致命错误:在不在对象上下文中时使用$ this

Question

如果有错误,这是部分.

致命错误:在第6行的/pb_events.php中不在对象上下文中时使用$ this

第6行是: $jpp = $this->vars->data["jpp"];

function DoEvents($this) {

    global $_CONF, $_PAGE, $_TSM , $base;

    $jpp = $this->vars->data["jpp"];

    $cache["departments"] = $this->db->QFetchRowArray("SELECT * FROM {$this->tables[job_departments]}");
    $cache["locations"] = $this->db->QFetchRowArray("SELECT * FROM {$this->tables[job_location]}");
    $cache["names"] = $this->db->QFetchRowArray("SELECT * FROM {$this->tables[job_names]}");
    $cache["categories"] = $this->db->QFetchRowArray("SELECT * FROM {$this->tables[job_categories]}");

非常感谢!欣赏!

Answer 1

$这只在方法中有意义,而不是在函数中

还行吧

class Foo {
     function bar() {
          $this->...

这不是

function some() {
    $this->

//编辑:没有注意到他将"$ this"作为参数传递

建议:简单地用"$ somethingElse"替换"$ this"

Answer 2

你不能传递$this给程序功能.$this是保留变量.

Answer 3

根据我的评论.你想使用$this传递变量而php不允许它在类方法体外.

function DoEvents($obj) {

    global $_CONF, $_PAGE, $_TSM , $base;

    $jpp = $obj->vars->data["jpp"];

    $cache["departments"] = $obj->db->QFetchRowArray("SELECT * FROM {$obj->tables[job_departments]}");
    $cache["locations"] = $obj->db->QFetchRowArray("SELECT * FROM {$obj->tables[job_location]}");
    $cache["names"] = $obj->db->QFetchRowArray("SELECT * FROM {$obj->tables[job_names]}");
    $cache["categories"] = $obj->db->QFetchRowArray("SELECT * FROM {$obj->tables[job_categories]}");