当表字段名称与类变量名称相同时，将来自MySQL select的变量存储到PHP类变量的效率更高? [英] Store variables from MySQL select to PHP class variables more efficiently when the table field names are identical to the class variable names?

问题描述

我正在从数据库中选择数据.数据库字段名称与类变量名称完全相同.有没有一种方法可以将这些数据存储到类变量中，而无需单独指定每个变量?

I am selecting data from a database. The database field names are exactly the same as the class variable names. Is there a way to store this data into the class variables without specifying each one individually?

    //gets info about a specified file.
    //chosen based on a supplied $fileId
    function getFileInfo($fileId)
    {
        //select info from database
        $sql = "SELECT id, companyId, url, name, contentType, fileSize, saved, retrieved
                FROM files
                WHERE id = $fileId";
        $results = $this->mysqli->query($sql);
        $results = $results->fetch_object();

        //store info into class variables
        $this->id = $results->id;
        $this->companyId = $results->companyId;
        $this->url = $results->url;
        $this->name = $results->name;
        $this->contentType = $results->contentType;
        $this->fileSize = $results->fileSize;
        $this->saved = $results->saved;
        $this->retrieved = $results->retrieved;
    }

推荐答案

我建议采用这种方法:

$nameMap = array(
   'id',
   'companyId',
   'url',
   'name',
   'contentType',
   'fileSize',
   'saved',
   'retrieved',
);

foreach( $nameMap as $attributeName ) {
   $this->$attributeName  = $results->$attributeName ;
}

虽然可以写

foreach($result as $var => $value) {
...
}

结果完全取决于支持表的结构.如果在表中添加其他属性，则代码可能会中断.

the outcome fully depends on backing table's structure. If you add further attributes to the table, your code might break.

使用$nameMap，该应用程序仍然可以正常工作.

Using $nameMap, the application still works.

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