本文實例講述了php實現的簡單數據庫操作Model類。分享給大家供大家參考,具體如下:
該數據庫模型類可實現數據庫的增刪改查,簡化數據庫操作。
1. config.php代碼:
<?php
define("HOSTNAME","127.0.0.1");
define("USERNAME","root");
define("PASSWORD","");
define("DATANAME","class");
2. 使用代碼:
<?php
/*
作者:shyhero
*/
require("./config.php");
class Model{
private $link;
//構造函數,初始化數據庫連接
public function __construct(){
$this -> link = mysqli_connect(HOSTNAME,USERNAME,PASSWORD,DATANAME) or die("數據庫連接失敗");
mysqli_set_charset($this -> link,"utf8");
}
//查找 1.表名 2.條件 3.值 如果不添加條件或者值,就全部查詢
public function find($table="",$key="",$value=""){
if(!$key||!$value){
$sql = "select * from {$table}";
}else{
$sql = "select * from {$table} where {$key} = '{$value}'";
}
$res = mysqli_query($this -> link,$sql);
$arr = mysqli_fetch_all($res,MYSQLI_ASSOC);
mysqli_free_result($res);
return $arr;
}
//增加 1.表名 2.需要插入的字段 3.值1
public function ins($table="",$zd="name,score",$value=""){
$arr = explode(",",$value);
$str = "";
foreach($arr as $k => $v){
$str .= "'".$v."'".",";
}
$str = rtrim($str,",");
$sql = "insert into {$table}({$zd})values({$str})";
$res = mysqli_query($this -> link,$sql);
return mysqli_insert_id($this -> link);
}
//修改 1.表名 2.修改字段 3.值 4.條件 5.值
public function upd($table="",$key="",$value="",$key2="",$value2=""){
$sql = "update {$table} set {$key}='{$value}' where {$key2}='{$value2}'";
$res = mysqli_query($this -> link,$sql);
return mysqli_affected_rows($this -> link);
}
//刪除 1.表名 2.條件 3.值
public function del($table="",$key="",$value=""){
$sql = "delete from {$table} where {$key}='{$value}'";
$res = mysqli_query($this -> link,$sql);
return mysqli_affected_rows($this -> link);
}
//析構函數
public function __destruct(){
if(isset($res))
mysqli_free_result($res);
mysqli_close($this -> link);
}
}
$m = new Model();
//var_dump($m -> find("stu","id"));
var_dump($m -> ins("stu","name","zhu"));
//var_dump($m -> upd("stu","name","dujianing","id","1"));
//var_dump($m -> del("stu","name","li"));
?>
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希望本文所述對大家PHP程序設計有所幫助。