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HomeBackend DevelopmentPHP Tutorialphp数组转换js数组操作及json_encode的用法详解_php实例

对于php,个人感觉能够熟练操作数组和字符串,基本上已经是入门了,php本身有很多操作数组和字符串的函数,今天在做一个功能时,需要用Js动态的创建门店信息,这些信息是要从后台添加的,想来想去,通过php读取数据库,得到数组,然后将数组转化成符合需求js数组,
php数组形式为:

复制代码 代码如下:

$newArray = array(array('地区'=>'北京地区','items'=>'10','detail'=>array(0=>array('店名'=>'旗舰店','url'=>'http://www.'),1=>array('店名'=>'jjjj','url'=>'http://www.fdd'))),
                  array('地区'=>'上海地区','items'=>'11','detail'=>array(0=>array('店名'=>'旗舰店','url'=>'http://www.'),1=>array('店名'=>'jjjj','url'=>'http://www.fdd'))),
                 );

这里用的一个很方便的函数:json_encode();
var jsarray = new Array();
jsarray = ;
用console.log();可以看到jsarray的结构。   

在这里如何构造这样一个符合要求的php数组呢?(我用的是phpcms系统)

复制代码 代码如下:

    $sql= 'select catid,catname,items from category where parentid=10';
    $result = mysql_query($sql);
    while ($row = mysql_fetch_array($result)) {
     $arrArea['地区'] = $row['catname'];
     $arrArea['items'] = $row['items'];
     unset($arrArea['detail']);//这一步很关键,要不得出的信息就会累加。
     $sql2 = 'select title,url from news where catid='.$row['catid'];
     $fendian = mysql_query($sql2);
     while ($re=mysql_fetch_assoc($fendian)) {
      $item['店名']=$re['title'];
      $item['url']=$re['url'];  
      $arrArea['detail'][] = $item; 
     }
     $newwarr[]=$arrArea;
    }   
    var_dump($newwarr);

json_encode:对变量进行 JSON 编码,该函数只能接受 UTF-8编码的数据 .

相对应的还有json_decode,第二个参数为true,返回数组,执行相逆的过程。

json只能用对象和数组。

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