Question

我有一张表，我正在绘制有关几个字段的信息。特别需要连接的两个字段是Firstname和Surname，但我不能在实际表中添加另一列。

我可以编辑我的PHP函数，以便在我调用它时为它连接它吗？

这是我的PHP函数

public function getCustomerinfoByCompanyID($itemID) {

    $stmt = mysqli_prepare($this->connection, "SELECT * FROM $this->tablename where CompanyID=?");
    $this->throwExceptionOnError();

    mysqli_stmt_bind_param($stmt, 'i', $itemID);        
    $this->throwExceptionOnError();

    mysqli_stmt_execute($stmt);
    $this->throwExceptionOnError();

    $rows = array();

    mysqli_stmt_bind_result($stmt, $row->CustID, $row->CompanyID, $row->FirstName, $row->Surname, $row->CellNo, $row->Email);

    while (mysqli_stmt_fetch($stmt)) {
      $rows[] = $row;
      $row = new stdClass();
      mysqli_stmt_bind_result($stmt, $row->CustID, $row->CompanyID, $row->FirstName, $row->Surname, $row->CellNo, $row->Email);
    }

    mysqli_stmt_free_result($stmt);
    mysqli_close($this->connection);

    return $rows;
}

Answer 1

您可以将查询更改为

SELECT *,CONCAT(FirstName,' ',Surname) as FullName FROM $this->tablename where CompanyID=?

然后你改变

mysqli_stmt_bind_result($stmt, $row->CustID, $row->CompanyID, $row->FirstName, $row->Surname, $row->CellNo, $row->Email);

类似

mysqli_stmt_bind_result($stmt, $row->CustID, $row->CompanyID, $row->FullName, $row->CellNo, $row->Email);

你有一栏

如何在PHP函数中连接？

1 个答案: