Question

我通过一个函数传递一个sql查询，该函数返回SELECT查询的结果数。结果是mysqli_num_rows（）方法的正确计数，但不将计数器变量设置为等于该数字，而是将计数器保持等于其原始值。

public function getNum($sql){
    $count = 0;  //set the count variable
    if($result = $this->query($sql)){
        //echo 'No results for criteria';
    $count = mysqli_num_rows($result); //count is properly set to number
    }                                  //does not modify outside count variable
   return $count;  //returning 0 or whatever count was originaly set to
}   //count does not change despite $count inside the if statement being changed

以下是$ this-＆gt;查询的内容

public function __construct(){
    $this->mysqli = new mysqli('localhost', 'root', '', 'mydb');
}

public function query($sql){
    return $this->mysqli->query($sql);
}

我确定查询正在执行。问题是变量没有被改变。

Answer 1

据我了解你的问题，你想在函数内部设置变量$count，但你应该知道函数内的变量不能在外部访问，反之亦然，这是因为variable scope in PHP。

你可以试试这个：

var $count;
function set_count($new_value) { 
  $this->count = $new_value; 
} 
function get_count() { 
  return $this->count; 
}
public function getNum($sql){
    $count = 0;  //set the count variable
    if($result = $this->query($sql)){
      $count = mysqli_num_rows($result);
    }
   //set `$count`
   $this->set_count($new_value);             
   return $count;
}
//and if you want to get the count
echo $this->get_count();

试图在php中传递结果数量变量

1 个答案: