如何防止数据重复？

Question

我想避免在我的数据库中插入名为username的现有数据，我的代码运行没有错误，但问题是我输入的数据是插入的，即使它已经存在于我的数据库中。请告诉我我的代码缺少什么。

if($_POST["controller"] == "add") {
        // validations here
        $result = query("SELECT * FROM tbl_employee WHERE surname ='$surname'");
        if(empty($_POST["surname"]) && (query($result)>0) ) {
            echo 'surname already taken'; die();
        }
        else {
            $age = getAge($_POST["dobMonth"], $_POST["dobDay"], $_POST["dobYear"]);
            $result = query("INSERT INTO tbl_employee (carid, surname, firstname, position, department, birthplace, sex, address, contact, codename, birthdate, age, dateAdded)  VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?)", $_POST["modelo"], $_POST["surname"], $_POST["firstname"], $_POST["position"], $_POST["department"], $_POST["birthplace"], $_POST["sex"], $_POST["address"], $_POST["contact"], $_POST["codename"], $_POST["dobYear"] . '-' . $_POST["dobMonth"] . '-' . $_POST["dobDay"], $age, date("Y-m-d"));
            if($result === false) {
                echo 'error in insert'; die();
            }
            else {
                $row = query("SELECT LAST_INSERT_ID() AS employee_id");
                $id = $row[0]["employee_id"];
                redirect("employee.php?profile=" . $id); //redirect("employee.php?action=view_all");
            }
        }
    }

Answer 1

试试这个：我希望它能帮助你。

if($result['surname']==$_POST["surname"]) {
            echo 'surname already taken'; die();
        }
        else {
            $age = getAge($_POST["dobMonth"], $_POST["dobDay"], $_POST["dobYear"]);
            $result = query("INSERT INTO tbl_employee (carid, surname, firstname, position, department, birthplace, sex, address, contact, codename, birthdate, age, dateAdded)  VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?)", $_POST["modelo"], $_POST["surname"], $_POST["firstname"], $_POST["position"], $_POST["department"], $_POST["birthplace"], $_POST["sex"], $_POST["address"], $_POST["contact"], $_POST["codename"], $_POST["dobYear"] . '-' . $_POST["dobMonth"] . '-' . $_POST["dobDay"], $age, date("Y-m-d"));
            if($result === false) {
                echo 'error in insert'; die();
            }

Answer 2

在检查为空之前，您可能需要先检查它是否为NULL

Answer 3

你需要改变你的代码。我希望这会对你的问题有所帮助。你的代码问题首先是它会检查姓氏值是否为空，如果你发送姓氏值，它将使条件为假，因为“和” “条件和你的其他声明将被执行

if($_POST["controller"] == "add") {
    // validations here
    $surname=(empty($_POST["surname"])?null:$_POST["surname"]);
    $result = query("SELECT * FROM tbl_employee WHERE surname ='$surname'");
    if(empty($_POST["surname"]) ){
        echo 'surname is Empty'; die();
    }
    elseif((query($result)>0) ) 
    {
        echo 'surname already taken'; die();
    }
    else {
        $age = getAge($_POST["dobMonth"], $_POST["dobDay"], $_POST["dobYear"]);
        $result = query("INSERT INTO tbl_employee (carid, surname, firstname, position, department, birthplace, sex, address, contact, codename, birthdate, age, dateAdded)  VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?)", $_POST["modelo"], $_POST["surname"], $_POST["firstname"], $_POST["position"], $_POST["department"], $_POST["birthplace"], $_POST["sex"], $_POST["address"], $_POST["contact"], $_POST["codename"], $_POST["dobYear"] . '-' . $_POST["dobMonth"] . '-' . $_POST["dobDay"], $age, date("Y-m-d"));
        if($result === false) {
            echo 'error in insert'; die();
        }
        else {
            $row = query("SELECT LAST_INSERT_ID() AS employee_id");
            $id = $row[0]["employee_id"];
            redirect("employee.php?profile=" . $id); //redirect("employee.php?action=view_all");
        }
    }
}