Question

我认为这会更简单，但我在网上找到一个简单的答案时遇到了一些困难。

我只是想从带有ajax的MySQL表中获取数据并将其转换为js数组。

我的桌子非常简单..它只是：

Table 1
 id     value
  1     1 
  2     2 
  3     3

这样做的最佳方式是什么？

到目前为止，我有php文件：

while($row = mysql_fetch_array($query)) {$array[] = $row;}

发出如下内容：

Array ( [0] => Array ( [0] => 7 [idGlobal] => 7 [1] => 4.9 [tc] => 4.9 ) ) Array ( [0]     => Array ( [0] => 7 [idGlobal] => 7 [1] => 4.9 [tc] => 4.9 ) [1] => Array ( [0] => 3     [idGlobal] => 3 [1] => 2.2 [tc] => 2.2 ) ) Array ( [0] => Array ( [0] => 7 [idGlobal] => 7 [1] => 4.9 [tc] => 4.9 ) [1] => Array ( [0] => 3 [idGlobal] => 3 [1] => 2.2 [tc] => 2.2 ) [2] => Array ( [0] => 5 [idGlobal] => 5 [1] => 1.9 [tc] => 1.9 ) ) Array ( [0] => Array ( [0] => 7 [idGlobal] => 7 [1] => 4.9 [tc] => 4.9 ) [1] => Array ( [0] => 3 [idGlobal] => 3 [1] => 2.2 [tc] => 2.2 ) [2] => Array ( [0] => 5 [idGlobal] => 5 [1] => 1.9 [tc] => 1.9 ) [3] => Array ( [0] => 6 [idGlobal] => 6 [1] => 1.9 [tc] => 1.9 ) ) Array ( [0] => Array ( [0] => 7 [idGlobal] => 7 [1] => 4.9 [tc] => 4.9 ) [1] => Array ( [0] => 3 [idGlobal] => 3 [1] => 2.2 [tc] => 2.2 ) [2] => Array ( [0] => 5 [idGlobal] => 5 [1] => 1.9 [tc] => 1.9 ) [3] => Array ( [0] => 6 [idGlobal] => 6 [1] => 1.9 [tc] => 1.9 ) [4] => Array ( [0] => 4 [idGlobal] => 4 [1] => 1.6 [tc] => 1.6 ) )

但我仍然不知道用ajax获取JS数组的最佳方法

Answer 1

$query = "SELECT id, value FROM table";
$data = array();

if ($result = $mysqli->query($query)) {
    while ($row = $result->fetch_assoc()) {
        $data[] = $row;
    }
}

header('Content-Type: application/json');
echo json_encode($data);

Answer 2

从mysql查询
获取结果oas associative array
json编码数组
将json字符串返回给javascript代码。

Answer 3

使用php函数json_encode创建包含json的字符串。稍后在javascript中检索字符串时，最简单的方法是将其转换为类似这样的数组。

var myArray = eval（'（'+ jsonStringRetrieved +'）'）;

并查看这些内容 how to use json_encode Safely turning a JSON string into an object

使用PHP和AJAX从MySQL表创建JS数组的最佳方法是什么？

3 个答案: