Question

我正在使用以下PHP在MySql数据库中为表值添加1，并在每次执行代码时更新数据库。

function create_update() {
    $user_id = $_SESSION['user_id'];
    $con = mysqli_connect(HOST, USER, PASSWORD, DATABASE);
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $numberofupdates = number_of_updates(); //'number_of_updates()' gets an INT from the database
    settype($numberofupdates, "integer"); //set the number to an INT if it already is not
    $spare_update_id = $numberofupdates + 1; //get number and add 1
    mysqli_query($con,"UPDATE members SET updates=$spare_update_id WHERE id = $user_id"); //update the value to the database
}

代码工作正常并更新到数据库，但似乎每次都将值更新为1。如果我手动将数据库值设置为4并运行代码，则它应该在5时将值设置回1。

感谢任何帮助！

Answer 1

如果我理解你的问题，那么你只需要1个mysql查询

function create_update() {
  $user_id = $_SESSION['user_id'];
  $con = mysqli_connect(HOST, USER, PASSWORD, DATABASE);
  if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
  mysqli_query($con,"UPDATE members SET updates = (updates + 1) WHERE id = $user_id");
}

Answer 2

您的代码未运行的原因是因为$numberofupdates设置为0，所以它总是评估为0 + 1
发生这种情况是因为number_of_updates()返回零或将其作为整数强制设置为零

而不是所有这些大惊小怪，您可以每次只是运行此查询以增加数量

UPDATE members SET updates = updates + 1 WHERE id = $user_id

使用PHP将数字添加到MySql值

2 个答案: