問題描述

這是錯誤致命錯誤:調用未定義的函數assign(
這是代碼，正如你所看到的，我顯然已經定義了函數，為什么它不起作用

this is the error Fatal error: Call to undefined function assign(
this is the code, as you can see i obviously have defined the function so why is it not working

class shades {
    function create($name, $shades, $slug, $shortDesc, $longDesc, $position){
        $name = sanitize_paranoid_string($name);
        $slug = slug($name);
        $shortDesc = sanitize_sql_string($shortDesc);
        $longDesc = sanitize_sql_string($longDesc);
        $query = mysql_query("INSERT INTO products (type, name, slug, shortDesc, htmlDesc, position)VALUES('shades','$name','$slug','$shortDesc','$longDesc','$position')")or die(mysql_error());  
        $ID = mysql_insert_id();
        assign($shades, $ID);
        if($query) {return true;}
        else {return false;};
    }
    function delassign($toID){
        mysql_query("DELETE FROM assign WHERE type='shades' AND toID='$toID'")or die(mysql_error());    
    }
    function assign($shades, $toID)
    {
        foreach($shades as $shade)
        {
            $result = mysql_query("INSERT INTO assign(type, typeID, toID)VALUES('shades','$shade','$toID')")or die(mysql_error());
            if($result){echo "Added!";}
            else{echo"Not Added!";}
        };  
    }
}

推薦答案

您沒有名為 assign() 的函數，但有一個具有此名稱的方法.PHP 不是 Java，在 PHP 中你必須明確，如果你想調用一個函數

You dont have a function named assign(), but a method with this name. PHP is not Java and in PHP you have to make clear, if you want to call a function

assign()

或方法

$object->assign()

在您的情況下，對函數的調用駐留在另一個方法中.$this 總是指對象本身，其中存在一個方法.

In your case the call to the function resides inside another method. $this always refers to the object, in which a method exists, itself.

$this->assign()

這篇關于“調用未定義的函數"調用類方法時出錯的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！