PHP - 不能使用标量作为数组警告
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PHP - cannot use a scalar as an array warning
提问by zozo
I have the following code:
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'];
}
}
}
The code SEEMS to work fine, but I get this warning:
代码似乎工作正常，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).
Can anyone tell me how to fix this?
谁能告诉我如何解决这个问题？
回答by brian_d
You need to set$final[$id]to an array before adding elements to it. Intiialize it with either
您需要先设置$final[$id]为数组，然后再向其添加元素。用任何一个初始化它
$final[$id] = array();
$final[$id][0] = 3;
$final[$id]['link'] = "/".$row['permalink'];
$final[$id]['title'] = $row['title'];
or
或者
$final[$id] = array(0 => 3);
$final[$id]['link'] = "/".$row['permalink'];
$final[$id]['title'] = $row['title'];
回答by Lan
A bit late, but to anyone who is wondering why they are getting the "Warning: Cannot use a scalar value as an array" message;
有点晚了，但是对于那些想知道为什么他们会收到“警告：不能将标量值用作数组”消息的人来说；
the reason is because somewhere you have first declared your variable with a normal integer or string and then later you are trying to turn it into an array.
原因是因为您首先使用普通整数或字符串声明变量，然后尝试将其转换为数组。
hope that helps
希望有帮助
回答by Peace Ngara
The Other Issue I have seen on this is when nesting arrays this tends to throw the warning, consider the following:
我在这方面看到的另一个问题是嵌套数组时这往往会引发警告，请考虑以下事项：
$data = [
"rs" => null
]
this above will work absolutely fine when used like:
上面的这个在使用时绝对可以正常工作：
$data["rs"] = 5;
But the below will throw a warning ::
但下面会发出警告::
$data = [
"rs" => [
"rs1" => null;
]
]
..
$data[rs][rs1] = 2; // this will throw the warning unless assigned to an array
回答by Allyn O
Also make sure that you don't declare it an array and then try to assign something else to the array like a string, float, integer. I had that problem. If you do some echos of output I was seeing what I wanted the first time, but not after another pass of the same code.
还要确保您没有将其声明为数组，然后尝试将其他内容分配给数组，例如字符串、浮点数、整数。我有那个问题。如果你做一些输出的回声，我第一次看到了我想要的东西，但不是在另一遍相同的代码之后。
回答by Benjamin
Make sure that you don't declare it as a integer, float, string or boolean before. http://php.net/manual/en/function.is-scalar.php
确保之前没有将其声明为整数、浮点数、字符串或布尔值。 http://php.net/manual/en/function.is-scalar.php
