Question

我在确定如何接收一组id或完整的节点信息并使用该数据将对应的行插入数据库时遇到了一定的麻烦。

为什么会这样？好吧，我有以下层次结构projeto＆gt; uc＆gt; ambiente＆gt; secao＆gt; med。在我的JS树中，我使用延迟加载，所以让我们说用户选择一个＆＃39; projeto＆＃39;他们提交的所有内容都是＆＃39; projeto＆＃39; id，这很容易，我知道我必须在数据库中插入所有它的孩子和他们的孩子。但是，让我们说用户选择一个特定的＆＃39; ambiente＆＃39;或者特定的“secao”＃39;我得到的只是一个id或单个节点数据，但要插入该信息，我需要插入所有它的父数据，然后才能将其插入数据库。

示例1单个＆＃39; projeto＆＃39;选定的数据。

[{"id":"projeto_1","text":"Pr\u00e9dios P\u00fablicos","icon":"fa fa-folder icon-lg icon-state-info","parent":"#","parents":["#"],"data":{"id_mobile":"1"},"state":{"loaded":"false","opened":"false","selected":"true","disabled":"false"},"li_attr":{"id":"projeto_1"},"a_attr":{"href":"#"},"original":{"id":"projeto_1","text":"Pr\u00e9dios P\u00fablicos","icon":"fa fa-folder icon-lg icon-state-info"}}]

示例2单身＆＃39; ambiente＆＃39;选中，可能有“secao＆＃39;孩子与否。

[{"id":"ambiente_4","text":"protocolo","icon":"fa fa-folder icon-lg icon-state-info","parent":"uc_1","parents":["uc_1","projeto_1","#"],"data":{"id_ambiente_mobile":"4"},"state":{"loaded":"false","opened":"false","selected":"true","disabled":"false"},"li_attr":{"id":"ambiente_4"},"a_attr":{"href":"#"},"original":{"id":"ambiente_4","text":"protocolo","icon":"fa fa-folder icon-lg icon-state-info","type":"ambiente"}}]

例3单身＆＃39; secao＆＃39;选定的数据。

[{"id":"secao_5","text":"1 Lumin\u00e1ria(s) LFT 1X40W","icon":"fa fa-folder icon-lg icon-state-info","parent":"ambiente_5","parents":["ambiente_5","uc_1","projeto_1","#"],"data":{"id_secao_mobile":"5"},"state":{"loaded":"false","opened":"false","selected":"true","disabled":"false"},"li_attr":{"id":"secao_5"},"a_attr":{"href":"#"},"original":{"id":"secao_5","text":"1 Lumin\u00e1ria(s) LFT 1X40W","icon":"fa fa-folder icon-lg icon-state-info","type":"secao"}},{"id":"ambiente_5","text":"Recep\u00e7\u00e3o","icon":"fa fa-folder icon-lg icon-state-info","parent":"uc_1","parents":["uc_1","projeto_1","#"],"children":["secao_5"],"children_d":["secao_5"],"data":{"id_ambiente_mobile":"5"},"state":{"loaded":"true","opened":"true","selected":"true","disabled":"false","loading":"false"},"li_attr":{"id":"ambiente_5"},"a_attr":{"href":"#"},"original":{"id":"ambiente_5","text":"Recep\u00e7\u00e3o","icon":"fa fa-folder icon-lg icon-state-info","type":"ambiente"}}]

上面的所有数据都是传递给php文件的数据。所以我只是jso_encoded并在此发布。

所以我需要的是在db中插入所选节点，但考虑到如果父节点没有加载到树上，它可能有子节点。当然，当我选择一个孩子并且需要迭代所有备份它们时，在插入孩子之前插入它的父依赖者（最后两个例子）。

希望你们能帮助我。如果需要任何澄清，请询问。

谢谢。

Answer 1

好的一半完成了。创建了以下代码：

//take the actual node.
for ($i = 0; $i < count($ids); $i++) {

    //if the actual node is loaded and opened.
    if (($ids[$i]['state']['loaded'] == true) && ($ids[$i]['state']['opened'] == true)) {
        //then the node is inserted in the db.
        checaNo ($ids[$i]);
        //and the iteration jump all it's selected children. This is because checaNo already insert all actual node children.
        $i = count($ids[$i]['children'])+1;

    }
    //the actual node has not any children or it's children is not loaded.
    else
    {
        //insert the node and it's children if they exists. Then go to the next node.
        checaNo($ids[$i]);
    }

}

现在的问题是，假设我选择了一个“ambiente”作为一个新的“projeto”，所以在我甚至可以插入我需要创建它的父母的“ambiente”数据之前。在这个例子中是“projeto”和“uc”，需要插入“projeto”，然后需要插入“uc”，只有这样我才能在db上插入“ambiente”。

编辑：好的，我在这里解决了这个问题。创造了以下功能

function checaPai ($no) {

    global $data;

    $nivel = count($no['parents'])-1;

    switch ($nivel) {
        case 0;
            break;
        case 1;
            $args_projeto = new stdClass();
            $id_projeto = explode("_", $no['parent']);
            $args_projeto->where = "data_hora_importacao = '$data' AND id_mobile = '" . $id_projeto[1]."'";

            $projeto = getMobileProjeto($args_projeto);

            $args_projeto_online = new stdClass();
            $args_projeto_online->where = "id = '" . $projeto[0]->id_online."'";
            $projeto_online = getOnlineProjeto($args_projeto_online);

            if (count($projeto_online) == 0) {

                $id_projeto = insereProjeto($args_projeto, false);

                return $id_projeto;
            }
            else {

                return $projeto_online[0]->id;

            }

            break;
        case 2;
            $args_uc = new stdClass();
            $id_uc = explode("_", $no['parent']);
            $args_uc->where = "data_hora_importacao = '$data' AND id_uc_mobile = '" . $id_uc[1]."'";
            $uc = getMobileUC($args_uc);

            $args_uc_online = new stdClass();
            $args_uc_online->where = "contrato = '" . $uc[0]->contrato."'";
            $uc_online = getOnlineUC($args_uc_online);

            if (count($uc_online) == 0) {


                $no_uc = array();
                $no_uc['parent'] = $uc->projeto;
                $id_uc = checaPai($no_uc);
                return $id_uc;
            }
            else {

                return $uc_online[0]->id;
            }

            break;
        case 3;
            break;
        case 4;
            break;
    }
}

上面的函数检查它的父存在，然后插入或创建它的所有子节点。当孩子有例如2或3个级别时，该函数调用自身并返回父ID。而已。

这是不完整和可怕的丑陋，但为了澄清，也许如果有人有同样的问题，他们可以看到我是如何理解的。

将数据从jstree插入数据库

1 个答案: