zoz*_*ozo 49 php arrays scalar warnings zend-framework
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'];
}
}
}
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代码SEEMS工作正常,但我收到此警告:
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).
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谁能告诉我如何解决这个问题?
bri*_*n_d 75
您需要在$final[$id]向其添加元素之前设置为数组.用其中任何一个来实现它
$final[$id] = array();
$final[$id][0] = 3;
$final[$id]['link'] = "/".$row['permalink'];
$final[$id]['title'] = $row['title'];
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要么
$final[$id] = array(0 => 3);
$final[$id]['link'] = "/".$row['permalink'];
$final[$id]['title'] = $row['title'];
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Lan*_*Lan 69
有点晚了,但是对于那些想知道他们为什么会得到"警告:不能使用标量值作为数组"的消息;
原因是因为你首先用普通的整数或字符串声明你的变量然后你试图将它变成一个数组.
希望有所帮助