php - 如何在没有 javascript 处理的情况下使用多插入内部函数 PDO PHP 显示成功

Question

我有一个这样的函数:

rangking.inc.php

function insert2(){     
    $query = "insert into ".$this->table_name." values(?,?,?,'','')";
    foreach ($this->ik as $key => $value){
        $stmt = $this->conn->prepare($query);
        $stmt->bindParam(1, $this->ia);
        $stmt->bindParam(2, $value);
        $stmt->bindParam(3, $this->nn[$key]);
        $stmt->execute();
    }
}

但我不知道如何计算已插入的值，我只想显示所有插入值的一条消息。

有人可以帮我怎么做吗？

这里是我的一段代码:

rangking.php

<?php
# include code

if($_POST){

    include_once 'path/rangking.inc.php';
    $eks = new rangking($db);

    $eks->ia = $_POST['ia'];
    $eks->ik = $_POST['ik'];
    $eks->nn = $_POST['nn'];

    # if($eks->insert2()){ using this code just throw me to false condition
?>
<!-- <div class="alert alert-success alert-dismissible" role="alert">
  <button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">&times;</span></button>
  <strong>Berhasil Tambah Data!</strong> Tambah lagi atau <a href="rangking.php">lihat semua data</a>.
</div> -->
<?php
    # }

    # else{
?>
<!-- <div class="alert alert-danger alert-dismissible" role="alert">
  <button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">&times;</span></button>
  <strong>Gagal Tambah Data!</strong> Terjadi kesalahan, coba lagi.
</div> -->
<?php
    # }
}
?>
# below is all html code to show for form

Answer 1

因为你的函数没有返回任何东西

function insert2(){     
    $i = 0;

    $query = "insert into ".$this->table_name." values(?,?,?,'','')";
    $stmt = $this->conn->prepare($query);

    foreach ($this->ik as $key => $value){
        $stmt->bindParam(1, $this->ia);
        $stmt->bindParam(2, $value);
        $stmt->bindParam(3, $this->nn[$key]);
        $stmt->execute();
        $i += 1;
    }

    return $i;

}