PHP MySQL准备的语句什么也不做,没有错误

时间:2020-11-03 07:40:41

标签: php mysql mysqli prepared-statement

我有一个MySQL插入语句,它不会返回错误,但不会执行任何操作。

  $sql = "INSERT INTO classes (ownerClasses, nameClasses, codeClasses) VALUES (?, ?, ?)";
    $stmt = mysqli_stmt_init($con);
    if (!mysqli_stmt_prepare($stmt, $sql)) {
      header("Location: ../createclass.php?error=sqlerror");
      exit();
    }
    else {
      mysqli_stmt_bind_param($stmt, "isi", $_SESSION['id'], $classname, $classcode);
      mysqli_stmt_execute($stmt);
      header("Location: ../createclass.php?n=".$classname."&c=".$classcode);
      exit();
    }
  mysqli_stmt_close($stmt);
  mysqli_close($con);

$_SESSION['id']$classname$classcode已全部定义。我想念什么?

0 个答案:

没有答案
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