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我想學習面向對象編程。 我現在的問題是如何在類中的mysqli_query()下添加鏈接。OPP mysqli_query()期望兩個參數...
在程序的風格我剛剛創建的連接細節$變量,然後使用它的mysqli_query($變量$ SQL)等
在我的課堂我有函數內部連接(),它會連接到數據庫並返回true,但唧可以在我的mysqli_query()中使用它嗎?請參閱此處的代碼:
<?php
class DB {
protected $db_name = 'OOP_forum';
protected $db_user = 'Marcel';
protected $db_pass = *****;
protected $db_host = 'localhost';
public function connect() {
$connection = mysqli_connect($this->db_host, $this->db_user, $this->db_pass);
mysqli_select_db($connection, $this->db_name);
return true;
}
public function processRowSet($rowSet, $singleRow = false) {
$resultArray = array();
while($row = mysqli_fetch_assoc($rowSet)) {
arraypush($resultArray, $row);
}
if($singleRow==true) {
return $resultArray[0];
}else {
return $resultArray; }
}
public function select ($table, $where, $column = '*') {
$sql = "SELECT $column FROM $table WHERE $where";
$result = mysqli_query($sql);
if(mysqli_num_row($result) == 1){
return $this->processRowSet($result, true);
} else
{ return $this->processsRowSet($result);
}
}
public function update ($data, $table, $where) {
foreach ($data as $column->$value) {
$sql = "UPDATE $table SET $column = $value WHERE $where";
mysqli_query($sql) or die(mysqli_error());
}
return true;
}
public function delete ($table, $column, $where) {
$sql = "DELETE FROM $table WHERE $column = $where";
if (query($sql)=== TRUE) {
echo "Record Deleted sucessfully";
}else {
echo "Error deleting record: " . $connection->error;
}
}
public function insert($data, $table) {
$columns = "";
$values = "";
foreach ($data as $column->$value) {
$columns .= ($columns == "") ? "": ", " ;
$columns .= $column;
$values .= ($values == "") ? "" : ", ";
$values .= $value;
}
$sql = "insert into $table ($columns) values ($values)";
mysqli_query($sql) or die(mysqli_error());
return mysqli_insert_id();
}
}
?>
爲什麼你使用過程式mysqli,如果你正在學習OOP語法? –
你必須傳遞你的連接對象作爲第一個參數: - '$ result = mysqli_query($ sql);' –
感謝您的評論。這就是我想要做的,將連接作爲第一個參數傳遞給mysqli_query(something,$ sql),但不知道如何從我的connect函數返回true來實現。將學習更多:) – Marcel