Question

有三个功能。其中两个函数使用一些完全相同的变量，因此我创建了第三个函数，但变量无法传递给这两个函数。

修改的我需要更详细的内容。两个函数是相同的search（）和report（），但当然也会有差异。

搜索（）

public function search(){
    if(($_POST['number'] != null) && (is_numeric($_POST['number']))){
        $number = trim($_POST['number']);
    }else{
        $number = "";
    }

    if (($_POST['name'] != null)) {
        $name = strtoupper(trim($_POST['name']));
    }else{
        $comments = "";
    }

    if($_POST['date'] != null){
        $date = trim($_POST['date']);
        $parseDate = explode("/", $date);
        $newDate = $parseDate[2].$parseDate[0].$parseDate[1];
    }else{
        $newDate = null;
    }

    //something will be done here//
    //will require using all 3 of the variables above
}

报告（）

  public function report(){
    if(($_POST['number'] != null) && (is_numeric($_POST['number']))){
        $number = trim($_POST['number']);
    }else{
        $number = "";
    }

    if (($_POST['name'] != null)) {
        $name = strtoupper(trim($_POST['name']));
    }else{
        $comments = "";
    }

    if($_POST['date'] != null){
        $date = trim($_POST['date']);
        $parseDate = explode("/", $date);
        $newDate = $parseDate[2].$parseDate[0].$parseDate[1];
    }else{
        $newDate = null;
    }

    //something will be done here//
    //will require using all 3 of the variables above
}

结合（）

public function combine(){
    //as can be seen.  The above two functions are totally the same but after that where I commented "something will be done here" different things would be done with different output.
}

search()和report()最终会使用json_encode将一些数据传回客户端。

将函数中的变量传递给外部函数 - php

0 个答案: