如何在特定索引的数组值之前回显文本标签答案

【问题标题】：How to echo text labels before array values for specific indexes如何在特定索引的数组值之前回显文本标签
【发布时间】：2020-07-27 23:33:44
【问题描述】：

我写了以下PHP代码，我想生成一个类似截图所示的表格：

该表还应在职位名称和电子邮件地址之前具有Job: 和Email: 之类的标题。谢谢

<?php

$company = array
  (

  array("John Smith","Accountant","johnsmith@example.com", 70),
  array("May Black","Business Analyst","mayblack@example.com", 40),
  array("Tom white","Plummer","tomwhite@example", 30),
  array("Susan Kim","Engineer","susankim@example.com", 80)

  );

  echo "--Customer List-- <br> <br>";

  for($i = 0; $i < 4; $i++) {
      for($j = 0; $j < 3; $j++) {
        echo $company[$i][$j]."<br>";  

      }
      echo "<br><br>";
  }

?>

【问题讨论】：

标签： php arrays for-loop label echo

【解决方案1】：

您也可以为标题定义一个数组，并（有条件地）首先回显这些标题（如果它们不是空字符串）。请注意，我使用foreach 循环不必调查数组中有多少员工。我仍然保持内部循环 ([0-3)) 不显示敏感的年龄信息（尽管可以以不同的方式控制）。

<?php
  $company = array(
    array("John Smith", "Accountant", "johnsmith@example.com", 70),
    array("May Black", "Business Analyst", "mayblack@example.com", 40),
    array("Tom white", "Plummer", "tomwhite@example", 30),
    array("Susan Kim", "Engineer", "susankim@example.com", 80)
  );

  $title = array("", "Job", "Email", "Age");

  echo "--Customer List--<br><br>\n";

  foreach($company as $employee) {
    for($j = 0; $j < 3; $j++) {
      if($title[$j]) echo $title[$j] . ": ";
      echo $employee[$j] . "<br>";
    }
    echo "<br><br>\n";
  }
?>

如果您想从一个地方控制员工的显示方式，另一个选项可能会很方便：通过创建class 和一个成员函数来显示（适当的）信息，您只需要更改显示函数想要更改员工信息在任何地方的显示方式。它还增加了其他优点，例如将处理特定类型对象的代码保持在定义该类型的代码附近。

<?php
    class Employee {
        public $Name;
        public $Job;
        protected $Email;
        private $Age;

        public function __construct($n, $j, $e, $a) {
            $this->Name = $n;
            $this->Job = $j;
            $this->Email = $e;
            $this->Age = $a;
        }

        public function display($delim = "<br>") {
            echo $this->Name . $delim;
            echo "Job: " . $this->Job . $delim;
            echo "Email: " . $this->Email . $delim;
            echo $delim . $delim;
            echo "\n";
        }
    }

    $employees = array(
        new Employee("John Smith","Accountant","johnsmith@example.com", 70),
        new Employee("May Black","Business Analyst","mayblack@example.com", 40),
        new Employee("Tom white","Plummer","tomwhite@example", 30),
        new Employee("Susan Kim","Engineer","susankim@example.com", 80)
    );

    echo "--Customer List--<br><br>\n";
    foreach($employees as $employee) {
        $employee->display();
    }
?>

【讨论】：