Javascript php中的编码和解码问题

我正在使用php，我需要将数组转换为javascript数组。如何更改？ 请帮助下面是我的输出值

我试图从很长一段时间调试，但没有得到任何线索

我的php代码：

<?php
  include "db_connection.php";
  $locations=array();
      $query =  $conn->query('SELECT `pg_address` FROM `tbl_master_property` limit 10');
       while ($row = $query->fetch_assoc()) {
         $locations[] = $row;
      }
      $locations = json_encode($locations);
      //echo "<pre>";print_r($locations);die;
?>

Javascript代码：

var locations = [
           '3/20, 1st Main, 1st Cross, Hosur Main Road, Adugodi, Bangalore',
           '24/3 Bazaar Street, Adugodi, Bangalore - 560030',
           '# 430, Koramangala 7th Block, Beside Sai Baba Temple, Bangalore',
           '# 41, 1st Cross, 2nd Main, Behind M R Granite, Adugodi, Bannerghatta Main Road, Bangalore',
           '# 27, 2nd Main, B cross, Nanjappa Layout, Adugodi, opp. to Vijaya Bank, Bangalore'  
  ];

---

在php中，while循环在第行下方更改

array_push($locations,json_encode($row['pg_address']));

---

试试这个检查是否有效？您的问题也不是很清楚，但这可能会对您有所帮助。

此代码可能适用于您：

<?php
  include "db_connection.php";
  $locations=array();
      $query =  $conn->query('SELECT `pg_address` FROM `tbl_master_property` limit 10');
       while ($row = $query->fetch_assoc()) {
         $locations[] = $row['pg_address'];
      }
?>

使用json_encode PHP函数对数据进行编码，并将其分配到Javascript变量中，该变量将为您提供所需的输出数组，更新后的代码如下所示

在javascript中

var位置=

---

希望这有帮助。

非常感谢，先生，工作正常，，，，我还需要传递名称，如何传递相同的数组。耶，工作正常，先生，我还需要谷歌标记弹出工具Kityaa的名称和地址，工作正常，先生，我还需要谷歌标记弹出工具的名称和地址

$locations[] = $row['pg_address'];

array_push($locations,json_encode($row['pg_address']));

<?php
  include "db_connection.php";
  $locations=array();
      $query =  $conn->query('SELECT `pg_address` FROM `tbl_master_property` limit 10');
       while ($row = $query->fetch_assoc()) {
         $locations[] = $row['pg_address'];
      }
?>