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What to do if var cannot be used in php

藏色散人
Release: 2023-03-17 08:28:01
Original
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Solution to the problem that var cannot be used in php: 1. Open the corresponding PHP file; 2. Check the "address.inc" code corresponding to the Address class; 3. Find "public" and remove it, or replace Just change "public" to "var".

What to do if var cannot be used in php

The operating environment of this tutorial: Windows 7 system, PHP version 8.1, Dell G3 computer.

What should I do if var cannot be used in php?

php var error, PHP exception Parse error: syntax error, unexpected T_VAR error solution

In fact, this is a very easy problem to solve. In my opinion, it seems familiar, haha, I recently learned JavaScript and learned to use var to declare variables.

In fact, there is no need to use var declaration in PHP, but when a variable is used as a member variable of a class, there is still no problem in using var.

When using var externally, an error will be reported. Parse error: syntax error, unexpected T_VAR in..., for example, my error message:

Parse error: syntax error, unexpected T_VAR in D:Apache2.2htdocsshirdrnpagep2pageUtil.inc on line 34
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I am testing: inside a class, use a An error occurred when using a self-defined class object as a member of this class.

The address.inc code corresponding to the Address class:

The code is as follows:

class Address {
var $road;
function Address(){}
function setRoad($road){
$this->road = $road;
}
}
?>
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The Person class and its test code for person.php are as follows:

Code As follows:

require("address.inc");
class Person {
var $name;
var $address;
function Person(){
}
function display(){
echo "Name : ".$this->name."
";
echo "Road : ".$this->address->road."
";
}
}
var $p = new Person();
$p->address = new Address();
$p->address->setRoad("Chagnchun Road");
$p->name = "Shirdrn";
$p->display();
?>
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The test output is abnormal:

Parse error: syntax error, unexpected T_VAR in D:Apache2.2htdocsshirdrnpagep2pageUtil.inc on line 34
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It is because var is used to declare variables in the person.php code. This cannot be done in PHP. Just use the "$" symbol to start. Indicates that what follows this character is a PHP variable. Haha:-)

Additional methods from other netizens:

Problem solving: syntax error, unexpected T_STRING, expecting T_OLD_FUNCTION or T_FUNCTION or T_VAR or I started my own PHP journey two days ago and made a very For an ordinary website, it turns out that our PHP is version 5.0 and the server is version 4.0. I am really confused. I had a busy day yesterday, and when I came back this morning, I found a solution in an article. Parse error: syntax error, unexpected T_STRING, expecting T_OLD_FUNCTION or T_FUNCTION or T_VAR or "}", if there is "public", remove "public". There will be no error. If "public" is a defined variable, change "public" to "var".

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