php 对象无法转换为字符串？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/3200899/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Object could not be converted to string?
提问by cactusbin
Why am I getting this error:
为什么我收到此错误：
Catchable fatal error: Object of class Card could not be converted to string in /f5/debate/public/Card.php on line 79
可捕获的致命错误：第 79 行的 Card 类对象无法转换为 /f5/debate/public/Card.php 中的字符串
Here is the code:
这是代码：
public function insert()
{
$mysql = new DB(debate);
$this->initializeInsert();
$query = "INSERT INTO cards
VALUES('$this->$type','$this->$tag','$this->$author->$last','$this->$author->$first',
'$this->$author->$qualifications','$this->$date->$year','$this->$date->$month',
'$this->$date->$day','$this->$title', '$this->$source', '$this->$text')";
$mysql->execute($query);
}
(Line 79 is the $queryand the function is part of class Card)
（第 79 行是$query并且函数是 class 的一部分Card）
All the declarations of Card:
的所有声明Card：
public $type;
public $tag;
public $title;
public $source;
public $text;
public function __construct() {
$this->date = new Date;
$this->author = new Author;
}
After changing line 79 to this:
将第 79 行更改为：
$query = "INSERT INTO cards
VALUES('$this->type','$this->tag','$this->author->last','$this->author->first',
'$this-$author->qualifications','$this->date->year','$this->date->month','$this->date->day',
'$this->title', '$this->source', '$this->text')";
I now get this error:
我现在收到此错误：
Catchable fatal error: Object of class Author could not be converted to string in /f5/debate/public/Card.php on line 79
可捕获的致命错误：无法在第 79 行的 /f5/debate/public/Card.php 中将类 Author 的对象转换为字符串
采纳答案by Felix Kling
Read about string parsing, you have to enclose the variables with brackets {}:
阅读有关字符串解析的内容，您必须将变量用括号括起来{}：
$query = "INSERT INTO cards VALUES('$this->type','$this->tag','{$this->author->last}',"
Whenever you want to access multidimensional arrays or properties of a property in string, you have to enclose this access with {}. Otherwise PHP will only parse the variable up to the first[i]or ->property.
每当您想访问多维数组或字符串属性的属性时，您必须将此访问包含在{}. 否则 PHP 只会解析变量到第一个[i]或->property.
So with "$this->author->last"instead of "{$this->author->last}", PHP will only parse and evaluate $this->authorwhich gives you the error as authoris an object.
因此，使用"$this->author->last"代替"{$this->author->last}"，PHP 只会解析和评估$this->author哪个会像author对象一样给出错误。
回答by uvgroovy
I don't think you need the $ sign when using arrow operator.
我认为使用箭头运算符时不需要 $ 符号。
回答by Sergey Eremin
you shouldn't put $ before property names when you access them:
访问它们时，不应将 $ 放在属性名称之前：
public function insert() {
$mysql = new DB(debate);
$this->initializeInsert();
$query = "INSERT INTO cards VALUES('$this->type','$this->tag','$this->author->last','$this->author->first','$this-$author->qualifications','$this->date->year','$this->date->month','$this->date->day','$this->title', '$this->source', '$this->text')";
$mysql->execute($query);
}
回答by Jacob Relkin
You are trying to echo an object itself, not a string property of it. Check your code carefully.
您正在尝试回显对象本身，而不是它的字符串属性。仔细检查您的代码。
回答by quantumSoup
You probably want to use:
您可能想要使用：
$query = "INSERT INTO cards VALUES('$this->type','$this->tag' // etc
回答by Alex Kleshchevnikov
I think one of the object doesn't have toString() method defined so it cannot be represented as string.
我认为其中一个对象没有定义 toString() 方法，因此它不能表示为字符串。
