dan
dan

Reputation: 563

IF/ELSE condition in sql and php

I'm inserting into my database id and hours, which is working successfully. However when i insert a new value for an id that has a previous value, the old val isn't replaced by the new, instead they're kept both. I'm trying to update the function with an if/else statement (commented part). But still the same result. The if/else statement must check first if hours (a column in the table) value is empty, if so then perform sql insert, else perform sql update. Any help please?

if (isset($_POST['submit'])) {
    $hours = $_POST['Hours'];
    $selectid = $_POST['SelectID'];
    $sql1 = "INSERT INTO `editedworkhours` (`id`,`H`) VALUES('$selectid','$hours')";

    $getResult = mysql_query($sql1);
    if (mysql_affected_rows() > 0) {

    } else {

    }

    $tempname = $row['Field'];
    $sql2 = "UPDATE editedworkhours SET H ='" . $_GET["hours"] . "' WHERE IDNumber='" . $_GET["selectid"] . "'";
    $result2 = mysqli_query($con, $sql2);
    if ($con->query($sql2) === TRUE) {
    } else {

    }


}
echo $menu;

Upvotes: 2

Views: 5111

Answers (1)

Abdulla Nilam
Abdulla Nilam

Reputation: 38670

Try this

<?php
    if(isset($_POST['submit']))
    {
        $addedhours = $_POST['AddedHours'];
        $selectaf = $_POST['SelectAF'];

        $sql1="SELECT * FROM editedworkhours WHERE AFNumber='$selectaf' and AddedWH ='$addedhours'";

        $getResult = mysql_query($sql1);
        $count = count($getResult);
        if(!empty($count) || $count==1)
        {
            $tempname = $row['Field'];
            $sql2 = "UPDATE editedworkhours SET AddedWH ='$addedhours' WHERE AFNumber='$selectaf'";
            $result2 = mysql_query($sql2);

            if (isset($result2))
            {
                //Data inserted
            }
            else
            {
                //Insert Failed
                echo '<script>swal("Error", "Something went wrong error");</script>';
            }
            echo '<script>swal("Success", "Changes have been saved", "success");</script>';
        }
        else
        {
            $sql3 = "INSERT INTO editedworkhours (AFNumber,AddedWH) VALUES('$selectaf','$addedhours')";
            $Result =   mysql_query($sql3);
            if(isset($Result))
            {
                echo 'Success';
            }
            else
            {
                echo 'Failed';
            }
        }
    }
    echo $menu;

Upvotes: 2

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