将一个函数变量作为参数传递给Class的另一个函数

Question

I am new to PHP. What i am trying i have performed insertion and i will have to grab the its id so i used ;insert_lastid(); function which is same as mysqli_last_insert_id ,so dont get confused with it .i tried to hold that last id on basis of last id i will perform select operation in another function Here is the code which will clear more my problem . This function is for insertion

public function cars_insert($_POST,$path) {
    $this->insert_query="insert into `_cars` (`_year`,`_make`,`_model`,`_version`,`_city`,`_color`,`_mileage`,`_engine_capacity`,                                                                                 `_engine_type`,`_assembly`,`_image`)VALUES('" .$_POST['year'] . "','" .$_POST['make'] . "','" .$_POST['model'] . "','" .$_POST['version'] . "','" .$_POST['city'] . "','" .$_POST['color'] . "','" .$_POST['mileage'] . "','" .$_POST['engine_capacity'] . "','" .$_POST['engine_type'] . "','" .$_POST['assembly'] . "','" .$path . "')";
    return $this->insert();
    global $car;
    $this->car=insert_lastid();
}

Here it is the second function it will Select on basis of insert_lastid();

 public function cars_id($this->car){
    $this->select_query="SELECT * FROM `_cars` WHERE `_id`='$this->car'";
    return $this->select();
}

Answer 1

The reason that the ID isn't being set is that you have a return statement before you set the variable. The moment you hit the return statement the function stops executing. Move your code so that the variable assignment is done before the function returns.

$insert = $this->insert();
$this->car=insert_lastid();

return $insert;

Your global $car statement and the parameter in cars_id are both unnecessary because $this->car is a member of the class and is accessible from any member function of that class without having to be passed in or global'ed as you would with a normal variable.

public function cars_id(){
    $this->select_query="SELECT * FROM `_cars` WHERE `_id`='$this->car'";
    return $this->select();
}

As a side note, your code is currently extremely vulnerable to SQL injection attacks unless you have some kind of code in $this->insert() that is checking your SQL statements before running them. Make sure you properly escape all $_POST variables before using them in a query, or use prepared statements.