我正在尝试将一个简单的 Javascript代码段写入CodeignITer链接.我正在使用该链接删除仪表板中所需的帖子.虽然我想学习它,但我对JS一无所知.

码

$js = 'onClick = "alert("Are you sure")"';

$this->table->set_heading('Date','Title','Delete','Update');

foreach($records as $row){
$row->title = ucwords($row->title);
$this->table->add_row($row->date,$row->title = ucwords($row->title),anchor("main/delete/$row->id",$row->id,$js),//this is the link in question
anchor("main/fill_form/$row->id",$row->id)
);
}
$table = $this->table->generate();
echo $table;

我的问题是如何为链接($js)编写JS.我想使用确认声明(是或否).我完全迷失了JS,以防止意外删除

谢谢

echo anchor('delete/something',array('onClick' => "return confirm('Are you sure you want to delete?')"));