Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/php/233.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181

Warning: file_get_contents(/data/phpspider/zhask/data//catemap/2/jquery/69.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Php 如何将关联数组传递给javascript_Php_Jquery - Fatal编程技术网

Php 如何将关联数组传递给javascript

Php 如何将关联数组传递给javascript,php,jquery,Php,Jquery,我得到了以下关联数组($the_数组): 如何将其传递给jQuery?到目前为止,这是我得到的,但它不起作用: 在服务器端: echo json_encode((object) [ 'uploaded' => true, 'message' => "test", 'match' => false,

我得到了以下关联数组(
$the_数组
):

如何将其传递给jQuery?到目前为止,这是我得到的,但它不起作用:

在服务器端:

 echo json_encode((object) [
                        'uploaded' => true,
                        'message' => "test",
                        'match' => false,
                        'ac_array' => $the_array,
                        'ac_array_count' => $count_not_matched
                 ]);
在客户端:

var arr = JSON.parse(result.ac_array);
在服务器端尝试

echo json_encode(array(
   'uploaded' => true,
   'message' => "test",
   'match' => false,
   'ac_array' => $the_array,
   'ac_array_count' => $count_not_matched
));
var arr = jQuery.parseJSON(result);

在客户端尝试

echo json_encode(array(
   'uploaded' => true,
   'message' => "test",
   'match' => false,
   'ac_array' => $the_array,
   'ac_array_count' => $count_not_matched
));
var arr = jQuery.parseJSON(result);

服务器端到底有什么?请注意,您需要脚本标记,并且需要将
json\u encode()
的结果分配给javascript变量。您不需要强制转换到对象。那么什么是
result.ac_array
?您没有向我们展示创建该对象或设置属性的任何位置。