check if the data exist make update query

"UPDATE request SET field name='$value' WHERE agency_id='$agency_id'";

otherwise something like this.

INSERT INTO forex_rates_new (`agency_id`,`currency_name`,`rate`,`logo`,`created_at`) 
VALUES('".$agency_id."','".$currency_name."','".$rate."','".$logo."', now())

Your code should be like this.

$qry=mysqli_query($con,"SELECT * FROM request WHERE agency_id='$agency_id' ");
$rowCheck=mysqli_num_rows($qry);
    if ($rowCheck>0) { // if data exist update the data
        $qry=mysqli_query($con,"UPDATE request SET field name='$value' WHERE agency_id='$agency_id'");  
    }
    else{ // insert the data if data is not exist
        $qry=mysqli_query($con,"INSERT INTO request (field name) VALUES('$value')");
    }

Remember given code is open to sql injection attack. this is logic only this is not full code.to avoiding sql injection attack use prepared statements.

Answer 2

Check true or false by IF and EXISTS

IF ( IF(EXISTS(SELECT `agency_id`,`currency_name` FROM forex_rates_new where `agency_id` = "$agency_id" AND  `currency_name` = "$currency_name"),1,0) ) THEN

BEGIN
    UPDATE forex_rates_new  
    SET
    `rate` = "$rate",
    `logo` = "$logo",
    `created_at` = now();
    WHERE `agency_id` = "$agency_id" AND `currency_name` = "$currency_name"
END;

ELSE

BEGIN
    INSERT INTO forex_rates_new 
    (`agency_id`,`currency_name`,`rate`,`logo`,`created_at`) 
    VALUES
    ("$agency_id","$currency_name","$rate","$logo", now());
END;

END IF;

Answer 3

Here you go!

<?php    
        $dupl = mysqli_query($con, "SELECT * FROM forex_rates_new WHERE currency_name='$agency_id'") or die(mysql_error($con));

        $row = mysqli_num_rows($dupl);
        if ($row > 0)
        {
            echo "<script language='javascript'>alert('Exists!!!')</script>";

        $stmt1 = $con->prepare("UPDATE forex_rates_new SET column_name=? WHERE column_name=?");
                        if($stmt1)
                        {
                            $stmt1->bind_param('is', $value, $value);
                            $stmt1->execute();
                        }
                      if($stmt1->affected_rows >0)
                      {
                        echo "<script language='javascript'>alert('Updated Sucessfully!!!')</script>";
                      }
                    else
                    {
                        echo "<script language='javascript'>alert('error!!!')</script>";
                    } 
        }

        $stmt = $con->prepare("INSERT into forex_rates_new VALUES(?,?,?,?, now())
        if($stmt)
        {
            $stmt->bind_param('isis', $currency_name, $agency_id, $rate, $logo);
            $stmt->execute();
        }
        if($stmt->affected_rows >0)
        {
            echo "<script language='javascript'>alert('Inserted Successfully!!!')</script>";
        }
        else{
            echo "<script language='javascript'>alert('An error occurred!!!')</script>";
        }
        } 
        ?>

Above is what you want using mysqli prepared statement.

i stands for integer s stands for string

Answer 4

You can add a UNIQUE KEY to agency_id , currency_name and than you can use "REPLACE" Syntax -> https://dev.mysql.com/doc/refman/8.0/en/replace.html .

No need to select and add condition for Update/Insert statement.

Answer 5

INSERT INTO t1 (a,b,c) VALUES (1,2,3)
  ON DUPLICATE KEY UPDATE c=c+1;

https://dev.mysql.com/doc/refman/8.0/en/insert-on-duplicate.html