Question

我刚开始使用jpgrid。我在php json_encode中遇到了这个问题。

根据jqgrid网站的例子，json字符串可以通过php代码生成，如下所示：

    $responce->page = "1";
    $responce->total = 1;
    $responce->records = "1";

    $i=1; $amttot=0; $taxtot=0; $total=0;
    while($i<=10) {
        $amttot += $row[amount];
        $taxtot += $row[tax];
        $total += $row[total];
        $responce->rows[$i]['id']=strval($i);
        $responce->rows[$i]['cell']=array("123","2012-10-15","name","10000","tax","9999","note");
        $i++;
    }
    $responce->rows = (array)$responce->rows;
    $responce->userdata['amount'] = $amttot;
    $responce->userdata['tax'] = $taxtot;
    $responce->userdata['total'] = $total;
    $responce->userdata['name'] = 'Totals:';

我遇到问题

    $responce->rows[$i]['id']=strval($i);
    $responce->rows[$i]['cell']=array("123","2012-10-15","name","10000","tax","9999","note");

当php json_encode生成json字符串时，rows数组被视为关联数组。因此结果json字符串看起来像：

{"page":"1","total":1,"records":"1",
 "rows":{"1":{"id":"1","cell":["123","2012-10-15","name","10000","tax","9999","note"]},
         "2":{"id":"2","cell":["123","2012-10-15","name","10000","tax","9999","note"]},
         "3":{"id":"3","cell":["123","2012-10-15","name","10000","tax","9999","note"]},
         "4":{"id":"4","cell":["123","2012-10-15","name","10000","tax","9999","note"]},
         "5":{"id":"5","cell":["123","2012-10-15","name","10000","tax","9999","note"]},
         "6":{"id":"6","cell":["123","2012-10-15","name","10000","tax","9999","note"]},
         "7":{"id":"7","cell":["123","2012-10-15","name","10000","tax","9999","note"]},
         "8":{"id":"8","cell":["123","2012-10-15","name","10000","tax","9999","note"]},
         "9":{"id":"9","cell":["123","2012-10-15","name","10000","tax","9999","note"]},
         "10":{"id":"10","cell":["123","2012-10-15","name","10000","tax","9999","note"]}},
 "userdata":{"amount":0,"tax":0,"total":0,"name":"Totals:"}}

这个json字符串无法由jqgrid处理。实际上jqgrid不会引发任何javascript错误。它只是不呈现表内容。我通过firebugs查看了jqgrid示例网站上的json字符串。它看起来像这样：

{"page":"1","total":2,"records":"13",
 "rows":[{"id":"13","cell":["13","2007-10-06","Client 3","1000.00","0.00","1000.00",null]},
         {"id":"12","cell":["12","2007-10-06","Client 2","700.00","140.00","840.00",null]},
         {"id":"11","cell":["11","2007-10-06","Client 1","600.00","120.00","720.00",null]},
         {"id":"10","cell":["10","2007-10-06","Client 2","100.00","20.00","120.00",null]},          
         {"id":"9","cell":["9","2007-10-06","Client 1","200.00","40.00","240.00",null]},
         {"id":"8","cell":["8","2007-10-06","Client 3","200.00","0.00","200.00",null]},
         {"id":"7","cell":["7","2007-10-05","Client 2","120.00","12.00","134.00",null]}, 
         {"id":"6","cell":["6","2007-10-05","Client 1","50.00","10.00","60.00",""]},
         {"id":"5","cell":["5","2007-10-05","Client 3","100.00","0.00","100.00","no tax at all"]},
         {"id":"4","cell":["4","2007-10-04","Client 3","150.00","0.00","150.00","no tax"]}],
        "userdata":{"amount":3220,"tax":342,"total":3564,"name":"Totals:"}}

我无法弄清楚如何使用关联数组的php示例代码在json字符串中生成这样的数组obj。我使用的是PHP 5.4

我可以在jqgrid或php中更改以使其工作吗？我很惊讶没有人有这个问题。我可能会遗漏一些相当明显的东西。

感谢

casbby

Answer 1

问题是由PHP中如何生成响应数组对象引起的。当我最初在循环中创建数组时，我从1而不是0开始索引。响应 - ＆GT;行[I] 这导致php将响应数组视为关联而非顺序。因此php将名称值对象放在响应对象中（当我尝试创建一个非命名对象的数组时）

解决方案是确保数组索引是连续的和连续的

jpgrid php json_encode（）关联数组

1 个答案: