Question

你好我想用我的mysql结果来构建json对象。我的第一个查询是获取父菜单而另一个查询正在获取子菜单，所以我将把我的子菜单结果放在父菜单下..我想知道我的代码有什么问题以及如何纠正

<?php
    $selectparentMenu=mysql_query("SELECT `id`,`item_name`,`menu_type`,`parent` FROM `epic_master_menu` where parent=0");
    //echo $myfetch=mysql_fetch_array($selectMenu);
        if(mysql_num_rows($selectparentMenu)>1) {
    while($fetchparentMenu=mysql_fetch_array($selectparentMenu)) {
         //echo 'parent id is' .$parentId=$fetchparentMenu['id']. '<br/>';
        // echo 'parent name is' .$parentId=$fetchparentMenu['item_name']. '<br/>';


         $selectchildMenu=mysql_query("SELECT `id` , `item_name` , `menu_type` , `parent` FROM `epic_master_menu`
    WHERE parent >0 AND `menu_type` = 'item' AND `parent` ='".$fetchparentMenu['id']."'");
    if(mysql_num_rows($selectchildMenu)>1) {
         while($fetchchildMenu=mysql_fetch_array($selectchildMenu)) {

             $textjson = '{
    "dataSource": [{
            "id": "", "text": "Select All", "expanded": "true", "spriteCssClass": "rootfolder", "items": [
                {
                    "id": "'.$fetchparentMenu["id"].'", "text": "'.$fetchparentMenu["item_name"].'", "expanded": true,"spriteCssClass": "folder", "items": [
                        { "id": "'.$fetchchildMenu["id"].'", "text": "'.$fetchparentMenu["item_name"].'", "spriteCssClass": "html" },
                        { "id": "'.$fetchchildMenu["id"].'", "text": "'.$fetchparentMenu["item_name"].'", "spriteCssClass": "html" },
                        { "id": "'.$fetchchildMenu["id"].'", "text": "'.$fetchparentMenu["item_name"].'", "spriteCssClass": "image" }
                    ]
                }
            ]
        }]
    }';

             }  }

         /* $fetchMenu['item_name'];
           $fetchMenu['menu_type'];
           $fetchMenu['parent'];*/
        //print_r($fetchMenu);
        } }

//我的json数据是

 $textjson = '{
"dataSource": [{
        "id": 1, "text": "My Documents", "expanded": "true", "spriteCssClass": "rootfolder", "items": [
            {
                "id": 2, "text": "Project", "expanded": true,"spriteCssClass": "folder", "items": [
                    { "id": 3, "text": "about.html", "spriteCssClass": "html" },
                    { "id": 4, "text": "index.html", "spriteCssClass": "html" },
                    { "id": 5, "text": "logo.png", "spriteCssClass": "image" }
                ]
            },
            {
                "id": 6, "text": "New Web Site", "expanded": true, "spriteCssClass": "folder", "items": [
                    { "id": 7, "text": "mockup.jpg", "spriteCssClass": "image" },
                    { "id": 8, "text": "Research.pdf", "spriteCssClass": "pdf" }
                ]
            },
            {
                "id": 9, "text": "Reports", "expanded": true, "spriteCssClass": "folder", "items": [
                    { "id": 10, "text": "February.pdf", "spriteCssClass": "pdf" },
                    { "id": 11, "text": "March.pdf", "spriteCssClass": "pdf" },
                    { "id": 12, "text": "April.pdf", "spriteCssClass": "pdf" }
                ]
            }
        ]
    }]
}';

Answer 1

您可以简单地使用json_encode()

，而不是创建自己的版本

while($fetchchildMenu=mysql_fetch_array($selectchildMenu))
 {
      $somearr[]=$fetchchildMenu;
 }

$jsonData = json_encode($somearr);
echo $jsonData; //<---- Prints your JSON data

Answer 2

使用PHP中包含的json函数要容易得多。在这里，您可以使用json_encode从数组创建json字符串。

$childArray = array();

while($fetchchildMenu=mysql_fetch_array($selectchildMenu)) {
   $childArray[] = array(
       'id' => $fetchchildMenu['id'],
       'text' => $fetchchildMenu['text']
   );
}

$jsonDataChilds = json_encode($childArray);
echo $jsonDataChilds;

使用mysql获取结果构建json

2 个答案: