PHP - 不能使用标量作为数组警告 [英] PHP - cannot use a scalar as an array warning

问题描述

我有以下代码:

 $final = array();
    foreach ($words as $word) {
        $query = "SELECT Something";
        $result = $this->_db->fetchAll($query, "%".$word."%");
        foreach ($result as $row)
        {
            $id = $row['page_id'];
            if (!empty($final[$id][0]))
            {
                $final[$id][0] = $final[$id][0]+3;
            }
            else
            {
                $final[$id][0] = 3;
                $final[$id]['link'] = "/".$row['permalink'];
                $final[$id]['title'] = $row['title'];
            }
        } 
    }

代码似乎工作正常，但我收到此警告:

The code SEEMS to work fine, but I get this warning:

Warning: Cannot use a scalar value as an array in line X, Y, Z (the line with: $final[$id][0] = 3, and the next 2).

谁能告诉我如何解决这个问题?

Can anyone tell me how to fix this?

推荐答案

在向数组中添加元素之前，您需要将$final[$id] 设置为数组.用其中之一初始化它

You need to set$final[$id] to an array before adding elements to it. Intiialize it with either

$final[$id] = array();
$final[$id][0] = 3;
$final[$id]['link'] = "/".$row['permalink'];
$final[$id]['title'] = $row['title'];

或

$final[$id] = array(0 => 3);
$final[$id]['link'] = "/".$row['permalink'];
$final[$id]['title'] = $row['title'];

这篇关于PHP - 不能使用标量作为数组警告的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！