php 如何将字符串/变量传递给函数？

发布于 2024-11-06 00:24:58 字数 676 浏览 0 评论 0原文

我有一个名为 function.php 的文件，其中包含如下函数：

public function issue_challenge () 
{
    $conn= fns_connect(13);
    mysql_select_db("et_games",$conn);
    $sql = "insert into challengers (challenger,defender,stake,matching_game_id) values (".$this->talentnum_self.",".$this->talentnum_other.",".$this->stake.",'".$this->matching_str()."')";
    $result = mysql_query($sql);
    echo $sql; mysql_error();

    if($result){
        echo "ok.\n";return true;                
    }else{
        return false;                    
    }
}

另一个名为 test.php 的文件，其中包含一个表单。

如何将值从该表单传递到该函数中？

我正在寻找有关如何将字符串/变量传递给函数的一些通用答案。

谢谢

原文

I have a file called function.php that has a function like this:

public function issue_challenge () 
{
    $conn= fns_connect(13);
    mysql_select_db("et_games",$conn);
    $sql = "insert into challengers (challenger,defender,stake,matching_game_id) values (".$this->talentnum_self.",".$this->talentnum_other.",".$this->stake.",'".$this->matching_str()."')";
    $result = mysql_query($sql);
    echo $sql; mysql_error();

    if($result){
        echo "ok.\n";return true;                
    }else{
        return false;                    
    }
}

and another file called test.php that has a form in it.

How do I pass the values from that form into that function?

I am looking for some generic answer on how to pass a string/variable to a function.

Thanks

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静若繁花 2024-11-13 00:24:58

以这种方式调用您的函数：

 if(isset($_GET['form_name'])) {
     $val = $_GET['form_name'];
     public function issue_challenge($val);
}

另外：将声明更改为：

public function issue_challenge($formValue) {

...function contents...

}

Call your function this way:

 if(isset($_GET['form_name'])) {
     $val = $_GET['form_name'];
     public function issue_challenge($val);
}

Also: Change the declaration to:

public function issue_challenge($formValue) {

...function contents...

}

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话少情深 2024-11-13 00:24:58

增加值的通用答案：

函数 inc($x) {
 echo "x = " . $x；
 $x++；
 返回$x； 
}
函数 main() {
<前><代码>$x = '1';
$y = $this->inc($x);
回声 $y;
}

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独﹏钓一江月 2024-11-13 00:24:58

确保表单提交调用该函数，然后从 POST 数组中获取值。如果使用 php 框架，这会更容易，许多框架会为您提供清理输入的方法。在这种情况下，由于您可能使用数据库查询的输入，因此您需要确保它被正确转义以避免 sql 注入。

mysql_real_escape_string 的 PHP 手册

<form action="issue_challenge" method="POST">
<input type="text" name="testInput" id="testInput"/>
<input type="submit" name="submit" value="submit"/>
</form>

public function issue_challenge () {
    $conn= fns_connect(13);
    $inputText = $_POST['testInput']
mysql_select_db("et_games",$conn);
    $sql = "insert into challengers (challenger,defender,stake,matching_game_id) values (".$this->talentnum_self.",".$this->talentnum_other.",".$this->stake.",'".$this->matching_str()."')";
    $result = mysql_query($sql);
    echo $sql; mysql_error();
        if($result){
            echo "ok.\n";return true;                
            }else{
                return false;                    
            }
}

Make sure the form submission calls the function, then grab the value from the POST array. This is easier if using a php framework, many will offer you methods to clean the input. In this case since you are possibly using the input for a db query you will need to make sure it is properly escaped to avoid sql injection.

PHP manaul for mysql_real_escape_string

<form action="issue_challenge" method="POST">
<input type="text" name="testInput" id="testInput"/>
<input type="submit" name="submit" value="submit"/>
</form>

public function issue_challenge () {
    $conn= fns_connect(13);
    $inputText = $_POST['testInput']
mysql_select_db("et_games",$conn);
    $sql = "insert into challengers (challenger,defender,stake,matching_game_id) values (".$this->talentnum_self.",".$this->talentnum_other.",".$this->stake.",'".$this->matching_str()."')";
    $result = mysql_query($sql);
    echo $sql; mysql_error();
        if($result){
            echo "ok.\n";return true;                
            }else{
                return false;                    
            }
}

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