Object could not be converted to string?
Why am I getting this error:
Catchable fatal error: Object of class Card could not be converted to
string in /f5/debate/public/Card.php on line 79
Here is the code:
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 $query
and the function is part of class Card
)
All the declarations of 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:
$query = "INSERT INTO cards
VALUES('$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:
Catchable fatal error: Object of class Author could not be converted
to string in /f5/debate/public/Card.php on line 79
php
add a comment |
Why am I getting this error:
Catchable fatal error: Object of class Card could not be converted to
string in /f5/debate/public/Card.php on line 79
Here is the code:
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 $query
and the function is part of class Card
)
All the declarations of 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:
$query = "INSERT INTO cards
VALUES('$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', '$this->text')";
I now get this error:
Catchable fatal error: Object of class Author could not be converted
to string in /f5/debate/public/Card.php on line 79
php
That's because of$this-$author->qualifications
– quantumSoup
Jul 8 '10 at 5:51
add a comment |
Why am I getting this error:
Catchable fatal error: Object of class Card could not be converted to
string in /f5/debate/public/Card.php on line 79
Here is the code:
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 $query
and the function is part of class Card
)
All the declarations of 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:
$query = "INSERT INTO cards
VALUES('$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:
Catchable fatal error: Object of class Author could not be converted
to string in /f5/debate/public/Card.php on line 79
php
Why am I getting this error:
Catchable fatal error: Object of class Card could not be converted to
string in /f5/debate/public/Card.php on line 79
Here is the code:
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 $query
and the function is part of class Card
)
All the declarations of 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:
$query = "INSERT INTO cards
VALUES('$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', '$this->text')";
I now get this error:
Catchable fatal error: Object of class Author could not be converted
to string in /f5/debate/public/Card.php on line 79
php
php
edited Sep 18 '15 at 9:36
hg8
7411124
7411124
asked Jul 8 '10 at 5:37
cactusbincactusbin
3711311
3711311
That's because of$this-$author->qualifications
– quantumSoup
Jul 8 '10 at 5:51
add a comment |
That's because of$this-$author->qualifications
– quantumSoup
Jul 8 '10 at 5:51
That's because of
$this-$author->qualifications
– quantumSoup
Jul 8 '10 at 5:51
That's because of
$this-$author->qualifications
– quantumSoup
Jul 8 '10 at 5:51
add a comment |
6 Answers
6
active
oldest
votes
Read about string parsing, you have to enclose the variables with brackets {}
:
$query = "INSERT INTO cards VALUES('$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
.
So with "$this->author->last"
instead of "{$this->author->last}"
, PHP will only parse and evaluate $this->author
which gives you the error as author
is an object.
add a comment |
I don't think you need the $ sign when using arrow operator.
add a comment |
you shouldn't put $ before property names when you access them:
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);
}
add a comment |
You are trying to echo an object itself, not a string property of it. Check your code carefully.
add a comment |
You probably want to use:
$query = "INSERT INTO cards VALUES('$this->type','$this->tag' // etc
add a comment |
I think one of the object doesn't have toString() method defined so it cannot be represented as string.
add a comment |
Your Answer
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
Read about string parsing, you have to enclose the variables with brackets {}
:
$query = "INSERT INTO cards VALUES('$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
.
So with "$this->author->last"
instead of "{$this->author->last}"
, PHP will only parse and evaluate $this->author
which gives you the error as author
is an object.
add a comment |
Read about string parsing, you have to enclose the variables with brackets {}
:
$query = "INSERT INTO cards VALUES('$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
.
So with "$this->author->last"
instead of "{$this->author->last}"
, PHP will only parse and evaluate $this->author
which gives you the error as author
is an object.
add a comment |
Read about string parsing, you have to enclose the variables with brackets {}
:
$query = "INSERT INTO cards VALUES('$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
.
So with "$this->author->last"
instead of "{$this->author->last}"
, PHP will only parse and evaluate $this->author
which gives you the error as author
is an object.
Read about string parsing, you have to enclose the variables with brackets {}
:
$query = "INSERT INTO cards VALUES('$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
.
So with "$this->author->last"
instead of "{$this->author->last}"
, PHP will only parse and evaluate $this->author
which gives you the error as author
is an object.
edited Jul 8 '10 at 5:52
answered Jul 8 '10 at 5:46
Felix KlingFelix Kling
557k130864924
557k130864924
add a comment |
add a comment |
I don't think you need the $ sign when using arrow operator.
add a comment |
I don't think you need the $ sign when using arrow operator.
add a comment |
I don't think you need the $ sign when using arrow operator.
I don't think you need the $ sign when using arrow operator.
answered Jul 8 '10 at 5:43
uvgroovyuvgroovy
43539
43539
add a comment |
add a comment |
you shouldn't put $ before property names when you access them:
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);
}
add a comment |
you shouldn't put $ before property names when you access them:
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);
}
add a comment |
you shouldn't put $ before property names when you access them:
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);
}
you shouldn't put $ before property names when you access them:
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);
}
answered Jul 8 '10 at 5:39
Sergey EreminSergey Eremin
9,13913244
9,13913244
add a comment |
add a comment |
You are trying to echo an object itself, not a string property of it. Check your code carefully.
add a comment |
You are trying to echo an object itself, not a string property of it. Check your code carefully.
add a comment |
You are trying to echo an object itself, not a string property of it. Check your code carefully.
You are trying to echo an object itself, not a string property of it. Check your code carefully.
answered Jul 8 '10 at 5:39
Jacob RelkinJacob Relkin
135k24316303
135k24316303
add a comment |
add a comment |
You probably want to use:
$query = "INSERT INTO cards VALUES('$this->type','$this->tag' // etc
add a comment |
You probably want to use:
$query = "INSERT INTO cards VALUES('$this->type','$this->tag' // etc
add a comment |
You probably want to use:
$query = "INSERT INTO cards VALUES('$this->type','$this->tag' // etc
You probably want to use:
$query = "INSERT INTO cards VALUES('$this->type','$this->tag' // etc
answered Jul 8 '10 at 5:40
quantumSoupquantumSoup
14.9k73453
14.9k73453
add a comment |
add a comment |
I think one of the object doesn't have toString() method defined so it cannot be represented as string.
add a comment |
I think one of the object doesn't have toString() method defined so it cannot be represented as string.
add a comment |
I think one of the object doesn't have toString() method defined so it cannot be represented as string.
I think one of the object doesn't have toString() method defined so it cannot be represented as string.
answered Jul 8 '10 at 6:34
Alex KleshchevnikovAlex Kleshchevnikov
121214
121214
add a comment |
add a comment |
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That's because of
$this-$author->qualifications
– quantumSoup
Jul 8 '10 at 5:51